Sullivan Microsite TE SAMPLE

Chapter 

1 

Limits and Continuity 

Overview 

In Chapter 1, we explore limits and continuity, two basic concepts 

of calculus. Limits are the first of the four Big Ideas in 

the 2016–2017 AP ® Calculus Curriculum Framework. 

Big Idea 1: Limits 

Big Idea 2: Derivatives 

Big Idea 3: Integrals and the Fundamental Theorem of 

Calculus 

Big Idea 4: Series (AP ® Calculus BC only) 

The concept of limits is central to calculus. To understand 

calculus, it is crucial for students to know what it means for a 

function to have a limit and how to find the limit, when it exists. 

In Chapter 1, students learn what a limit is, how to find the 

limit of a function, and how to prove that limits exist using the 

definition of a limit. The chapter begins with numerical and 

graphical approaches to identifying limits, and students learn 

that these methods sometimes fail. Students then explore a variety 

of analytical techniques for finding limits. The precise 

definition of a limit, the so-called e–∂ (epsilon-delta) definition, 

is introduced at the end of the chapter. 

Building on what they’ve learned about limits, students 

learn the precise definition of continuity. Students often have 

an intuitive sense of the concept of continuity, because they 

have likely been tracing letters, numbers, and shapes from a 

young age. You might help students to make connections between 

these intuitive concepts and the formal definition of 

continuity in Section 1.3. 

As students work through Chapter 1, they will encounter 

the familiar concepts of horizontal and vertical asymptotes 

in the context of limits. From Chapter P, or their precalculus 

course, students should understand how to determine the domain 

and range of a function when viewing it graphically as 

well as when they are given an expression. They should also be 

able to identify horizontal and vertical asymptotes graphically 

and be able to find all asymptotes of a function analytically. 

Building on this knowledge, in this chapter, students will use 

the concept of a limit to find and verify values at which a function 

has an asymptote. This skill will be revisited when they 

learn about curve sketching in Chapter 4. 

Relationship to the AP® Calculus 

Curriculum Framework 

Chapter 1 covers topics from Big Idea 1: Limits. Here is a table 

that indicates the Essential Knowledge statements and Learning 

Objectives from the 2016–2017 AP ® Calculus Curriculum 

Framework that are covered in this chapter. 

Essential Knowledge 

Statements Learning Objectives Coverage 

EK 1.1A1 LO 1.1A(a) Section 1.1 

EK 1.1A2 LO 1.1A(b) Section 1.1 

EK 1.1A3 LO 1.1A(b) Section 1.1 

EK 1.1B1 LO 1.1B Section 1.1 

EK 1.1C1 LO 1.1C Section 1.2 

EK 1.1C2 LO 1.1C Section 1.2 

EK 1.1D1 LO 1.1D Section 1.5 

EK 1.1D2 LO 1.1D Section 1.5 

EK 1.2A1 LO 1.1D Section 1.3 

EK 1.2A2 LO 1.2A Section 1.3 

EK 1.2A3 LO 1.2A Section 1.3 

EK 1.2B1 LO 1.2 B Section 1.3 

1-2 

Chapter 1 • Limits and Continuity 

TE_Sullivan_Chapter01_PART 0.indd 1 

11/01/17 9:51 am

Chapter 1 Section Topics 

Section 1.1: Limits of Functions Using 

Numerical and Graphical Techniques 

Students are introduced to the slope of a tangent line to a 

graph. Students should already be familiar with calculating 

the slope of a line given two distinct points. The concept of a 

limit is introduced through calculating the slope of the secant 

line on a curve based upon two points (x, f(x)) and (c, f(c)). If 

we continue to calculate the slope of the secant line with points 

that become closer and closer together (as x → c), the slope of 

the secant line approaches the slope of the tangent line at a 

given point. This work lays a foundation for the calculation of 

the slope of the tangent line as well as introduces the concept 

of the limit. 

Students investigate the concept of a limit first through 

the use of a table of numbers and then by using a graph. Both 

methods are handy for determining a limit, but both have 

drawbacks. Several examples are offered so that the students 

can gain an understanding of the pros and cons of these two 

methods for finding a limit of a function. Stress to students 

that a limit is the value that a function approaches as x approaches 

a particular value. Students often misinterpret the 

limit of a function as x → c with the value of f(c). 

Section 1.2: Limits of Functions Using 

Properties of Limits 

Through a series of examples, students explore finding limits using 

properties of limits and see that in many cases, a limit can 

be found by direct substitution or by substitution after some algebraic 

manipulation. The examples at the end of the section are 

of particular importance, as they introduce alternative strategies 

that can be used when substitution will not work. Students also 

are introduced to the average rate of change of a function in this 

section. This will provide the foundation they will use to learn 

about derivatives in Chapter 2. 

Section 1.3: Continuity 

Section 1.3 introduces the concept of continuity. Students are 

presented a formal definition of continuity at a point and use 

that definition to determine if a function is continuous at a 

given value. The section also covers classifications of discontinuities, 

in particular, removable, jump, and infinite discontinuities. 

The section concludes with the introduction of the 

Intermediate Value Theorem, which is applicable for functions 

that are continuous on a given closed interval. 

Section 1.4: Limits and Continuity of 

Trigonometric, Exponential, and 

Logarithmic Functions 

In Section 1.4, the first concept covered is finding a limit using 

the Squeeze Theorem. Students then explore finding limits 

and continuity of trigonometric, exponential, and logarithmic 

functions. 

Section 1.5: Infinite Limits; Limits at Infinity; 

Asymptotes 

We extend the language of limits in Section 1.5 to allow c to 

be ∞ or −∞ (limits at infinity) and to allow L to be ∞ or −∞ 

(infinite limits). These limits can then be used for locating asymptotes 

that aid in graphing certain functions. Sometimes 

when we think about finding asymptotes, we find ourselves 

turning quickly to rational functions. Yet other functions (like 

exponential functions, logarithmic functions, and trigonometric 

functions) have asymptotes as well. See Example 10 for an application 

of finding a limit at infinity of an exponential function. 

Section 1.6: The e–d Definition of a Limit 

(Optional) 

As the chapter stresses throughout, we can be sure a limit is 

correct only if it was based on the e–d definition of a limit. 

In this section, we examine this definition and how to use it 

to prove that a limit exists, to verify the value of a limit, and 

to show if a limit does not exist. Although the content in this 

section is not directly tested on either the AB or BC version of 

the exam, students may benefit from coverage of this section. 

Promoting Good Habits and Skills 

Students may not have a great deal of experience in justifying 

their answers. This skill is fundamental to the course and to applying 

calculus to future fields of study. It is also a key part of the 

AP ® Calculus Curriculum Framework as addressed in MPAC 6. 

1. It is important to be able to justify solutions and to do so 

succinctly and completely. It helps to teach this skill by 

example. Write out solutions that are just enough. 

Explain to students why each step is sufficient, specific, 

and on point. 

2. Get in the habit of asking your students to explain 

their reasoning as they state answers to questions 

that you pose or problems you assign. You can build 

a justification step into assignments so that students 

expect to have to do so. 


1-3 


11/01/17 9:51 am

Chapter 1: Resources 

TRM Teacher’s Resource Materials 

The following resources can be found by clicking on the links in the 

Teacher’s e-Book (TE-book), logging into the Teacher’s Resource 

Center, or opening the Teacher’s Resource Flash Drive (TRFD). 

• Chapter 1 PD Video 

• Chapter 1 Bell Ringers The prepared Bell Ringer slide set is 

a great tool to help you make good use of the first few minutes 

of class. Designed to be used every day, each Bell Ringer set 

contains a set of short mental math questions that review prerequisite 

topics and keep the student engaged while practicing 

basic skills. The number and type of questions vary by chapter. 

• AP ® Calculus AB Exam Prep Flashcards Flashcards are 

a great tool for students to use in learning the essentials 

of calculus. Prepped to be printed and cut into 2 × 3–inch 

cards, the flashcards are designed to be used in class or on 

the go. Each card is labeled with the chapter number in 

which the topic is first introduced so they can be sorted and 

used throughout the course as a study aid. 

• Chapter 1 Alternate Examples All of the Chapter 1 Alternate 

Examples are also provided in PDF document format. 

Use these as additional examples in class, as the basis for assessments, 

or as additional practice for students. 

• Chapter 1 AP ® Calc Skill Builders All of the Chapter 1 

AP ® Calc Skill Builders are also provided in PDf format. 

Use these to provide instant practice of skills that are essential 

for success on the AP ® Exam. 

• Chapter 1 Skill Building Worksheets There are printable 

worksheets for each section, with solutions. Each one is 

referenced as point of use in the wraparound pages. 

Section 1.1 Worksheet 1 














Section 1.6 Worksheet 

• Chapter 1 Prepared Tests (Forms A and B) No need to worry 

about your students sharing exam information when you 

have two tests in two parallel versions to use for your various 

sections or as a makeup exam. Each test is 4 pages long and is 

designed to be scored out of a maximum of 50 points, making 

percentages simple. 

• Chapter 1 Teacher’s Solutions Manual Complete worked solutions 

to every problem in the book are found in the Teacher’s 

Solutions Manual, which may be downloaded as a PDF. The 

solutions for each set of Section Problems and for the Chapter 

Review Problems, AP ® Review Problems, and Tests at the ends 

of chapters are referenced at point of use in the Teacher’s Edition 

and may be downloaded as a smaller chunk of material for 

ease of use. 

• Additional Chapter 1 Resources We have created a list of 

third-party videos, Web sites, and other resources to support 

the content in this chapter. The Word document includes clickable 

URLs to help you access this external content. (Note: all 

of the URLs were live when this book was published.) 

Free-Response Questions from 

Previous AP® Exams 

Chapter 1 is too early to have students work with previous 

free-response questions. Limit and continuity problems seldom 

appear directly in free-response questions on the exam. It 

is possible, however, that a subpart of a free-response question 

may ask about a concept related to the limit or continuity. 

Students are more likely to encounter limit and continuity 

problems in the multiple-choice questions of the exam. On 

average, students can expect about 3 or 4 multiple-choice questions 

that cover limits. However, it’s likely that multiple-choice 

questions on limits will be framed so that they also require 

knowledge gained later in the year. For example, the exam 

may have questions on limits that incorporate more advanced 

content from later chapters, such as the limit of the difference 

quotient (Section 2.1) or L’Hôpital’s Rule (Section 4.5) 

or mixed in with problems related to the derivative, covered in 

Chapters 3 and 4. 

It is a good idea to start exploring the College Board 

Web site to become familiar with what is available to teachers. 

Released free-response questions can be found on the 

AP ® Central Web site: http://apcentral.collegeboard.com/apc/ 

members/exam/exam_information/232050.html 

1-4 Chapter 1 • Limits and Continuity 


11/01/17 9:51 am

Chapter 

1 

Free-response questions often draw on content that spans 

more than one chapter. As students prepare during the year, 

you can introduce students to the skills required to solve full 

FRQ on the exam. If there are appropriate questions or subparts, 

look under this headline in future chapters to see what 

might be helpful to use in class. 

As an AP ® Calculus teacher, you can gain access to complete 

released exams by logging into your Course Audit Account 

on the Course Home Page. 

AB Calculus: 

http://apcentral.collegeboard.com/apc/public/courses/ 

teachers_corner/2178.html 

BC Calculus: 

http://apcentral.collegeboard.com/apc/public/courses/ 

teachers_corner/2118.html 

You may want to keep these old exams on hand to select multiple-choice 

questions related to limits to present to students 

for further practice. Likewise, you may want to make a chart 

or find some other organizing tool that allows you to quickly 

group multiple-choice questions by topic so they may be used 

at appropriate times throughout the year. 

You may also want to direct your students to the following 

Kahn Academy site. College Board has partnered with Kahn 

Academy to provide support to students. Many released FRQs 

from previous exams are reviewed in these worked example 

videos: 

https://www.khanacademy.org/math/calculus-home/apcalc-topic 

Chapter 1: Pacing Guides, Objectives, 

and Suggested Assignments 

These pacing guides are based on a schedule with 125 standard 

45-minute classroom sessions before the exam. This 125- 

day course includes assessment days and allows about 3 weeks 

for review before the AP ® Calculus exam. 

The suggested homework assignments list odd-numbered 

problems, whenever possible, so students can check their answers 

against the back-of-the-book answers. If you would rather 

students not have access to the answers while doing homework, 

adding 1 to the exercise numbers usually will do the trick, because 

the homework problems typically are paired. The answers 

for the even-numbered problems are not in the Answer appendix. 

The authors observe pairing strictly in the Skill Building 

category. The Applications and Extensions usually are paired, 

although more advanced problems tend not to be paired. The 

AP ® Practice Problems are not paired. 

Calculus AB Pacing Guide 

Only topics that appear on the AP ® Exam are included. 

Day Topic Sullivan/Miranda Chapter Objectives Suggested Assignment 

1 Section 1.1 

1. Discuss the slope of the tangent line to the graph 

2. Investigate a limit using a table 

7, 9, 11, 17–28 

2 Section 1.1 3. Investigate a limit using a graph 

31, 35, 37, 38, 41, 42 

All AP ® Practice Problems 

3 Section 1.2 

1. Find the limit of a sum, a difference, and a product 

2. Find the limit of a power and the limit of a root 

3. Find the limit of a polynomial 

35, 37, 39, 41, 43, 47, 48, 51–59 odd 

4. Find the limit of a quotient 

4 Section 1.2 

5. Find the limit of an average rate of change 

61–65 odd, 67, 71, 73–79 odd 

6. Find the limit of a difference quotient 


1. Determine whether a function is continuous at a number 

5 Section 1.3 2. Determine intervals on which a function is continuous 

13–18, 19–35 odd 

3. Use properties of continuity 

6 Section 1.3 4. Use the Intermediate Value Theorem 59–63 odd, All AP ® Practice Problems 

7 Section 1.4 

1. Use the Squeeze Theorem to find a limit 

9–45 odd 

2. Find limits involving trigonometric functions 

(continued next page) 


1-5 


11/01/17 9:51 am

Sullivan AP˙Sullivan˙Chapter01 October 8, 2016 17:4 

Sullivan 

88 Chapter 1 • Limits and Continuity 

33. 

2x 2 if x < 1 

f (x) = 

3x 2 − 1 if x > 1 

at c = 1 

34. f (x) = 

x 2 − 1 if x > −1 

x 3 if x < −1 

at c =−1 

 

35. f (x) = 

at c = 0 

2x + 1 if x > 0 

⎧ 

⎨ x 2 if x < 1 

x 2 if x ≤ 0 

36. f (x) = 2 if x = 1 at c = 1 

⎩ 

−3x + 2 if x > 1 

Calculus AB Pacing Guide (continued) 

Applications and Extensions 

In Problems 37–40, sketch a graph of a function with the given 

properties. Answers will vary. 

37. lim f (x) = 3; 

x→2 

lim f (x) = 3; 

x→3− lim f (x) = 1; 

x→3 + 

f (2) = 3; f (3) = 1 

38. lim f (x) = 0; 

x→−1 lim f (x) =−2; 

x→2− lim f (x) =−2; 

x→2 + 

f (−1) is not defined; f (2) =−2 

39. lim f (x) = 4; 

x→1 

lim f (x) =−1; 

x→0− lim f (x) = 0; 

x→0 + 

f (0) =−1; f (1) = 2 

f (x) = 1; 

x→1 

In Problems 41–50, use either a graph or a table to investigate 

each limit. 

|x − 5| 

|x − 5| 

41. lim 

42. lim 

43. lim 

x→5 + x − 5 

x→5 − x − 5 

 

x→ 12 

2x 

− 

44. lim 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. lim 

− 

x→ 23 

2x 

+ 

 

47. lim |x|−x 48. lim |x|−x 

x→2 + x→2 − 

3 3 

49. lim x−x 50. lim x−x 

x→2 + x→2 − 

51. Slope of a Tangent Line For f (x) = 3x 2 : 

(a) Find the slope of the secant line containing the points (2, 12) 

and (3, 27). 

(b) Find the slope of the secant line containing the points (2, 12) 

and (x, f (x)), x = 2. 

(c) Create a table to investigate the slope of the tangent line to the 

graph of f at 2 using the result from (b). 

(d) On the same set of axes, graph f , the tangent line to the graph 

of f at the point (2, 12), and the secant line from (a). 


and (3, 27). 


and (x, f (x)), x = 2. 





53. Slope of a Tangent Line For f (x) = 1 2 x2 − 1: 


8 Section 1.4 

9 Section 1.5 

10 Section 1.5 

3. Determine where the trigonometric functions are continuous 

4. Determine where an exponential or a logarithmic function is 

continuous 

1. Investigate infinite limits 

2. Find the vertical asymptotes of a graph 

3. Investigate limits 40. at lim f (x) = 2; lim f (x) = 0; lim 

x→2 

infinity 

x→−1 

4. Find the horizontal asymptotes f (−1) = 1; f (2) of = a 3graph 

5. Find the asymptotes of the graph of a rational function 

(a) Find the slope m sec of the secant line containing the 

points P = (2, f (2)) and Q = (2 + h, f (2 + h)). 

(b) Use the result from (a) to complete the following table: 

h −0.5 −0.1 −0.001 0.001 0.1 0.5 

m sec 

(c) Investigate the limit of the slope of the secant line found in (a) 

as h → 0. 

(d) What is the slope of the tangent line to the graph of f at the 

point P = (2, f (2))? 

(e) On the same set of axes, graph f and the tangent line to f at 

P = (2, f (2)). 

54. Slope of a Tangent Line For f (x) = x 2 − 1: 



points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)). 


2–26, 27–41 odd 

h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1 

m sec 

43–59 odd, 67–71 odd, All AP ® Practice 

Problems 

(c) Investigate the limit of the slope of the secant line found 

in (a) as h → 0. 


point P = (−1, f (−1))? 

(e) On the same set of axes, graph f and the tangent line to f 

at P = (−1, f (−1)). 

11 Review Chapter 1 Review Exercises Chapter 1 AP ® Review Problems 

12 Test Chapter 1 Test AP ® Practice Exam, Big Idea 1: Limits 

Calculus BC Pacing Guide 

PAGE 

85 55. (a) Investigate lim cos π by using a table and evaluating the 

x→0 x 

function f (x) = cos π x at 


1 Section 1.1 

2 Section 1.2 

3 Section 1.2 

4 Section 1.3 

5 Section 1.4 

6 Section 1.5 

1. Discuss the slope of the tangent line to the graph 

2. Investigate a limit using a table 

3. Investigate a limit using a graph 

1. Find the limit of a sum, a difference, and a product 

2. Find the limit of a power and the limit of a root 

3. Find the limit of a polynomial 

52. Slope of a Tangent Line For f (x) = x 3 : 

4. Find the limit of a quotient 

5. Find the limit of an average rate of change 

6. Find the limit of a difference quotient 

1. Determine whether a function is continuous at a number 

2. Determine intervals on which a function is continuous 

3. Use properties of continuity 

4. Use the Intermediate Value Theorem 

1. Use the Squeeze Theorem to find a limit 

2. Find limits involving trigonometric functions 

3. Determine where the trigonometric functions are continuous 

4. Determine where an exponential or a logarithmic function is 

continuous 

1. Investigate infinite limits 

2. Find the vertical asymptotes of a graph 

3. Investigate limits at infinity 

4. Find the horizontal asymptotes of a graph 

5. Find the asymptotes of the graph of a rational function 

x =− 1 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1 

12 , 1 10 , 1 8 , 1 4 , 1 2 . 

(b) Investigate lim cos π by using a table and evaluating the 

x→0 x 


9, 11, 13, 15, 17, 19, 31, 37, 39, 43, 47, 

51, 55, 59, 63–66, AP ® Practice Problems 

3–8 

x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 3 , 1. 

(c) Compare the results from (a) and (b). What do you conclude 

about the limit? Why do you think this happens? What is 

your view about using a table to draw a conclusion about 

limits? 

(d) Use technology to graph f . Begin with the x-window 

[−2π, 2π] and the y-window [−1, 1]. If you were finding 

lim f (x) using a graph, what would you conclude? Zoom in 

x→0 

on the graph. Describe what you see. (Hint: Be sure your 

calculator is set to the radian mode.) 

10, 13, 19, 35, 37, 39, 43, 47, 51, 55, 59 

63, 66, 74, 75, 83, 93, 96 

All AP ® Practice Problems (especially 5) 

56. (a) Investigate lim cos π by using a table and evaluating the 

x→0 x2 function f (x) = cos π at x =−0.1, −0.01, −0.001, 

x2 −0.0001, 0.0001, 0.001, 0.01, 0.1. 

13–18, 24, 29, 45, 51–56, 59, 62, 63, 

73, 79 


11,15,19, 23, 24, 25, 35, 38, 43, 45, 55, 

70 


17–26, 30–35, 40, 45, 50, 53, 55, 60, 63, 

70, 73, 74, 75, 78, 86 


7 Review Chapter 1 Review Exercises Chapter 1 AP ® Review Problems 

8 Test Chapter 1 Test AP ® Practice Exam, Big Idea 1: Limits 

1-6 Chapter 1 • Limits and Continuity 


11/01/17 9:51 am

Sullivan AP˙Sullivan˙Chapter01 September October 8, 20, 2016 2016 17:416:7 

1 Limits and Continuity 


x→0 x2 function f (x) = cos π x 2 at 

x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 3 . 


about the limit? Why do you think this happens? What is your 

view about using a table to draw a conclusion about limits? 




x→0 



PAGE 

x − 8 

85 57. (a) Use a table to investigate lim . 

x→2 2 

(b) How close must x be to 2, so that f (x) is within 0.1 of the 

limit? 

(c) How close must x be to 2, so that f (x) is within 0.01 of the 

limit? 

58. (a) Use a table to investigate lim(5 − 2x). 

x→2 


limit? 


limit? 

59. First-Class Mail As of April 

2016, the U.S. Postal Service 

charged $0.47 postage for 

first-class letters weighing up to 

and including 1 ounce, plus a flat 

fee of $0.21 for each additional 

or partial ounce up to and 

including 3.5 ounces. First-class 

letter rates do not apply to letters 

weighing more than 3.5 ounces. 

Source: U.S. Postal Service Notice 123 

(a) Find a function C that models the first-class postage charged, 

in dollars, for a letter weighing w ounces. Assume w>0. 

(b) What is the domain of C? 

(c) Graph the function C. 

(d) Use the graph to investigate lim C(w) and lim C(w). Do 

w→2− w→2 + 

1.1 Limits of Functions Using 

Numerical these suggest and that Graphical lim C(w) exists? 

w→2 

(e) Techniques Use the graph to investigate lim C(w). 

w→0 + 

1.2(f) Limits Use the ofgraph Functions to investigate Using lim 

Properties of Limits 

C(w). 

w→3.5 − 

60. 1.3 First-Class Continuity Mail As of April 2016, the U.S. Postal Service 

charged $0.94 postage for first-class large envelope weighing up to 

1.4 Limits and Continuity of 

and including 1 ounce, plus a flat fee of $0.21 for each additional 

or partial Trigonometric, ounce up toExponential, and includingand 

13 ounces. First-class rates do 

notLogarithmic apply to largeFunctions 

envelopes weighing more than 13 ounces. 

1.5 Source: Infinite U.S. Limits; Postal Service LimitsNotice at Infinity; 123 

Asymptotes 


1.6 The in dollars, ε-δ Definition for a largeofenvelope a Limitweighing w ounces. Assume 

Chapter w>0. Review 

(b) 

Chapter 

What is 

Project 

the domain of C? 

Kathryn Sidenstricker /Dreamstime.com 



w→1− w→1 + 

these suggest that lim C(w) exists? 

w→1 

(e) Use the graph to investigate lim C(w) and lim C(w). 

w→12− w→12 + 

Do these suggest that lim C(w) exists? 

w→12 

(f) Use the graph to investigate lim C(w). 

w→0 + 

(g) Use the graph to investigate lim C(w). 

w→13 − 

61. Correlating Student Success to Study Time Professor Smith 

claims that a student’s final exam score is a function of the time t 

(in hours) that the student studies. He claims that the closer to 

seven hours one studies, the closer to 100% the student scores 

on the final. He claims that studying significantly less than seven 

hours may cause one to be underprepared for the test, while 

studying significantly more than seven hours may cause 

“burnout.” 

(a) Write Professor Smith’s claim symbolically as a limit. 

(b) Write Professor Smith’s claim using the ε-δ definition 

of limit. 

Source: Submitted by the students of Millikin University. 

62. The definition of the slope of the tangent line to the graph of 

f (x) − f (c) 

y = f (x) at the point (c, f (c)) is m tan = lim 

. 

x→c x − c 

Another way to express this slope is to define a new variable 

h = x − c. Rewrite the slope of the tangent line m tan using h and c. 

63. If f (2) = 6, can you conclude anything about lim f (x)? Explain 

x→2 

your reasoning. 

64. If lim f (x) = 6, can you conclude anything about f (2)? Explain 

x→2 


65. The graph of f (x) = x − 3 is a straight line with a point punched 

3 − x 

out. 

(a) What straight line and what point? 

(b) Use the graph of f to investigate the one-sided limits of f as 

x approaches 3. 

AP Photo 

(c) Does the graph suggest that lim f (x) exists? If so, what is it? 

x→3 

Oil Spills and Dispersant Chemicals 

66. (a) Use a table to investigate lim(1 + x) 1/x . 

x→0 

On April 20, 2010, the Deepwater (b) Horizon Use graphing drillingtechnology rig exploded toand graph initiated g(x) = the (1 worst + x) 1/x marine . oil 

spill in recent history. Oil gushed (c) from What thedo well (a) for and three (b) suggest monthsabout and released lim(1 + millions x) 1/x ? of gallons 

of crude oil into the Gulf of Mexico. One technique used to help clean x→0 

up during and after the 

CAS 

spill was the use of the chemical (d) dispersant Find lim(1 + x) 1/x . 

x→0 

Corexit. Oil dispersants allow the oil particles to 

spread more freely in the water, thus allowing the oil to biodegrade more quickly. Their use is 

debated, however, because some of their ingredients are carcinogens. Further, the use of oil 

dispersants can increase toxic Challenge hydrocarbon Problems levels affecting sea life. Over time, the pollution 

caused by the oil spill and the dispersants will eventually diminish and sea life will return, more 

or less, to its previous condition. 

For Problems 

In the short 

67–70, 

term, 

investigate 

however, 

each of the following limits. 

{ 

pollution raises serious questions 

about the health of the local sea life and the safety of fish 1 and if shellfish x an for integer human consumption. 

f (x) = 

Explore a hypothetical situation of pollution in a lake in the Chapter 0 1 Project if x isonnot p. 156. an integer 

67. lim f (x) 68. lim f (x) 69. lim f (x) 70. lim f (x) 

x→2 x→1/2 x→3 x→0 

77 

PD Chapter 1 Overview 

Watch the chapter overview video for 

expert advice on teaching the content in 

this chapter, anticipating likely problem 

spots, and guidance on staying on pace. 

Teaching Tip 

Limits are foundational to the study of 

calculus. This chapter opens with a 

practical application of limits by modeling 

the level of contamination in Clear Lake. 

Students will have an opportunity to 

apply what they have learned through 

this chapter by completing the Chapter 1 

Project, Pollution in Clear Lake, at the end 

of this unit of study. 

TRM Chapter 1 Bell Ringers 

Each day, the students review and 

evaluate 5 trigonometric functions and 2 

logarithmic functions without the use of 

a calculator. Each problem is presented 

on a PowerPoint slide, and the problems 

transition on a timer, allowing you time to 

take attendance or check homework while 

your students work the problems. At the 

end of the Bell Ringer file is a 20-question 

quiz that can be given on the day that you 

review the chapter. 

WEB SITE 

Mathisfun: The Math Is Fun Web site is a 

helpful resource for the student who would 

benefit from learning about limits in simple 

terms with lots of illustrations. A link to this 

resource is available on the Additional 

Chapter 1 Resources document, available 

for download. 

Section 1.1 • Assess Your Understanding 89 

Where to Find the 

u Teacher Resources? 

t 

All of the Teacher Resource Materials listed in 

the blue pages for this chapter and referenced 

through the PD and TRM icons may be 

found by clicking on the links in the Teacher’s 

e-Book (TE-book), logging in to LaunchPad 

(password required) highschool.bfwpub.com/ 

launchpad/apsullivan2e, or opening the 

Teacher’s Resource Flash Drive (TRFD). 

TRM AP® Calc AB Exam Prep Flashcards 

You may want to give your students the AP ® Calc 

AB Exam Prep Flashcards now so that they may 

use them throughout the year. Have them separate 

the Chapter 1 cards by referring to the chapter 

number in the bottom corner. 


77 


11/01/17 9:51 am

Sullivan AP˙Sullivan˙Chapter01 October8, 8, 2016 17:4 


TRM Alternate Examples Section 1.1 

You can find the Alternate Examples for 

this section in PDF format in the Teacher’s 

Resource Materials. 

TRM AP® Calc Skill Builders 

Section 1.1 

You can find the AP ® Calc Skill Builders for 




Many Web sites provide help for teaching 

and learning calculus. Some provide 

advice and concept development videos 

or applets, while others offer step-by-step 

support in working calculus problems. 

Three very useful sites: 

• Khan Academy is a well-known source 

of good tutorials, and the College Board 

has partnered with Khan to provide 

instructional videos on released AP ® 

Exam problems. https://www. 

khanacademy.org/math/calculus-home/ 

ap-calc-topic 

• Lin McMullin’s Teaching Calculus blog 

offers an extensive set of resources for 

teachers and students, including 

instructional videos for most topics, 

suggestions for addressing the MPACs, 

and other helpful information. 

https://teachingcalculus.com/ 

• The Wolfram|Alpha site offers help with 

solving many types of problems by “doing 

dynamic computations based on a vast 

collection of built-in data, algorithms, and 

methods.” It offers much helpful guidance 

to students when they are stuck and 

looking for quick specific help. 

https://www.wolframalpha.com/ 

Links to these resources, and others, 

are found in the Additional Chapter 1 

Resources document, available for download. 

88 78 Chapter 1 • Limits and Continuity 

T 

2x 2 if x < 1 

33. f (x) = 

3x 2 at c = 1 

he concept of a limit 53. is Slope central of a to Tangent calculus. Line To understand For f (x) = calculus, 1 

− 1 if x > 1 

2 x2 − 1: it is essential to 

know what it means for a function to have a limit, and then how to find a limit of a 

x 3 function. Chapter 1 explains (a) what Finda the limit slope is, shows m 

if x < −1 

sec ofhow the secant to findline a limit containing of a function, the and 

34. f (x) = 

x 2 at c =−1 demonstrates how to prove thatpoints limitsP exist = (2, using f (2)) the anddefinition Q = (2 + of h, limit. f (2 + h)). 

− 1 if x > −1 

We begin the chapter using (b) Use numerical the result and from graphical (a) to complete approaches the following to explore table: the idea 

x 2 if x ≤ 0 

35. f (x) = 

at c = 0 of a limit. Although these methods seem to work well, there are instances in which they 

2x + 1 if x > 0 

⎧ 

fail to identify the correct limit. h −0.5 −0.1 −0.001 0.001 0.1 0.5 

⎨ x 2 if x < 1 

In Section 1.2, we provide analytic m sec techniques for finding limits. Some of the proofs 

36. f (x) = 2 if x = 1 at c = 1 

⎩ 

of these techniques are found in Section 1.6, others in Appendix B. A limit found by 

−3x + 2 if x > 1 

correctly applying these analytic (c) Investigate techniques theis limit precise; of thethere slopeis ofno thedoubt secantthat lineit found is correct. in (a) 

as h → 0. 

In Sections 1.3–1.5, we continue to study limits and some ways that they are used. 



For example, we use limits to define continuity, an important property of a function. 

point P = (2, f (2))? 

In Problems 37–40, sketch a graph of a function with the Section given 1.6 provides a(e) precise On the definition same set of ofaxes, limit, graph the so-called f and the tangent ε-δ (epsilon-delta) 

line to f at 


definition, which we use to showP = when (2, f a(2)). 

limit does, and does not, exist. ■ 

37. lim f (x) = 3; lim f (x) = 3; lim f (x) = 1; 

x→2 x→3− x→3 + 

f (2) = 3; f (3) = 1 


1.1 Limits of Functions Using Numerical 

38. lim f (x) = 0; lim f (x) =−2; lim f (x) =−2; 


x→−1 x→2− x→2 + and Graphical Techniques 



39. lim f (x) = 4; lim f (x) =−1; lim f 

OBJECTIVES 

(x) = 0; 

When you(b) finish Usethis the result section, from you (a) to should complete be able the following to: table: 

x→1 x→0− x→0 + 

AP® 

f 

EXAM 

(0) =−1; 

INSIGHT 

f (1) 

Limits 

= 2 

is the first 1 Discuss the slope of a tangent line to a graph (p. 78) 

h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1 

Big Idea in the AP® Calculus curriculum. 

40. lim f (x) = 2; lim f (x) = 0; lim 2 Investigate a limit using a table (p. 80) 

f (x) = 1; 

m sec 

x→2 x→−1 x→1 3 Investigate a limit using a graph (p. 82) 

f (−1) = 1; f (2) = 3 


Calculus can be used to solve certain (a) asfundamental h → 0. questions in geometry. Two of these 

In Problems 41–50, use either a graph or a table toquestions investigate are: 


each limit. 

point P = (−1, f (−1))? 

|x − 5| 

|x − 5| 

41. lim 

42. lim 

43. 

• Given 

lim 

a 

x→5 + x − 5 

x→5 − x − 5 

 

x→ 12 

2x 

function f and(e) a point On theP same on its set graph, of axes, what graph is the f and slope the tangent of the line 

to f 

tangent − 

to the graph of f at atP? P = See (−1, Figure f (−1)). 1. 

• Given a nonnegative function f whose domain 

PAGE 

44. lim 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. lim 

− 

x→ 23 

2x 

+ 

85 55. (a) Investigate lim cos π is the closed interval [a, b], what is 

the area of the region enclosed by the by using a table evaluating the 

x→0 

graph of x f , the x-axis, and the vertical lines 

x = a and x = b? See Figure 2. 

function f (x) = cos π 

47. lim |x|−x 48. lim |x|−x 

x at 

x→2 + x→2 − 

x =− 1 

3 3 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1 

12 , 1 10 , 1 8 , 1 4 , 1 y 

y f (x) 

y 

2 . 


x→2 + x→2 − l T 

51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π y f (x) 

by using a table and evaluating the 

: 

Tangent x→0 x 

P 

line 



and (3, 27). 


x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 x 

a 

3 , 1. b x 

and (x, f (x)), x = 2. 


(c) Create a table to investigate the slope of the DFtangent Figureline 1 to the 

Figure 2 





These questions, traditionally limits? called the tangent problem and the area problem, 


were solved by Gottfried Wilhelm von Leibniz and Sir Isaac Newton during the late 


52. Slope of a Tangent Line For f (x) = x 3 : seventeenth and early eighteenth [−2π, centuries. 2π] and The thesolutions y-window to[−1, the1]. twoIfseemingly you were finding different 

problems are both based on the idea 


lim f of (x) a limit. using Their a graph, solutions what would not you onlyconclude? are related Zoom to each in 

x→0 

other, but are also applicable to many other problems in science and geometry. Here, 

and (3, 27). 


we begin to discuss the tangent problem. The discussion of the area problem begins in 



Chapter 5. 

and (x, f (x)), x = 2. 



x→0 x2 graph of f at 2 using the result from (b). 1 Discuss the Slope offunction a Tangent f (x) = Line cos π to at x a=−0.1, Graph −0.01, −0.001, 


x2 NEED TO REVIEW? The slope of a line 

Notice that the line 

is discussed of finatAppendix the point A.3, (2, 8), p. A-18. 

T in Figure 1 just touches the graph of f at the point P. This unique 

and the secant line from (a). 

−0.0001, 0.0001, 0.001, 0.01, 0.1. 

line is the tangent line to the graph of f at P. But how is the tangent line defined? 

∑ Mathematical Practices Tip 

MPAC 4: Connecting Multiple Representations 

Because limits can be presented numerically, 

graphically, and algebraically, you will have many 

opportunities to reinforce MPAC 4. Students 

may find that one method is more useful or 

appealing than another for a given function, but 

it is important that they have experience with 

multiple approaches. Show students how to solve 

the same problem by creating a table, examining 

a graph, and manipulating the problem using 

algebra. By constructing one representational form 

from another, you will help to solidify the concept 

in the students’ minds as they see that the various 

methods all lead to the same result. 

Teaching Tip 

In the AB pacing guide, two days have been 

allocated to Section 1.1 and two days have been 

allocated to Section 1.2. If you teach limits with all 

three techniques together (tabular, graphical, and 

analytical), then you may spend 4 days working 

on limits and still adhere to the schedule. The 

suggested pacing guide will allow you to cover all 

concepts on the AP ® Calculus Exam and still have 

approximately 3 weeks for review before the test. 

78 



11/01/17 9:51 am


Section 1.1 • Limits of Functions Using Section Numerical 1.1 • Assess and Graphical Your Understanding Techniques 89 79 

(b) Investigate lim cos π by using a table and evaluating In planethe 

geometry, a tangent (c) Graph linethe tofunction a circleC. 

is defined as a line having exactly one 

x→0 x2 function f (x) = cos π point in common with the circle, as shown in Figure 3. However, this definition does 

x 2 at 


not work for graphs in general. For example, in Figure 4, w→1 the − three lines w→1 1 + , 2 , and 3 

x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 contain the point P and have exactly these suggest one point thatinlimcommon C(w) exists? with the graph of f , but they 

3 . 

w→1 

P 

do not meet the requirement of just touching the graph at P. On the other hand, 

(e) Use the graph to investigate lim C(w) and lim C(w). the line 

T just touches the graph of f at P, but it intersects thew→12 graph − at other points. w→12 


+ It is the 

slope of the tangent line 

about the limit? Why do you think this happens? What is your T that Do distinguishes these suggestithat fromlimallC(w) otherexists? 

lines containing P. 

w→12 

Figure 3 Tangent line to a circle at the 

view about using a table to draw a conclusion about So before limits? defining a tangent (f) Use line, thewe graph investigate to investigate its slope, lim which C(w). we denote by m 

point P. 

tan . 

w→0 

We begin with the graph of a function f , a point P on its graph, + 


(g) Use the graph to investigate lim C(w). and the tangent line T 

to f at P, as shown in Figure 5. 

w→13 


− y 

lim f (x) l using l a graph, what would you conclude? Zoom in 

2 1 

The tangent line T 61. to Correlating the graph ofStudent f at P Success must contain to Study theTime point P. Professor We denote Smiththe 

x→0l y f (x) 

3 

on the graph. Describe what you see. (Hint: 

coordinates 

Be sure your 

of P by (c, f (c)). claims Since that afinding student’s a final slopexam requires score two is a function points, of and thewe time have t 

calculator is set to the radian mode.) only one point on the tangent (in hours) line T that , wethe proceed studentas studies. follows. He claims that the closer to 

PAGE 

x −l 8 T 


Suppose we choose any seven point hours Q one = studies, (x, f (x)), the closer othertothan 100% P, the onstudent the graph scoresof f , 

x→2 2 


Tangent as shown in Figure 6. (Q can be to the left or to the right of P; we chose Q to be to the 

(b) How close must x be to 2, soline 

that f (x) is within 0.1 of the hours may cause one to be underprepared for the test, while 

limit? 

right of P.) The line containing studying thesignificantly points P = more (c, f than (c)) seven and Q hours = (x, mayf cause (x)) is called a 

P 

secant line of the graph of 

(c) How close must x be to 2, so that f (x) is within 0.01 of the “burnout.” f . The slope m sec of this secant line is 

limit? 

(a) Write Professor f (x) Smith’s − f (c) 

m claim symbolically as a limit. 

sec = (1) 

58. (a) Use a table to investigate lim(5 − 2x). x 

x→2 (b) Write Professorx Smith’s − c claim using the ε-δ definition 


of limit. 

Figure 4 

Figure 7 shows three different points Q 1 , Q 2 , and Q 3 on the graph of f that are 

limit? 

successively closer to pointSource: P, andSubmitted three associated by the students secantof lines Millikin 1 , University. 

2 , and 3 . The closer 

(c) How close must x be to 2, so that f (x) isthe within point 0.01 Q is ofto the the point 62. P, The thedefinition closer theofsecant the slope lineof isthe to the tangent tangent line line to the T graph . Theof 

line T , 

limit? 

the limiting position of these secant lines, is the tangent line to the graph 

f (x) 

of 

− f 

(c) 

at P. 



. 

y 

y 

y 

x→c 


x − c 

y f (x) 

y f (x) 


Another Secant 

y f (x) 

way to express this slope is to define a new variable l 1 Secant 


h = x 

line 

l2l3 lines 

Q (x, f (x)) − c. Rewrite the slope of the tangent line m tan Q 1 

using h and c. 



fee of $0.21 for each additional l x→2 

T 


l Q T 

2 l T 

Q 


3 

Tangent 

64. If limTangent 

f (x) = 6, can you conclude anything about f (2)? Tangent Explain 

including P 3.5 (c, f ounces. (c)) First-class 

P (c, f (c)) x→2 

P (c, f (c)) 

line 

line 

line 





Source: U.S. Postal c Service Notice 123 x 

c 

3 − x 

out. x 

x 

c x x 

3 x 2 x 1 

x 

Figure (a) 5Find m tan 

a = function slope of C the thattangent modelsline. 

the first-class postage charged, 

Figure 6 m sec = slope of a(a) secant What line. straight line and DF what Figure point? 7 




If the limiting position of xthe approaches secant lines 3. is the tangent line, then the limit of the 


slopes of the secant lines should (c) Does equal the graph the slope suggest of the thattangent lim f (x) exists? If so, what is it? 


x→3 

line. Notice in Figure 7 

w→2− thatw→2 as + the point Q moves closer to the point P, the numbers x get closer to c. So, 

66. (a) Use a table to investigate lim(1 + x) 

these suggest that lim C(w) exists? equation (1) suggests that 

1/x . 

x→0 

w→2 

(e) Use the graph to investigate lim C(w). 

m (b) Use graphing technology to graph g(x) = (1 + x) 1/x tan = [Slope of the tangent line to f at P] 

. 

[ 

] 

Tangent line at P w→0 + (c) What f do (x) (a) −and f (c) (b) suggest about lim(1 + x) 1/x ? 


= Limit of 

as x gets closer x→0 to c 

w→3.5 − x − c 

CAS (d) Find lim(1 + x) 1/x . 

x→0 

60. First-Class Mail As of April 2016, the U.S. InPostal symbols, Service we write 

P 

f (x) − f (c) 


m tan = lim 


x→c x − c 

or partial ounce up to and including 13 ounces. The First-class notation rates limdois read, Challenge “the limit Problems as x approaches c.” 

x→c 

not apply to large envelopes weighing more than 13 ounces. 

The tangent line to Forthe Problems graph of 67–70, a function investigate f at each a point of theP following = (c, f limits. (c)) is the line 


{ 

containing the point P whose slope is 

1 if x is an integer 


f (x) = 


f (x) − f (c) 

in dollars, for a large envelope weighing w ounces. Assume 

m tan = lim 

w>0. 

67. lim f (x) x→c 68. limx −f (x) c 69. lim 

x→2 x→1/2 x→3 

(b) What is the domain ofSecant 

C? 

provided the limit exists. 

Figure 8 

lines 


Mathscoop: The Mathscoop Web site has a 

secant and tangent line applet that you may want 

to demo in class. You can drag one point of the 

secant line to the point of tangency to show the 

relationship between the two lines. The applet 

allows you to click Play to show the animation 

to the students. This can help to show how x 

approaches c when demonstrating how the slope 

of the secant line approaches that of the tangent 

line. A link to this resource is available on the 

Additional Chapter 1 Resources document, 

available for download. 

0 if x is not an integer 

f (x) 70. lim 

x→0 f (x) 

As Figure 8 illustrates, this new idea of a tangent line is consistent with the traditional 

definition of a tangent line to a circle. 

Teaching Tip 

Students should be familiar with finding the slope 

of a line given two points. When those two points 

fall on the graph of a function, we call the line that 

connects them the secant line. The word “secant” 

comes from the Latin word secare, meaning to 

cut. As the two points that create the secant line 

become closer and closer together on the curve, 

the slope of the secant line approaches the slope 

of the tangent line to the point upon which they are 

converging. 

common error 

Students may think that a tangent line 

must only touch the function at one point 

and cannot cross or touch the function 

at any other point. However, the tangent 

line is a localized concept. The tangent 

line may intersect the function at another 

point on the function. See in the figure that 

the graph of y = sin x, with a tangent line 

drawn at x = π / 3. This line intersects the 

function again in quadrant III. 

2π 

y 

1 

21 

Tangent line 

Teaching Tip 

Remind the students of the definition of 

slope with which they are familiar when 

naming the two points (x 1 

, y 1 

) and (x 2 

, y 2 

): 

m − 

= 

y y 2 1 

x − x 

2 1 

Ask the students to compare this to the 

formula for the slope of a secant line. 

fx ( ) − fc () 

msec 

= 

x− 

c 

The only difference between these two 

formulas is the way the two points are 

named. In this case, the points are named 

(c,f(c)) and (c,f(x)). 

Then, point out that as the two points get 

closer and closer together (as x approaches 

c), the slope of the secant line approaches 

the slope of the tangent line. Hence the 

formula: 

fx fc 

m lim ( ) − () 

x c 

. 

tan = 

x→c 

− 

Teaching Tip 

The word “tangent” comes from the Latin 

tangere, meaning to touch. Students may 

be familiar with the term “tangent line” from 

their geometry course, where they learned 

that a line that touches a circle at just one 

point is tangent to the circle. Let them know 

that a tangent line can be found for any 

function, not just circles. 

π 

x 

Section 1.1 • Limits of Functions Using Numerical and Graphical Techniques 79 


11/01/17 9:51 am



88 80 Chapter 1 •• Limits and Continuity 

Teaching Tip 

Consider explaining limits in this way to the 

students: The limit of a function is the value 

that the function approaches as the x value 

gets very close to some value of interest. 

Practically speaking, the easiest way to find 

a limit is to just plug in the number, and if 

the result is an indeterminate form such as 

0 

, this means that you must now do more 

0 

work. 

Teaching Tip 

Clarify that the limit of a function as x 

approaches some number (let’s call that 

number c) is not necessarily the same as 

f (c). The limit of a function is the y value, 

f (c), that the function gets close to as the 

x value gets close to (but not equal to) 

the value, c. In the figure below, the value 

of f (c) = 3, while the limit of f (x) as x 

approaches c is 1. 

y 

3 

2 

1 

c 

(c, 3) 

y f (x) 

AP® CaLC skill builder 

for example 1 

Interpreting a Limit Expressed 

Symbolically 

Express the limit lim 5x 

= 10 in words and 

x→2 

interpret it. 

Solution 

In words, lim 5x 

= 10 is stated: The limit 5x 

x→2 

as x approaches 2 is equal to 10. 

The limit is interpreted as: The value 

of the function f(x) = 5x can be made as 

close as we please to 10 by choosing x 

sufficiently close to but not equal to 2. 

On the graph of the function y = 5x 

shown below, explain that as you choose 

values of x closer and closer to the 2 points 

on the graph close in on the value y = 10. 

y 

16 

14 

12 

10 

8 

6 

4 

2 

22 22 

2 

4 

6 

x 

x 

2x 2 if x < 1 

We have begun to 

33. f (x) = 

3x 2 at c = 1 

53. 

answer 

Slope 

the 

of a 

tangent 

Tangent 

problem 

Line For 

by 

f 

introducing 

(x) = 1 

− 1 if x > 1 

2 x2 − 

the 

1: 

idea of a limit. 

Now we describe the idea of a limit in more detail. 

x 3 (a) Find the slope m 

if x < −1 

sec of the secant line containing the 

34. f (x) = 

x 2 at c =−1 The Idea of a Limit points P = (2, f (2)) and Q = (2 + h, f (2 + h)). 

− 1 if x > −1 

We begin by asking a question: (b) Use What the result doesfrom it mean (a) tofor complete a function the following f to have table: a limit L 

x 2 if x ≤ 0 

35. f (x) = 

at c = 0 as x approaches some fixed number c? To answer the question, we need to be more precise 

2x + 1 if x > 0 

⎧ 

about f , L, and c. To have a limit h at −0.5 c, the function −0.1 −0.001 f must be defined 0.001 everywhere 0.1 0.5 in 

⎨ x 2 if x < 1 

an open interval containing themnumber sec c, except possibly at c, and L must be a number. 

36. NEED f (x) TO= 

REVIEW? 2 If a < if b, x the = 1open 

at c = 1 Using these restrictions, we introduce the notation 

interval (a, b) ⎩consists −3x + 2of all if numbers x > 1 x 


for which a < x < b. Interval notation is 

lim 

as h → 0. 

f (x) = L 

x→c 

discussed in Appendix A.1, p. A-5. 



which is read, “the limit as x point approaches P = (2, cf (2))? of f (x) is equal to the number L.” The 

In Problems 37–40, sketch a graph of a function with notation the given lim f (x) = L can(e) be described On the same as 

x→c set of axes, graph f and the tangent line to f at 


37. lim f (x) = 3; lim f (x) = 3; lim 

The value f (x) P = can (2, be f (2)). made as close as we please to L, 

f (x) = 1; 

x→2 x→3− x→3 + for x sufficiently close to c, but not equal to c. 

f (2) = 3; f (3) = 1 


Figure 9 shows that as x gets closer to c, the value f (x) gets closer to L. In 

38. lim f (x) = 0; lim f (x) =−2; lim Figure f (x) =−2; 9(a), lim f (x) = L(a) and Find f (c) the slope = L, mwhile sec of the in secant Figureline 9(b), containing lim f (x) the = L, but 

x→−1 x→2− x→2 + x→c x→c 



f (c) = L. In Figure 9(c) lim f (x) = L, but f is not defined at c. 

x→c 

39. lim f (x) = 4; lim f (x) =−1; lim 


f (x) = 0; 

x→1 x→0− x→0 + 

y 

y 

f (0) =−1; f (1) = 2 

(c, f(c)) 

y 

h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1 


m sec 

x→2 x→−1 x→1 

L f (−1) = 1; f (2) = 3 

L 

(c) Investigate the limit L of the slope of the secant line found 

(c, L) 

in (a) as h → 0. 



each limit. 

point P = (−1, f (−1))? 

|x − 5| 

y 5 |xf (x) − 5| 

y 5 f (x) 

y 5 f (x) 

41. lim 

42. lim 

43. lim 

x→5 + x − 5 

x→5 − x − 5 

 

x→ 12 

2x 


− 

at P = (−1, f (−1)). 

a c b x 

a c b x 

PAGE 

44. lim 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. lim 

− 

x→ 23 

2x 

+ 

85 55. (a) Investigate lim cos π a c b 

x 


(a) f is defined at c; f(c) 5 L (b) f is defined at c; f(c) Þ L x→0 x 

function f (x) = cos π (c) f is not defined at c 

Figure 9 lim f (x) = L 

47. lim |x|−x 48. lim |x|−x 

x at 

x→c 

x→2 + x→2 − 

x =− 1 

3 3 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1 

12 , 1 10 , 1 8 , 1 4 , 1 2 . 


x→2 + x→2 − EXAMPLE 1 Interpreting a Limit Expressed 

51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π Symbolically 


: Express the limit 

x→0 x 


function f x(x) = cos π 2 − 1 

lim 

x at 

and (3, 27). 

x→−1 x 


x =−1, − 1 + 

3 , − 1 =−2 

1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 in words and interpret it. 

3 , 1. 

and (x, f (x)), x = 2. 


(c) Create a table to investigate the slope of the tangent line to the x 2 − 1 

Solution In words, lim about =−2, the limit? is stated, Why do“The you think limitthis as happens? x approaches What is −1 of 


x→−1 x + 1 


(d) On the same set of axes, graph f , the tangent x 2 −line 1 to the graph 

limits? 

of f at the point (2, 12), and the secant line from (a). is equal to the number −2.” The limit is interpreted “The value of the function 

x + 1 



f (x) = x 2 − 1 

can be made [−2π, as close 2π] asand wethe please y-window to −2 [−1, by1]. choosing If you were x sufficiently finding 

x + 1 



x→0 

close to, but not equal to, −1.” ■ 

and (3, 27). 



calculator is set to theNOW radianWORK mode.) AP® Practice Problem 7. 

and (x, f (x)), x = 2. 



x→0 x2 graph of f at 2 using the result from (b). 2 Investigate a Limit Using functiona f Table (x) = cos π at x =−0.1, −0.01, −0.001, 


x2 We can use a table to better understand what it means for a function to have a limit as x 


−0.0001, 0.0001, 0.001, 0.01, 0.1. 

approaches a number c. 

AP® Exam Tip 

Questions about limits that appear on the 

exams often require algebraic manipulation. 

Students must become comfortable with 

factoring the difference of squares and 

trinomials. 

Teaching Tip 

Consider teaching the students about limits 

using graphical, tabular, and algebraic methods 

together. For example, first estimate the limit of 

a function graphically. Then solidify that estimate 

using tabular techniques. Finally, use algebraic 

manipulation (when necessary) and substitution to 

show that all three methods give the same answer. 


Connecting Multiple Representations 

To find the limit as x approaches a, we choose x 

sufficiently close to a. Ask students to define the 

term “sufficiently.” Observe that as we choose 

any x as close to a as we want, the values of 

f(x) approach a number b. Draw the graph of 

a function and have students describe how the 

points on the graph of f move along the curve 

as x approaches b. Explain that the limit is an 

important concept that will be used to define the 

continuity of a function. 

On the graph that follows, note that the curve 

is smooth and unbroken so that the points on 

80 



11/01/17 9:51 am




x→0 x2 EXAMPLE 2 Investigating (c) Grapha Limit the function Using C. a Table 

function f (x) = cos π x 2 at 

Investigate lim(2x + 5) using (d) aUse table. the graph to investigate lim C(w) and lim C(w). Do 

w→1− w→1 + x→2 

x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 these suggest that lim C(w) exists? 

3 . Solution We create Table 1 by evaluating f (x) w→1 = 2x +5 at values of x near 2, choosing 

numbers x slightly less than(e) 2 and Use numbers the graph to x slightly investigate greater lim than C(w) 2. and lim C(w). 

w→12− w→12 


+ Do these suggest that lim C(w) exists? 


w→12 

TABLE 1view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim C(w). 

w→0 

(d) Use technology to graph numbers f . Begin x slightly with the less x-window than 2 

+ (g) Use the graph to investigate numbers limx slightly C(w). greater than 2 

−−−−−−−−−−−−−−−−−−−−−→ 

←−−−−−−−−−−−−−−−−−−−−−−−− 

w→13 


− 

x 1.99 1.999 1.9999 1.99999 → 2 ← 2.00001 2.0001 2.001 2.01 

lim f (x) using a graph, what would you conclude? Zoom in 61. Correlating Student Success to Study Time Professor Smith 

f (x) = x→0 2x + 5 8.98 8.998 8.9998 8.99998 f (x) 


claims approaches that a student’s 9 9.00002 final exam score 9.0002 is a function 9.002 of the time 9.02 t 



PAGE 

x − 8 Table 1 suggests that the value sevenof hours f (x) one = studies, 2x + 5the can closer be made to 100% “asthe close student as we scores please” 


x→2 2 


to 9 by choosing x “sufficiently close” to 2. This suggests that lim(2x + 5) = 9. ■ 

(b) How close must x be to 2, so that f (x) is within 0.1 of the hours may cause one to be underprepared x→2 for the test, while 

limit? 


NOW WORK Problem 9. 

(c) How close must x be to 2, so that f (x) is within 0.01 of the “burnout.” 

limit? 


In creating Table 1, first we used numbers x close to 2 but less than 2, and then we 


x→2 used numbers x close to 2(b) but Write greater Professor than 2. Smith’s When claim x < 2, using we the say, ε-δ “xdefinition 

is approaching 


of limit. 

2 from the left,” and the number 9 is called the left-hand limit. When x > 2, we say, 

limit? 

“x is approaching 2 from the Source: right,” Submitted and the bynumber the students 9 is ofcalled Millikinthe University. right-hand limit. 

(c) How close must x be to 2, so that f (x) isTogether, within 0.01 these of the are called 

62. 

the 

The 

one-sided 

definition of 

limits 

the slope 

of f 

of 

as 

the 

x approaches 

tangent line to 

2. 

the graph of 

limit? 

One-sided limits are symbolized as follows. The left-hand limit, written f (x) − f (c) 



. 

x→c 


lim f (x) = L x − c 

left 

x→c 


Another way − 

to express this slope is to define a new variable 




is read, “The limit as x approaches c from the left of f (x) equals L 

63. If f (2) = 6, can you conclude anything about left .” It means that 

lim f (x)? Explain 


the value of f can be made as close as we please to the number L left by x→2 choosing x < c 



and sufficiently close to c. 



Similarly, the right-hand x→2 limit, written 




lim 

65. The graph of f f (x) (x) = = xL− right 3 

x→c + 

is a straight line with a point punched 


3 − x 

out. 

is read, “The limit as x approaches c from the right of f (x) equals L right .” It means that 

(a) Find a function C that models the first-class 

the 

postage 

value of 

charged, 

f can be made(a) as close Whatas straight we please line and to the what number point? L right by choosing x > c 


and sufficiently close to c. 




EXAMPLE 3 Investigating (c) Does a Limit the graph Using suggest a Table that lim f (x) exists? If so, what is it? 


x→3 

w→2− w→2 + e x − 1 

Investigate lim 66. using (a) a Use table. a table to investigate lim(1 + x) 


1/x . 

x→0 x 

x→0 

w→2 


(b) Use graphing technology to graph g(x) = (1 + x) 1/x . 

w→0 + Solution The domain of f (c) (x) What = ex − 1 

do (a) and is {x|x (b) suggest = 0}. So, about f lim is defined (1 + x) 1/x everywhere ? in 


x 

x→0 

w→3.5 − an open interval containing CAS (d) theFind number lim(1 0, + except x) 1/x . for 0. 

x→0 

60. First-Class Mail As of April 2016, the U.S. Postal Service 

e x − 1 

We create Table 2, investigating the left-hand limit lim and the right-hand 


x→0 − x 

and including 1 ounce, plus a flat fee of $0.21 for each additional e x − 1 

limit lim . First, 

or partial ounce up to and including 13 ounces. First-class rates do Challenge we evaluate Problems f at numbers less than 0, but close to zero, and 

x→0 + x 

not apply to large envelopes weighing more than then13at ounces. numbers greaterFor than Problems 0, but close 67–70, toinvestigate zero. each of the following limits. 


{ 1 if x is an integer 

TABLE f (x) = 

(a) 2Find a function C that models the first-class postage charged, 


in dollars, for a large envelope x approaches weighing0w from ounces. theAssume 

left 

x approaches 0 from the right 

w>0. 

−−−−−−−−−−−−−−−−−−−−→ 67. lim f (x) 68. lim f (x) 69. lim f (x) 70. lim f (x) 

x→2 

←−−−−−−−−−−−−−−−−−−−−− 

x→1/2 x→3 x→0 



(b) What x is the domain −0.01 of C? −0.001 −0.0001 −0.00001 → 0 ← 0.00001 0.0001 0.001 0.01 

f (x) = ex − 1 

0.995 0.9995 0.99995 0.999995 f (x) approaches 1 1.000005 1.00005 1.0005 1.005 

x 

∑ MPAC Tip continued 

the graph gradually close in on the limiting value 

3 

y = as the values of x approach 1. 

2 

y 

3 

2 

1 

1 

2 

3 

x 

Teaching Tip 

Remember to model proper notation for your 

students when working with one-sided limits. 

• One-sided limit of f(x) as x approaches a from 

the left: lim fx ( ) 

x→a 

− 

• One-sided limit of f(x) as x approaches a from 

the right: lim fx ( ) 

x→ a 

+ 

Calculator Tip 

Most calculators have table settings 

that default to display a starting value 

(usually zero) and additional values with a 

consistent incremental change thereafter 

(like 0, 1, 2, 3…). Table settings can 

be changed (usually from Auto to Ask 

so students can input any independent 

variable they want. The table will then 

display the y-values that correspond to 

the selected x-values. 

TRM Section 1.1: Worksheet 1 

This worksheet includes 3 problems for 

which a student completes a premade 

table to find the limit of a function by using 

a table. 


for example 3 

Investigating a Limit Using a Table 

2 

x −1 

Investigate lim using a table. 

x→1 

x −1 

Solution 

2 

x −1 

The domain of fx ( ) = is {x | x ≠ 1}. So 

x −1 

f is defined everywhere in an open interval 

containing 1, except for 1. We create a 

table by evaluating the function at values of 

x near 1, choosing values that are slightly 

less than 1 and slightly greater than 1. 

x 

f(x) 

0.9 1.9 

0.99 1.99 

0.999 1.999 

1 Not defined 

1.001 2.001 

1.01 2.01 

1.1 2.1 

This table suggests that the value of 

2 

x −1 

fx ( ) = gets as close as we please 

x −1 

to 2 as x gets very close to 1. That is, there 

2 

x −1 

is evidence indicating that lim 

− = 2. 

x→1 

x 1 

In Section 1.2, students will verify 

2 

x −1 

lim algebraically by factoring the 

x→1 

x −1 

numerator and reducing the expression 

before substituting the value of 1. 

Section 1.1 • Limits of Functions Using Numerical and Graphical Techniques 

81 


11/01/17 9:51 am




AP® Exam Tip 

It is helpful to be familiar with limits of 

ax 

the form lim sin( ) before taking the 

x→0 

exam. 

x 

ax 

lim sin( ) = a 

x→0 

x 

Consider working with variations of this 

problem as well, such as 

ax a 

lim sin( ) = 

x→0 

bx b 

These limits can be found using the 

Squeeze Theorem, but familiarity will 

save time during the exam. 

On completion of Section 4.5, the 

student will be able to solve problems 

of this form using L’Hôpital’s Rule, but 

in the meantime, students may prefer 

to rely on these two forms rather than 

finding these limits using the Squeeze 

Theorem. 


for example 4 

Investigating a Limit Using a Table and 

Technology 

x 

Investigate lim sin(5 ) using a table. 

x→0 

x 

Solution 

sin(5 x) 

The domain of the function fx ( ) = 

x 

is {x | x ≠ 0}. So f is defined everywhere 

on the open interval containing 0, except 

at 0. We create a table by evaluating the 

function at values of x near 0, choosing 

values that are slightly less than 0 and 

slightly greater than 0. 

2x 2 if x < 1 

33. f (x) = 

3x 2 at c = 1 

53. Slope 

e x of 

− 

a 

1 

Tangent Line For 

e x − 

f (x) 

1 

Table 2 suggests that lim = 1 and lim = 1. 

1 − 1 if x > 1 

2 x2 − 1: 

x→0 

− x 

x→0 + x 

x 3 if x < −1 

e x − 1 (a) Find the slope m sec of the secant line containing the 

34. f (x) = 

x 2 at c =−1 This suggests lim = 1. points ■ 

− 1 if x > −1 

x→0 

P = (2, f (2)) and Q = (2 + h, f (2 + h)). 

x 


x 2 if x ≤ 0 

NOW WORK Problem 13 and AP® Practice Problem 6. 

35. f (x) = 

at c = 0 

2x + 1 if x > 0 

⎧ 

h −0.5 −0.1 −0.001 0.001 0.1 0.5 

⎨ x 2 if x < 1 

m sec 

36. f (x) = 2 if x = 1 at c = 1 EXAMPLE 4 Investigating a Limit Using a Table and Technology 

⎩ 

−3x + 2 if x > 1 

sin x (c) Investigate the limit of the slope of the secant line found in (a) 

Investigate lim using a table. 

x→0 x 

as h → 0. 



Solution The domain of the function point P = f (x) (2, f = (2))? 

sin x is {x | x = 0}. So, f is defined 


x 



everywhere in an open interval containing 0, except for 0. 

P = (2, f (2)). 


x→2 x→3− x→3 + We investigate the one-sided limits of sin x as x approaches 0 by using a graphing 

f (2) = 3; f (3) = 1 

54. Slope of a Tangent 

x 

Line For f (x) = x 2 − 1: 

calculator to set up a table. When making the table, we choose numbers x (in radians) 

38. lim f (x) = 0; lim f (x) =−2; lim slightly f (x) =−2; less than 0 and numbers (a) Find slightly the slope greater m sec than of the 0. secant See Figure line containing 10. the 

x→−1 x→2− x→2 + 


Figure f (−1) 10 

The table in Figure 10 suggests that lim f (x) = 1 and lim f (x) = 1. This 

is not defined; f (2) =−2 

x→0 − 

x→0 + 



f (x) = 0; sin x 

x→1 x→0− x→0 + suggests that lim = 1. ■ 

NOTE f (x) = sin x is an even function, 

f (0) =−1; 

x 

x→0 x 

f (1) = 2 

h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1 

so the Y 1 values in Figure 10 are 


m sec 


symmetric x→2 

about x = 0. 

x→−1 x→1 

f (−1) = 1; f (2) = 3 


in (a) as h → 0. 

In Problems 41–50, use either a graph or a table to 3investigate 

Investigate a Limit (d) Using What isa the Graph slope of the tangent line to the graph of f at the 

each limit. 

The graph of a function can alsopoint helpP us = investigate (−1, f (−1))? limits. Figure 11 shows the graphs 

|x − 5| 

|x − 5| 

41. lim 

42. lim 

43. of three lim 

x→5 + x − 5 

x→5 − x − 5 

 

different 

x→ 12 

2x functions (e) f , g, Onand theh. same Observe set of axes, that in graph eachf function, and the tangent as x line getstocloser 

f 

− 

to c, whether from the left or from at P the = (−1, right, f the (−1)). value of the function gets closer to the 

number L. This is the key idea of a limit. 

PAGE 

44. lim 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. lim 

− 

x→ 23 

2x 

+ 


Notice in Figure 11(b) that the value x→0 of g atxc does not affect the limit. Notice in 

 

Figure 11(c) that h is not defined function at c, but f (x) the= value cos π of 

47. lim |x|−x 48. lim |x|−x 

x at h gets closer to the number L for 

x sufficiently close to c. This suggests 

x→2 + x→2 − 

x =− 1 that 

3 3 2 , − 1 the limit 

4 , − 1 of 

8 , − 1 a function 

10 , − 1 as 

12 ,..., 1 x approaches 

12 , 1 10 , 1 8 , 1 4 , 1 c does 

not depend on the value, if it exists, of the function at c. 

2 . 


x→2 + x→2 − 

51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π y 


: 

y 

y x→0 x 

(c, g(c)) 

(a) Find the slope of the secant y 5 line f (x) containing the points (2, 12) 

y function 5 g(x) f (x) = cos π x at 

y 5 h(x) 

and (3, 27). 

y 

y 


x =−1, − 1 3 , − 1 y 

5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 (c, L) 

3 , 1. 

L 

L 

L 

and (x, f (x)), x = 2. 

y 

y 

(c) Compare the results y from (a) and (b). What do you conclude 






limits? 


c 

x 


52. Slope of a Tangent Line For f (x) = x 3 x c x 

x 

x c x 

x 

: 


(a) f (c) 5 L (b) g(c) Þ L (c) h(c) is not defined 



x→0 

DF Figure and 11 (3, 27). 




and (x, f (x)), x = 2. 



x→0 x2 graph of f at 2 using the result from (b). EXAMPLE 5 Investigatingfunction a Limit f (x) Using = cos a Graph 

NEED TO REVIEW? Piecewise-defined 

{ π at x =−0.1, −0.01, −0.001, 


x2 functions are discussed in Section P.1, 

3x + 1 if x = 2 

p. 8. of f at the point (2, 8), and the secant line 

Use 

from 

a graph 

(a). 

to investigate lim f 

x→2 −0.0001, (x) if f (x) 0.0001, = 0.001, 10 0.01, 0.1. if x = 2 . 

This table suggests that the value of 

sin(5 x) 

fx ( ) = gets closer and closer to 

x 

5 as x gets very close to 0. That is, there is 

sin(5 x) 

evidence indicating that limx→0 

= 5. 

x 

Teaching Tip 

Students should have become familiar with 

graphing constant, linear, and basic quadratic, 

cubic, radical, rational, and trigonometric functions 

before taking this course. If they are not, consider 

how to remediate these concepts, because basic 

graphing skills are necessary to complete the 

exam on time. There are many videos on the Khan 

Academy Web site and others that can guide a 

student to a stronger understanding of the basic 

functions. Consult the Additional Chapter 1 

Resources document for suggested URLs. 


This worksheet includes 19 problems that have 

students determine limits using graphs. 


This worksheet contains 3 problems in which the 

student is given a piecewise function. They are 

asked to find the one- and two-sided limits for 

a particular x-value and then verify the limits by 

graphing the function. 

82 



11/01/17 9:51 am



(b) Investigate lim cos π by using a table and Solution evaluating Thethe 

function f is a(c) piecewise-defined Graph the function function. C. Its graph is shown in Figure 12. 

x→0 x2 y 

function f (x) = cos π Observe that as x approaches 

x 2 at 

(d) 

2 

Use 

from 

the graph 

the left, 

to investigate 

the value 

lim 

of f 

C(w) 

is close 

and 

to 

lim 

7, 

C(w). 

and as 

Do 

x 

w→1− w→1 + 

10 

approaches 2 from the right, the value of f is close to 7. In fact, we can make the value 

(2, 10) 

x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 of f as close as we please to 7 these by choosing suggest that x sufficiently lim C(w) exists? 

3 . 

w→1 

close to 2 but not equal to 2. 

This suggests lim f (x) = 7. ■ 

(e) Use the graph to investigate lim C(w) and lim x→2 

w→12− w→12 


7 

If we use a table to investigate lim f (x), the result is the same. See Table 3. 

Do these 


x→2 suggest that lim C(w) exists? 

w→12 

view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim 

TABLE 3 

C(w). 

w→0 + 


(g) Use the graph to investigate lim w→13 

[−2π, 2π] and the y-window [−1, 1]. If you were x approaches finding 2 from the left 

x approaches − 2 from the right 

−−−−−−−−−−−−−−−−−−−−−→ 

←−−−−−−−−−−−−−−−−−−−−−− 


x→0 x 1.99 1.999 1.9999 1.99999 → 2 ← 2.00001 2.0001 2.001 2.01 



f (x) 6.97 6.997 6.9997 6.99997 


(in hours) that f (x) the approaches student studies. 7 7.00003 He claims 7.0003 that the7.003 closer 7.03 to 

PAGE 

x − 8 


85 57. (a) Use a table 1 to investigate 2 3 limx 

. 

x→2 2 


(b) How close must { x be to 2, so that f (x) is within 

Figure 

0.1 of 

12 

the 

shows that f hours (2) = may 10cause but that one to this bevalue underprepared has no impact for the test, on the while limit as 

3x + 1 if x = 2 

Figure 12 limit? f (x) = 

x approaches 2. In fact, f (2) studying can equal significantly any number, more than and seven it would hours have mayno cause effect on the 

10 if x = 2 

(c) How close must x be to 2, so that f (x) islimit within as0.01 x approaches of the 2. “burnout.” 

limit? 


x→2 

We make the following(a) observations: Write Professor Smith’s claim symbolically as a limit. 

• The limit L of a function (b) y Write = f (x) Professor as x approaches Smith’s claima number using thec ε-δ does definition not depend 

(b) How close must x be to 2, so that f (x) is within on the 0.1 of value the of f at c. of limit. 

limit? 

• The limit L of a function Source: y = Submitted f (x) as xbyapproaches the students aofnumber Millikin University. c is unique; that is, 

(c) How close must x be to 2, so that f (x) is within a function 0.01 of the cannot have more than one limit. (A proof of this property is given in 


limit? 

Appendix B.) 

f (x) − f (c) 


• If there is no single number y = f that (x) at the the value point of (c, f f approaches (c)) is m tan = aslim 

. 

x→c 

x gets 


xclose − c to c, we 

say that f has no limit as x approaches c, or more simply, that the limit of f does 



not exist at c. 






x→2 

your reasoning. NOW WORK AP® Practice Problem 4. 




x→2 

letter rates do not apply to letters Examples 6 and 7 that follow your illustrate reasoning. situations in which a limit does not exist. 


65. The graph of f (x) = x − 3 

CALC 


Source: U.S. Postal Service Notice 123 EXAMPLE 6 Investigating 3 − x 

out. a Limit Using a Graph 

CLIP 

{ 

(a) Find a function C that models the first-class postage charged, (a) What straight line and x 

what if 

point? x < 0 

Use a graph to investigate lim f (x) if f (x) = 


x→0 1 if x > 0 . 


(b) What is the y domain of C? 

Solution Figure 13 shows the graph x approaches of f . We 3. first investigate the one-sided limits. The 

(c) Graph the2 

function C. 

graph suggests that, as x approaches (c) Does the 0 from graphthe suggest left, that lim f (x) exists? If so, what is it? 


x→3 

1 

w→2− w→2 + lim 

66. (a) Use a table to investigate x→0 lim(1 + x) 


. 

x→0 

w→2 

(e) Use the graph to investigate lim and as x approaches 0 from(b) theUse right, graphing technology to graph g(x) = (1 + x) 1/x . 

2 1 1 2 3 x 

w→0 + (c) What lim do (a) and (b) suggest about lim(1 + x) 1/x ? 

1 

(f) Use the graph to investigate lim x→0 + x→0 

w→3.5 − CAS (d) Find lim(1 + x) 1/x . 

2 

Since there is no single number that x→0 

the values of f approach when x is close to 0, we 

60. First-Class Mail As of April 2016, the U.S. conclude Postal Service that lim f (x) does not exist. ■ 

charged $0.94 postage { for first-class large envelope weighing x→0 

x if x < 0 

up to 

Figure and13 including f (x) = 1 ounce, plus a flat fee of $0.21 for each Tableadditional 

4 uses a numerical approach to support the conclusion that lim f (x) does not 

1 if x > 0 

or partial ounce up to and including 13 ounces. First-class rates do Challenge Problems 

x→0 

exist. 

not apply to large envelopes weighing more than 13 ounces. For Problems 67–70, investigate each of the following limits. 


{ 

TABLE 4 

1 if x is an integer 

f (x) = 

(a) Find a function C that models the first-class postage x approaches charged, 0 from the left 

0 if xis approaches not an integer 0 from the right 

in dollars, for a large envelope weighing w ounces. −−−−−−−−−−−−−−−−−−−−→ 

Assume 

←−−−−−−−−−−−−−−−−−−−−− 

w>0. 

x −0.01 −0.00167. −0.0001 lim f (x) x→2 

→68. 0 lim f ←(x) x→1/2 

69. 0.0001 lim f (x) 0.001 70. 

x→3 

lim 0.01 


f (x) −0.01 −0.001 −0.0001 no single number 1 1 1 

Teaching Tip 

In Section 1.3, students will learn the formal 

definition of continuity. Problems about continuity 

on the exams often involve piecewise functions. 

When investigating the limits of piecewise functions 

graphically, consider checking the limits using 

direct substitution into the appropriate portions 

of the piecewise function. Doing so will lay the 

groundwork for Section 1.3. Here you want to focus 

heavily on getting students to write the one-sided 

limits correctly. A graphic organizer can be useful 

here, and a teacher modeling this correctly when 

students first see it will save many mistakes later. 

This is the template. 

lim ( what function ?) = ( what yvalue ?) 

x → 1 

+ 

( x approaches what value ?) 

The student can then set up the organizer with 

blanks for the parenthetical expressions. 


Examples 5 and 6 lead to the following result. 


for example 6 

Investigate a Limit Using a Graph 

Use a graph to investigate lim fx ( ) if 

x→2 

⎧⎪ 

2 

fx ( ) = 

− x + 4 x ≠ 2 

⎨ 

⎩⎪ 2 x = 2 

Solution 

The function f is a piecewise-defined function. 

The graph is shown. y 

24 

22 

4 

2 

22 

24 

2 

4 

x 

The graph suggests that as x approaches 2 

from the left, 

lim fx ( ) = 0. 

x→2 

− 

and as x approaches 2 from the right, 

lim fx ( ) = 0. 

x→ 2 

+ 

Because the right- and left-hand limits 

approach the same number, the graph 

suggests that 

lim fx ( ) = 0. 

x→2 

Note: If the left- and right-hand limits 

agree, then the limit exists. It does not 

matter that the value f (2) does not equal 0. 


for example 6 

Analyzing a Limit Using a Graph 

Suppose the function f has the graph 

shown. 

f (x) 

a b c d e 

Investigate whether or not each of the 

following limits exists. 

(a) lim fx ( ) 

(e) lim fx ( ) 

x→a 

− x→ b 

+ 

(b) lim fx ( ) (f) lim fx ( ) 

x→ a 

+ x→b 

(c) lim fx ( ) (g) lim fx ( ) 

x→a 

x→d 

(d) lim fx ( ) (h) lim fx ( ) 

x→b 

− x→e 

Solution 

(a) The graph suggests lim fx ( ) exists. 

x→a 

− 

(b) The graph suggests lim fx ( ) exists. 

x→ a 

+ 

(c) The graph suggests lim fx ( ) exists. 

x→a 

(d) The graph suggests lim fx ( ) exists. 

x→b 

− 

(e) The graph suggests lim fx ( ) exists. 

x→ b 

+ 

(f) The graph suggests lim fx ( ) does not 

x→b 

exist because the left-hand and righthand 

limits are not equal. 

(g) The graph suggests lim fx ( ) exists. 

(h) The graph suggests 

x→d 

x→e 

x 

lim fx ( ) exists. 

x→0 f (x) 83 



11/01/17 9:52 am


Sullivan SullivanAP 


Alternate Example 

Investigating a Limit 

x 

Investigate lim | − 1| . 

x→1 

x −1 

Solution 

We can investigate a limit such as this one 

in three ways: graphically, algebraically, and 

numerically. 

Graphically 

The graph suggests that lim fx ( ) does not 

x→1 

exist because lim fx ( ) ≠ lim fx ( ). 

24 

Algebraically 

− + 

x→1 x→1 

22 

y 

4 

2 

22 

24 

| x −1| 

fx ( ) = is equivalent to the 

x −1 

piecewise-defined function. 

⎧ −( x −1) 

=−1 

x < 1 

⎪ 

fx ( ) = 

x −1 

⎨ 

⎪ ( x −1) 

− = 1 x > 1 

⎩ 

⎪ x 1 

lim fx ( ) does not exist because 

x→1 

lim fx ( ) ≠ lim fx ( ). 

− + 

x→1 x→1 

Using a Table 

x f(x) 

0.9 −1 

0.99 −1 

0.999 −1 

1.001 1 

1.01 1 

1.1 1 

The table suggests lim fx ( ) does not exist 

x→1 

because lim fx ( ) ≠ lim fx ( ). 

− + 

x→1 x→1 

2 

4 

x 

2x 2 if x < 1 

33. f (x) = 

3x 2 at c = 1 THEOREM 53. Slope of a Tangent Line For f (x) = 1 

− 1 if x > 1 

2 x2 − 1: 

The limit L of a function y = f (x) as x approaches a number c exists if and only if 


if x < −1 


both one-sided limits exist at c and both one-sided limits are equal. That is, 

34. f (x) = 

x 2 at c =−1 


− 1 if x > −1 

lim f (x) = L(b) ifUse andthe only result if from lim(a) f to (x) complete = limthe following x 2 x→c table: 

x→c 

if x ≤ 0 

x→c + 

35. f (x) = 

at c = 0 

2x + 1 if x > 0 

NOW 

⎧ 

h WORK −0.5Problems −0.1 25, −0.001 31, and AP® 0.001 Practice 0.1 Problem 0.5 5. 

⎨ x 2 if x < 1 

m sec 

36. f (x) = 2 if x = 1 at c = 1 A one-sided limit is used to describe the behavior of functions such as 

⎩ 

−3x + 2 if x > 1 

f (x) = √ √ x − 1 near x = (c) 1. Since Investigate the √domain the limit of of ftheis slope {x|xof ≥the 1}, secant the left-hand line foundlimit, 

in (a) 

lim x − 1 makes no sense. But as h lim → 0. x − 1 = 0 suggests how f behaves near and 

x→1 − x→1 + 

Applications y and Extensions 

to the right of 1. See Figure(d) 14 and What Table is the5. slope They of suggest the tangent limlinef to(x) the= graph 0. of f at the 

point P = (2, f (2))? x→1 + 

In Problems 2 37–40, sketch a graph of a function with the given 



TABLE 5 

1 

P = (2, f (2)). 



x→2 x→3− x→3 + 

−−−−−−−−−−−−−−−−−−−−−→ 

f (2) = 3; f (3) = 1 


1 2 3 4 5 6 x 

x 1.009 1.0009 1.00009 1.000009 1.0000009 → 1 


=−2; 


Figure x→−1 14 f (x) = √ f (x) = √ x − 1 0.0949 0.03 0.00949 0.003 0.000949 f (x) approaches 0 

xx→2 − 1− x→2 + 



39. lim f (x) = 4; 

x→1 

lim f (x) =−1; 

x→0− lim f (x) = 0; 

x→0 + 

(b) Use the result from (a) NOW to complete WORK the AP® following Practicetable: 

Problem 1. 

f (0) =−1; f (1) = 2 

Using numeric tables and/orhgraphs −0.1gives −0.01us−0.001 an idea−0.0001 of what a0.0001 limit might 0.001 0.01 be. That 0.1 

40. lim f (x) = 2; 

x→2 

lim f (x) = 0; 

x→−1 lim is, these methods suggest a limit, 

f (x) = 1; 

m sec 

but there are dangers in using these methods, as the 

x→1 following example illustrates. 

f (−1) = 1; f (2) = 3 


EXAMPLE 7 Investigatingin a(a) Limit as h → 0. 



each limit. 

Investigate lim sin π 

x→0 x . 2 point P = (−1, f (−1))? 

|x − 5| 

|x − 5| 

41. lim 

42. lim 

43. lim 

x→5 + x − 5 

x→5 − x − 5 

 

x→ 12 

2x 


Solution The − 

domain of the function at P = (−1, f (x) f = (−1)). sin π is {x|x = 0}. 

x 

2 

PAGE 

44. lim 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. Suppose lim 

− 

x→ 23 

2x we let 

+ 

85 x approach 55. (a) Investigate zero in thelimfollowing cos π byway: 

using a table and evaluating the 

x→0 x 

function f (x) = cos π TABLE 47. lim6 

|x|−x 48. lim |x|−x 

x at 

x→2 + x→2 − 

x approaches 0 from the left 

x =− 1 

3 3 2 , − 1 4 , − 1 8 , − 1 x 

10 , approaches − 1 12 ,..., 1 

0 from 

12 , 1 10 , the 1 8 , 1 right 

4 , 1 2 . 

49. lim x−x 50. lim 

−−−−−−−−−−−−−−−−−−−−→ 

x−x 

←−−−−−−−−−−−−−−−−−−−−− 

x→2 + x→2 

x − 1 

− 1 − 1 − 1 

51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π 1 1 1 1 

→ 0 ← 

10 100 


: 1000 10,000 

x→0 x10,000 

1000 100 10 

f (x)(a) = Find sin π the slope of0the secant line 0 containing0the points (2, 12) 0 f (x) approaches function f (x) 0 = cos π x at 0 0 0 0 

x 2 

and (3, 27). 


x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 Table 6 suggests that lim sin π 3 , 1. 

and (x, f (x)), x = 2. 

x→0 x = 0. 2 


(c) Create a table to investigate the slope of the tangent 

Now suppose 

line to the 

we let x approach zero as follows: 



TABLE your view about using a table to draw a conclusion about 

(d) 7On the same set of axes, graph f , the tangent line to the graph 

limits? 

of f at the point (2, 

x 

12), 

approaches 

and the secant 

0 from 

line 

the 

from 

left 

(a). 


−−−−−−−−−−−−−−−−−−−−→ 


52. Slope of a Tangent Line For f (x) = x 3 ←−−−−−−−−−−−−−−−−−−−−− 

: 


x − 2 − 2 − 2 − 2 − 2 2 2 2 2 2 

→ 0 


lim 

← 

3 5 7 9 11 

f (x) using a11 

graph, what 9 would you 7 conclude? 5 Zoom3in 

x→0 

f (x) = sin and (3, 27). 0.707 0.707 0.707 0.707 0.707 f (x) approaches on the 0.707 graph. Describe 0.707 what 0.707 you see. 0.707 (Hint: Be 0.707 sure your 0.707 

(b) Findx the slope of the secant line containing the points (2, 8) 


and (x, f (x)), x = 2. 



x→0 x2 Table 7 suggests that lim sin π √ 

2 


function f (x) = cos π x→0 x = 2 2 ≈ 0.707. at x =−0.1, −0.01, −0.001, 


x2 In fact, by carefully selecting x, we can make f appear to approach any number in 


−0.0001, 0.0001, 0.001, 0.01, 0.1. 

the interval [−1, 1]. 

Teaching Tip 

Have students create their own piecewise 

functions on graph paper. Ask them to write 3 or 

4 limit questions based upon their (well-labeled) 

graph. Point out to students that being able to 

write out and explain piecewise functions so 

that another student can follow means that they 

have to be very clear. Reinforce good habits and 

good explanations. Then they can trade with one 

another, find the limits, and discuss the solutions 

with each other. Khan Academy offers explanatory 

videos, as well as a practice video, on piecewise 

functions. 


Graphfree: The Graphfree site is an easy tool 

for making piecewise graphs to create additional 

examples. Make sure you change the plot type to 

Piecewise. A link to this resource is available on 

the Additional Chapter 1 Resources document, 

available for download. 

R 

N 

D 

84 



11/01/17 9:52 am

Sullivan SullivanAP˙Sullivan˙Chapter01 October October 8, 2016 8, 201617:4 

17:4 

Section 1.1 • Limits of Functions Using Numerical Section 1.1 and• Assess Graphical Your Techniques Understanding 85 89 

(b) Investigate lim cos π by using a table and evaluating x→0 x2 Now look at the graphs of (c) f (x) Graph = sin the function π shown C. in Figure 15. In Figure 15(a), 

function f (x) = cos π x 

2 

x 2 at 


the choice of lim sin π = 0 seems reasonable. But in Figure w→115(b), − it appears w→1 + that 

x→0 x 2 


3 . 

w→1 

lim sin π 

x→0 x =−1 . Figure 15(c) (e) illustrates Use the graph that the to investigate graph of flimoscillates C(w) and rapidly lim as C(w). x 

2 2 

w→12− w→12 

(c) Compare the results from (a) and (b). approaches What do you 0. This conclude suggests that the + Do value these ofsuggest f doesthat not approach lim C(w) aexists? 

single number, and 


w→12 

view about using a table to draw a conclusion that lim sin π about limits? does not exist. (f) Use the graph to investigate lim C(w). 

x→0 x 

2 

w→0 + 



w→13 


− 


x→0 



PAGE 

x − 8 


x→2 2 


limit? 


limit? 

58. (a) Use a table to investigate lim 

Figure 

(5 − 2x). 

15 

x→2 


limit? 


limit? 




EXAMPLE 8 







RECALL 

weighing 

On the number 

more than 

line, 

3.5 

the 

ounces. 

distance between two points with 

coordinates Source: a andU.S. b is Postal |a − b|. Service Notice 123 



NEED TO REVIEW? Inequalities 


involving absolute values are discussed 

in Appendix 

(c) 

A.1, 

Graph 

p. A-7. 

the function C. 


w→2− w→2 + 


w→2 


w→0 + 

y 


w→3.5 − 

10 

60. 9 First-Class Mail As of April 2016, the U.S. Postal Service 

8 charged $0.94 postage for first-class large envelope weighing up to 


6 or partial ounce up to and including 13 ounces. First-class rates do 


4 



So, how do we find a limit 62. with The definition certainty? ofThe the slope answer of the liestangent in giving line atovery the graph precise of 

definition of limit. The next example helps explain the definition. f (x) − f (c) 


. 

x→c x − c 

Analyzing a Limit Another way to express this slope is to define a new variable 

In Example 2, we claimed that lim 

h = 

(2x 

x − 

+ 

c. 

5) 

Rewrite 

= 9. 

the slope of the tangent line m tan using h and c. 

63. x→2 If f (2) = 6, can you conclude anything about lim f (x)? Explain 

x→2 

(a) How close must x be to 2, so your that reasoning. f (x) = 2x + 5 is within 0.1 of 9? 

(b) How close must x be to64. 2, so If that lim 

6, can you conclude anything about f (2)? Explain 

x→2 

f (x) = 2x + 5 is within 0.05 of 9? 


Solution (a) The function f (x) = 2x + 5 is within 

65. The graph of f (x) = x − 0.1 3 of9, if the distance between 

f (x) and 9 is less than 0.1 unit. That is, if | f (x) − 9| is a straight line with a point punched 

3 −≤0.1. 

x 

out. 

|(2x + 5) − 9| ≤0.1 


|2x − 4| ≤0.1 


|2(xx− approaches 2)| ≤0.13. 


|x − 2| ≤ 0.1 

x→3 

2 = 0.05 

66. (a) −0.05 Use a table ≤ xto− investigate 2 ≤ 0.05 lim(1 + x) 1/x . 

x→0 

(b) Use 1.95 graphing ≤ x ≤technology 2.05 to graph g(x) = (1 + x) 1/x . 

(c) What do (a) and (b) suggest about lim(1 + x) 1/x ? 

x→0 

So, if 1.95 ≤ x ≤ 2.05, then f (x) will be within 0.1of9. 


(b) The function f (x) = 2x + 5 is within x→0 0.05 of 9 if | f (x) − 9| ≤0.05. That is, 

|(2x + 5) − 9| ≤0.05 

Challenge |2x −Problems 

4| ≤0.05 

For Problems 67–70, investigate each of the following limits. 

|x − 2| ≤ 0.05 = { 0.025 

2 1 if x is an integer 

2 

f (x) = 

(a) Find a function C that models the first-class So, postage if 1.975charged, 

≤ x ≤ 2.025, then f (x) will be within0 0.05ifof x is 9. not ■ an integer 


w>0. 1 2 x 


x→2 x→1/2 x→3 x→0 

Notice that the closer we require f to be to the limit 9, the narrower the interval for 


DF Figure 16 f (x) = 2x + 5 

x becomes. See Figure 16. 










(a) 24p ≤ x ≤ 4p (b) 2p ≤ x ≤ p (c) 21 ≤ x ≤ 1 


■ 


of limit. 



Optional/Enrichment 

If you are pressed for time, you might 

skip Example 8, because problems like 

this usually do not appear on the exam. 

However, if you have time, this example 

may deepen students’ conceptual 

understanding of limits. 

Teaching Tip 

Sometimes students come to calculus 

class without having developed good 

homework habits. Emphasize the message 

that practice is necessary. Point out the 

benefits of daily practice. If you observe 

students who are not completing their 

homework, it is critical to speak with them 

as soon as you can. 


85 


11/01/17 9:52 am




Must-Do Problems for 

Exam Readiness 

AB: 1–6, 17–28, 31, 35, 37, 38, 41, 42, 

AP ® Practice Problems 

BC: 11, 13, 17, 19, 31, 37, 39, 51, 59, 

63–65, AP ® Practice Problems 3–8 

TRM Full Solutions to Section 

1.1 Problems and AP® Practice 

Problems 

Answers to Section 1.1 

Problems 

1. (c) 

2. True. 

3. False. 

4. False. 

5. False. 

6. False. 

7. lim 2x 

= 2. For table values see TSM. 

x→1 

8. lim ( x + 3) = 5. For table values see 

x→2 

TSM. 

9. lim ( x 

2 + 2) = 2. For table values see 

x→0 

TSM. 

10. lim ( x 

2 − 2) =−1. For table values 

x→−1 

see TSM. 

2 

x − 9 

11. lim =−6. For table values 

x→−3 

x + 3 

see TSM. 

3 

x + 1 

12. lim 

+ = 3. For table values see 

x→−1 

x 1 

TSM. 

2x 2 if x < 1 

The discussion in Example 

33. f (x) = 

3x 2 at c = 1 

53. Slope8offorms a Tangent the basis Line ofFor thef definition (x) = 1 of 

− 1 if x > 1 

2 x2 − 1: a limit. We state 

the definition here, but postpone the details until Section 1.6. It is customary to use the 

x 3 Greek letters ε (epsilon) and(a) δ (delta) Find the in slope the definition, m 

if x < −1 

sec of thesosecant we call lineitcontaining the ε-δ definition the of a 

34. f (x) = 

x 2 at c =−1 limit. 


− 1 if x > −1 


x 2 if x ≤ 0 

35. f (x) = 

at c = 0 DEFINITION ε-δ Definition of a Limit 

2x + 1 if x > 0 

⎧ 

Let f be a function defined everywhere h −0.5 in an open −0.1interval −0.001 containing 0.001 c, except 0.1possibly 

0.5 

⎨ x 2 if x < 1 

at c. Then the limit of the function m sec f as x approaches c is the number L, written 

36. f (x) = 2 if x = 1 at c = 1 

⎩ 

−3x + 2 if x > 1 

(c) Investigate lim f the (x) limit = L of the slope of the secant line found in (a) 

x→c 

as h → 0. 


if, given any number ε>0, (d) there What is is a number the slopeδ>0 of the tangent so that line to the graph of f at the 

point P = (2, f (2))? 

In Problems 37–40, sketch a graph of a function with the given whenever(e) 0 < On|x the −same c|


Section 1.1 • Assess Your Understanding 89 87 

In Problems (b) Investigate 13–16, use limtechnology cos π by using to complete a tablethe andtable evaluating and the In Problems (c) Graph 21–28, the function use the graph C. to investigate lim f (x). If the limit 

x→0 x2 x→c 

investigate the limit. 

function f (x) = cos π does 

x 2 at 

(d) 

not exist, 

Use the 

explain 

graph 

why. 

to investigate lim C(w) and lim C(w). Do 

w→1− w→1 + 

PAGE 

2 − 2e x 

82 13. lim x→0 x 

x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 3 . 

21. these suggest that lim C(w) w→1 22. exists? 

y (e) Use the graph to investigate 

y 

x approaches 0 x approaches 0 

lim C(w) and lim C(w). 

w→12− w→12 

(c) Compare the resultsfrom the (a) left and (b). What from do youthe conclude right 

+ 

about the limit? Why 

−−−−−−−−−−→ 

do you think this happens? 

←−−−−−−−−−− 

Do these suggest that lim C(w) exists? 

What is your 3 

w→12 y f (x) 

viewx about using −0.2 a table −0.1to −0.01 draw a conclusion → 0 ← about 0.01limits? 

0.1 0.2 

y f (x) 

2 


w→0 + 

(d) f (x) Use = 2 technology − 2 

2ex 

1 

to graph f . Begin with the x-window 

1 (g) Use the graph to investigate lim 

x 

C(w). 

w→13 


− (c, 1) 

(c, 1) 

lim ln xf (x) using a graph, what would you conclude? Zoom in 

c 

x 

61. Correlating c Student Success x to Study Time Professor Smith 

14. lim x→0 

x→1 on x −the 1 graph. Describe what you see. (Hint: Be sure your 





PAGE 

x − 8 

23. seven hours one studies, the closer 24. to 100% the student scores 

85 57. (a) Use a table to investigate from the lim left . from the right y 

x→2 2 

on the final. He claims that studying y 

−−−−−−−−−−→ ←−−−−−−−−−− 

significantly less than seven 

(b) How x close must 0.9 x0.99 be to0.999 2, so that → f 1(x) ←is within 1.001 0.11.01 of the1.1 


limit? 


(c) f (x) How = 

ln close x 

3 (c, 3) 3 

must x be to 2, so that f (x) is within 0.01 of the 2 “burnout.” 

2 

x − 1 

limit? 

y f (x) 

y f (x) 

1 (a) Write Professor Smith’s claim1 

symbolically as a limit. 


PAGE 

1 − cos x 

(c, 1) 

x→2 (b) Write Professor Smith’s claim using the ε-δ definition 

82 15. lim , where x is measured in radians 

(b) x→0 

How 

x 

close must x be to 2, so that f (x) is within 0.1 of the 

ofclimit. 

x 

c 

x 

limit? x approaches 0 x approaches 0 


(c) How close must x from be to 2, thesoleft 

that f (x) is within from the 0.01right 

of the 

PAGE 

limit? −−−−−−−−−−→ ←−−−−−−−−−− 84 25. 62. The definition of the slope of the26. 

tangent line to the graph of 

59. First-Class 

x (in radians) 

Mail 

−0.2 

As of 

−0.1 

April 

−0.01 → 0 ← 0.01 0.1 0.2 

y f (x) − f (c) 


. 

x→c 


x − c 

f (x) = 1 − cos x 

y 

2 

charged $0.47 x postage for 

Another way y f to(x) 

express this slope 3is to define a new variable 

1 


h = x − c. Rewrite the slope of the tangent (c, 2) line my tan using f (x) 

sin x 

h and c. 

2 

16. and lim including , 1where ounce, xplus is measured a flat in radians 

x→0 1 + tan x 

63. If f (2) c = 6, can you conclude x anything about lim f (x)? Explain 


1 

x→2 


or partial ounce up to and x approaches 0 x approaches 0 2 



c 

x 

from the left from the right 

x→2 

letter rates do not apply−−−−−−−−−−→ 

to letters 

←−−−−−−−−−− your reasoning. 

weighing x (in radians) more than−0.2 3.5 ounces. −0.1 −0.01 → 0 ← 0.01 0.1 0.2 



3 − x 

f (x) = 

sin x 

27. 28. 

y out. 

y 

1 + tan x 

(a) Find a function C that models the first-class postage charged, (a) What straight line and what point? 

in dollars, for a letter weighing w ounces. Assume w>0. 3 

3 

In Problems 17–20, use the graph to investigate 


(a) lim 

(b) What 

f (x), 

is 

(b) 

the 

lim 

domain 

f (x), 

of 

(c) 

C? 

lim 



f (x). 

(c, 2) 

y f (x) 

2 

y f (x) 

2 (c, 2) 

x→2− x→2 + x→2 

1 (c) Does the graph suggest that lim1 

f (x) exists? If so, what is it? 


x→3 

17. 18. 

w→2− w→2 + 

y 

y 

66. (a) Use c a table to investigate x lim(1 + x) 


1/x . c 

x 

x→0 

4 

w→2 

(e) Use the graph to investigate lim 4 C(w). 


w→0 + (c) What do (a) and (b) suggest about lim(1 + x) 1/x ? 


x→0 

2 y f (x) 

w→3.5 2 − y f (x) 

CAS In Problems (d) Find29–36, lim(1 use + x) a 1/x graph . to investigate lim f (x) at the 

(2, 2) 

x→c x→0 

60. First-Class Mail As of April 2016, the U.S. Postal Service number c. 


2x + 5 if x ≤ 2 

2 4 x 

2 4 x 29. f (x) = 

at c = 2 


4x + 1 if x > 2 

or partial ounce up to and including 13 ounces. First-class rates do Challenge Problems 2x + 1 if x ≤ 0 

19. not apply to large envelopes weighing 20. more than 13 ounces. 30. For Problems f (x) = 67–70, investigate eachatof c = the0 

2x if x > 0 following limits. 

y 


y 

⎧ { 1 if x is an integer 

6 

8 

⎨ 3x −f (x) 1 = if x < 1 

(a) (2, Find 6) a function C that models the first-class postage y charged, f (x) PAGE 


in dollars, y for f a(x) 

large envelope6 

84 31. f (x) = 4 if x = 1 at c = 1 

weighing w ounces. Assume 

⎩ 

4x if x > 1 

3 w>0. 

4 


x→2 ⎧ x→1/2 x→3 x→0 

(b) What is the domain of C? 2 

⎨ x + 2 if x < 2 

32. f (x) = 4 if x = 2 at c = 2 

⎩ 

2 4 x 

2 4 x 

x 2 if x > 2 


− e 

13. lim 2 2 x 

=−2. For table values 

x→0 

x 

see TSM. 

x 

14. lim ln − = 

→ x 1 1 . For table values see 

x 1 

TSM. 

− x 

15. lim 1 cos = 0. For table values 

x→0 

x 

see TSM. 

x 

16. lim 

sin = 0. For table values 

x→0 

1+ 

tanx 

see TSM. 

17. (a) 2 

(b) 2 

(c) 2 

18. (a) 4 

(b) 4 

(c) 4 

19. (a) 3 

(b) 6 

(c) Limit does not exist. 

20. (a) 4 

(b) 2 


21. 1 

22. 1 

23. 1 

24. Limit does not exist. 

25. Limit does not exist, because the two 

one-sided limits are not equal. 


27. Limit does not exist, because the two 

one-sided limits are not equal. 


29. 9 



32. 4 

Section 1.1 • Assess Your Understanding 

87 


11/01/17 9:52 am




33. 2 




37. Answers will vary. Sample answer: 

y 

4 

3 

2 

1 

1 2 3 4 


y 

2 

1 

22 21 1 2 3 x 

21 

22 


y 

21 

4 

3 

2 

1 

21 

1 2 


y 

3 

2 

1 

22 21 1 2 3 4 x 

21 

41. 1 42. −1 

43. 0 44. 1 

45. 1 46. 1 

47. 0 48. 0 

49. 0 50. −1 

51. (a) m sec 

= 15 

(b) msec = 3( x+ 

2) 

(c) lim m = 12. For sample table 

x→2 

see TSM. 

(d) 

y 

30 

25 

20 

15 

10 

5 

sec 

(2, 12) 

(3, 27) 

1 2 3 

x 

x 

Tangent line 

y 12x 12 

Secant line 

y 15x 18 

52. (a) msec 

= 19 

2 

(b) msec 

= x + 2x 

+ 4 

(c) lim msec 

= 12. For sample table 

x→2 

see TSM. 

x 

33. 

2x 2 if x < 1 

f (x) = 

3x 2 − 1 if x > 1 

at c = 1 

34. 

x 3 if x < −1 

f (x) = 

x 2 − 1 

if x > −1 

at c =−1 

35. 

x 2 if x ≤ 0 

f (x) = 

at c = 0 

2x + 1 if x > 0 

⎧ 

⎨ 

x 2 if x < 1 

36. f (x) = 

2 if x = 1 

at c = 1 

⎩ 

−3x + 2 if x > 1 




37. lim 

f (x) = 3; 

x→2 lim 

f (x) = 3; 

x→3 − lim 

f (x) = 1; 

x→3 + f (2) = 3; f (3) = 1 

38. lim 

f (x) = 0; 

x→−1 lim 

f (x) =−2; 

x→2 − lim 

f (x) =−2; 

x→2 + f (−1) is not defined; f (2) =−2 

39. lim 

f (x) = 4; 

x→1 lim 

f (x) =−1; 

x→0 − lim 

f (x) = 0; 

x→0 + f (0) =−1; f (1) = 2 

40. lim 

f (x) = 2; 

x→2 lim 

f (x) = 0; 

x→−1 lim 

f (x) = 1; 

x→1 f (−1) = 1; f (2) = 3 


each limit. 

|x − 5| 

|x − 5| 

41. lim 

42. lim 

43. lim 

2x 

x→5 + x − 5 

x→5 − x − 5 

x→ 12 

2x 

− 

44. lim 2x 2x 2x 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. lim 

− 

x→ 23 

2x 

+ 

47. lim 

|x|−x 48. lim 

|x|−x 

x→2 + x→2 − 

3 3 49. lim x−x x−x 50. lim x−x x−x 

x→2 + x→2 − 51. Slope of a Tangent Line For f (x) = 3x 2 : 


and (3, 27). 


and (x, f (x)), x = = 2. 






(d) 


and (3, 27). 


and (x, f (x)), x = = 2. 





y 

40 

30 

(3, 27) 

Secant line 

20 y 5 19x 2 30 

10 

(2, 8) 

1 2 3 

Tangent line 

y 5 12x 2 16 

1 

53. (a) msec = 2+ 

h for h ≠ 0 

2 

(b) For table see TSM. 

(c) lim m = 2 

h→0 

sec 

(d) m = 2 

tan 

x 

53. Slope of a Tangent Line For f (x) = 1 2 x2 − 1: 




h −0.5 −0.1 −0.001 0.001 0.1 0.5 

m sec 


as h → 0. 


point P = (2, f (2))? 


P = (2, f (2)). 





h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1 

m sec 


in (a) as h → 0. 


point P = (−1, f (−1))? 


at P = (−1, f (−1)). 

PAGE 


x→0 x function f (x) = cos π x at 

x =− 1 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1 

12 , 1 10 , 1 8 , 1 4 , 1 2 . 

(b) Investigate lim 

cos π by using a table and evaluating the 

x→0 x function f (x) = cos π x at 

x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 3 , 1. 




limits? 



lim 

f (x) using a graph, what would you conclude? Zoom in 

x→0 on the graph. Describe what you see. (Hint: Be sure your 


56. (a) Investigate lim 


x→0 x 2 function f (x) = cos π at x =−0.1, −0.01, −0.001, 

x 2 −0.0001, 0.0001, 0.001, 0.01, 0.1. 

(e) 

3 

2 

1 

1 

y 

1 2 3 

y 2x 3 

54. (a) msec =− 2+ 

h for h ≠ 0 

(b) For table see TSM. 

(c) lim msec 

=−2 

h→0 

(d) m =−2 

(e) 

tan 

y 5 22x 2 2 

22 21 

y 

3 

2 

1 

21 

x 

1 

x 

Answers continue on p. 89 

88 



11/01/17 9:52 am



(b) Investigate lim 


x→0 x 2 function f (x) = cos π x 2 at 

x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 3 . 



view about using a table to draw a conclusion about limits? 



lim 

f (x) using a graph, what would you conclude? Zoom in 

x→0 on the graph. Describe what you see. (Hint: Be sure your 


PAGE 

x − 8 

85 57. (a) Use a table to investigate lim 

. 

x→2 2 


limit? 


limit? 

58. (a) Use a table to investigate lim 

(5 − 2x). 

x→2 (b) How close must x be to 2, so that f (x) is within 0.1 of the 

limit? 


limit? 
















(d) Use the graph to investigate lim 

C(w) and lim 

C(w). Do 

w→2 − w→2 + these suggest that lim 

C(w) exists? 

w→2 (e) Use the graph to investigate lim 

C(w). 

w→0 + (f) Use the graph to investigate 

lim 

C(w). 

w→3.5 − 60. First-Class Mail As of April 2016, the U.S. Postal Service 



or partial ounce up to and including 13 ounces. First-class rates do 





w>0. 


55. (a) For table see TSM. Values in table 

π 

suggest lim cos = 1. 

x→0 

x 

(b) For table see TSM. Values in table 

π 

suggest lim cos =− 1. 

x→0 

x 

(c) Limit does not exist (see TSM). 

(d) 

y 

1 

0.5 

23 22 21 1 2 3 x 

20.5 

21 

y 

1 

0.5 



(d) Use the graph to investigate lim 

C(w) and lim 

C(w). Do 

w→1 − w→1 + these suggest that lim 

C(w) exists? 

w→1 (e) Use the graph to investigate 

lim 

C(w) and 

lim 

C(w). 

w→12 − w→12 + Do these suggest that lim 

C(w) exists? 

w→12 (f) Use the graph to investigate lim 

C(w). 

w→0 + (g) Use the graph to investigate 

lim 

C(w). 

w→13 − 











of limit. 



f (x) − f (c) 


. 

x→c x − c 



63. If f (2) = 6, can you conclude anything about lim 

f (x)? Explain 

x→2 your reasoning. 

64. If lim 

f (x) = 6, can you conclude anything about f (2)? Explain 

x→2 your reasoning. 

65. The graph of f (x) = x − 3 


3 − x out. 




(c) Does the graph suggest that lim 


x→3 66. (a) Use a table to investigate lim 

(1 + x) 1/x . 

x→0 (b) Use graphing technology to graph g(x) = (1 + x) 1/x . 

(c) What do (a) and (b) suggest about lim 

(1 + x) 1/x ? 

x→0 CAS 

(d) Find lim 

(1 + x) 1/x . 

x→0 

Challenge Problems 



f (x) = 


67. lim 

f (x) 68. lim 

f (x) 69. lim 

f (x) 70. lim 

f (x) 

x→2 x→1/2 x→3 x→0 

56. (a) For table see TSM. Values in the table 

π 

suggest lim cos = 

→ x 

1 . 

x 0 

2 

(b) For table see TSM. Values in table 

π 2 

suggest. lim cos = . 

x→0 

2 

x 2 

(c) Limit does not exist (see TSM). 

(d) 

y 

1 

0.5 

24 23 22 21 1 2 3 4 x 

20.5 

21 

y 

1 

0.5 

21 20.5 0.5 1 x 

20.5 

21 

x − 8 

57. (a) lim =− 3. For sample table 

x→2 

2 

see TSM. 

(b) 1.8≤ x ≤2.2 

(c) 1.98 ≤ x ≤2.02 

58. (a) lim (5− 2 x) = 1. For sample 

x→ 2 

table see TSM. 

(b) 1.95 ≤ x ≤2.05 

(c) 1.995 ≤ x ≤2.005 

⎧0.47 if 0< w ≤1 

⎪ 

⎪0.68 if 1< w ≤2 

59. (a) Cw ( ) = ⎨ 

⎪0.89 if 2 < w ≤3 

⎪ 

⎩⎪ 

1.10 if 3 < w ≤3.5 

(b) { w|0< w ≤3.5} 

(c) 

C(w) 

1.2 

1.0 

0.8 

0.6 

0.4 

0.2 

1 2 3 3.5 w 

(d) lim Cw ( ) = 0.68, lim Cw ( ) = 0.89, 

w→2 

− + 

w→2 w→2 

lim Cw ( )does not exist. 

(e) 

(f) 

60. (a) 

⎧ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

Cw ( ) = ⎨ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎩ 

lim Cw ( ) = 0.47 

w → 0 

+ 

lim Cw ( ) = 1.10 

w →3.5 

− 

0.94 if 0 < w ≤1 

1.15 if 1< w ≤2 

1.36 if 2 < w ≤3 

1.57 if 3 < w ≤4 

1.78 if 4 < w ≤5 

1.99 if 5 < w ≤6 

2.20 if 6 < w ≤7 

2.41 if 7 < w ≤8 

2.62 if 8 < w ≤9 

2.83 if 9 < w ≤10 

3.04 if 10< w ≤11 

3.25 if 11< w ≤12 

3.46 if 12< w ≤13 

20.1 20.05 0.05 0.1 x 

20.5 

21 Answers continue on p. 90 


89 


11/01/17 9:52 am




(b) { w|0< w ≤13} 

(c) 

C(w) 

4 

3 

2 

1 

1 2 3 4 5 6 7 8 9 10 11 12 13 w 

(d) lim Cw ( ) = 0.94; lim Cw ( ) = 1.15; 

w →1 

− w → 1 

+ 

lim Cw ( ) does not exist. 

w →1 

(e) lim Cw ( ) = 3.25; 

w →12 

− 

lim Cw ( ) = 3.46; lim Cw ( ) does 

+ 

w → 12 w →12 

not exist. 

(f) lim Cw ( ) = 0.94 

w → 0 

+ 

(g) lim Cw ( ) = 3.46 

− 

w →13 

61. (a) lim St () = 100 

t→7 

(b) Given any ε > 0, there is a 

number δ > 0 such that whenever 

0 < | t − 7| < δ. 

| St ( − 100)|


Section 1.2 • Limits of Functions Using Properties of Limits 91 

y 

A 

f (x) 5 A 

THEOREM The Limit of a Constant 

If f (x) = A, where A is a constant, then for any real number c, 

lim f (x) = lim A = A (1) 

x→c x→c 

The theorem is proved in Section 1.6. See Figure 17 and Table 8 for graphical and 

numerical support of lim 

x→c 

A = A. 

Teaching Tip 

Depending on the function given, a 

graphical approach may be helpful. 

Consider verifying the limits using graphical 

and analytical (that is, algebraic) methods. 

Figure 17 For x close to c, the value 

of f remains at A; lim x→c 

A = A. 

y 

c 

c 

c 

f (x) 5 x 

Figure 18 For x close to c, the 

value of f is just as close to c; lim x→c 

x = c. 

x 

x 

TABLE 8 

x approaches c from the left 

−−−−−−−−−−−−−−−−−−−−→ 

x approaches c from the right 

←−−−−−−−−−−−−−−−−−−−−− 

x c− 0.01 c − 0.001 c − 0.0001 → c ← c + 0.0001 c + 0.001 c + 0.01 

f (x) = A A A A f(x) remains at A A A A 

For example, 

lim 

x→5 2 = 2 

lim 1 

3 = 1 3 

x→ √ 2 

THEOREM The Limit of the Identity Function 

If f (x) = x, then for any number c, 

lim(−π) =−π 

x→5 

lim f (x) = lim x = c (2) 

x→c x→c 

This theorem is proved in Section 1.6. See Figure 18 and Table 9 for graphical and 

numerical support of lim 

x→c 

x = c. 

TABLE 9 

x approaches c from the left 

−−−−−−−−−−−−−−−−−−−−→ 

x approaches c from the right 

←−−−−−−−−−−−−−−−−−−−−− 

x c− 0.01 c − 0.001 c − 0.0001 → c ← c + 0.0001 c + 0.001 c + 0.01 

f (x) = x c− 0.01 c − 0.001 c − 0.0001 f (x) approaches cc+ 0.0001 c + 0.001 c + 0.01 

For example, 

lim 

x→−5 x =−5 

lim 

x→ √ x = √ 3 

3 

lim x = 0 

x→0 

1 Find the Limit of a Sum, a Difference, and a Product 

Many functions are combinations of sums, differences, and products of a constant 

function and the identity function. The following properties can be used to find the 

limit of such functions. 

Teaching Tip 

Students do not have to be able to state 

the theorems presented in this section. 

However, they must be able to apply them. 

Consider presenting these general rules: 

• When finding the limit of a function, first 

try to substitute the value of interest. If 

that results in a finite value, or something 

in the undefined form N for a nonzero N 

0 

(in which case the limit doesn’t exist), then 

you’re done. 

• If the substitution results in an 

indeterminate expression (like 0 0 or ∞ ∞ ), 

try to simplify the expression. If the 

expression is rational, try to factor the 

numerator and denominator and simplify. 

If the rational expression contains a 

binomial and one of the terms is a radical, 

try multiplying by the conjugate. 

• Remember, the ultimate goal is to 

determine what value the function is 

approaching as x gets very close to the 

given value of interest. 

IN WORDS The limit of the sum of two 

functions equals the sum of their limits. 

THEOREM Limit of a Sum 

If f and g are functions for which lim f (x) and lim g(x) both exist, 

x→c x→c 

then lim[ f (x) + g(x)] exists and 

x→c 

A proof is given in Appendix B. 

lim[ f (x) + g(x)] = lim f (x) + lim g(x) 

x→c x→c x→c 


This worksheet contains 6 questions for 

finding limits by direct substitution. Graphs 

are provided for answer verification. 

Teaching Tip 

As you get into the meat of this chapter, it may 

be time to refer to the pacing guide and adjust 

pacing. There are two schools of thought to 

consider: (1) teaching to mastery level initially, 

leaving only 3 weeks to review for the end-of-term 

exam, or (2) moving more quickly through the 

material, which then will leave 4 to 5 weeks for 

review. The rationale for the second approach 

is that many of the topics that students found 

difficult at the beginning of the fall semester are 

ones with which they are very comfortable by the 

midterm in December. Here in Chapter 1, they still 

don’t yet have the background and experience 

to ask the good questions needed to push their 

understanding of the concepts because they are 

still focused on the nuts and bolts of getting the 

correct answer. 

Section 1.2 • Limits of Functions Using Properties of Limits 

91 


11/01/17 9:52 am




common error 

Remind students that when they use 

substitution to find the limit, they are not 

evaluating the function at the x-value 

of interest. Rather, they are trying to 

determine the value of the function when 

we approach the given value of interest. 

For continuous functions, the function value 

and the limit close to the value of interest 

will be the same. 

2x 2 if x < 1 

33. f (x) = 

3x 2 at c = 1 EXAMPLE 1 Finding 53. the Slope Limit of a Tangent of a Sum Line For f (x) = 1 

− 1 if x > 1 

2 x2 − 1: 

Find lim (x + 4). 

x 3 x→−3 (a) Find the slope m 

if x < −1 


34. f (x) = 

x 2 at c =−1 


− 1 if x > −1 

Solution F(x) = x + 4 is the sum of two functions f (x) = x and g(x) = 4. 


x 2 if x ≤ 0 

From the limits given in (1) and (2), we have 

35. f (x) = 

at c = 0 

2x + 1 if x > 0 

⎧ 

lim f (x) = lim x =−3 h −0.5and−0.1 lim −0.001 g(x) = lim 0.001 4 = 0.1 4 0.5 

⎨ x 2 x→−3 x→−3 x→−3 x→−3 

if x < 1 

m sec 

36. f (x) = 2 if x = 1 at c = 1 Then, using the Limit of a Sum, we have 

⎩ 

−3x + 2 if x > 1 

lim (x + (c) 4) Investigate = lim x the + limit of 4 the =−3 slope + of 4 the = secant 1 line found in (a) 

x→−3 as h x→−3 → 0. x→−3 

■ 



THEOREM Limit of a Difference point P = (2, f (2))? 

In Problems 37–40, sketch a graph of a function with If the f and given g are functions for(e) which On the limsame f (x) setand of axes, lim g(x) graphboth f andexist, 

the tangent line to f at 


x→c x→c 

then lim[ f (x) − g(x)] exists and P = (2, f (2)). 

37. lim f (x) = 3; lim f (x) = 3; lim f (x) = 1; x→c 

x→2 x→3− x→3 + 

IN WORDS f (2) = 3; The limit f (3) of = the 1 difference 

lim 54. [ f Slope (x) −ofg(x)] a Tangent = limLine f (x) For − lim f (x) g(x) = x 2 − 1: 


of two functions equals 

38. f (x) = 0; lim 

the difference of 

their limits. 

f (x) =−2; lim f (x) =−2; 


x→−1 x→2− x→2 + 


f (−1) is not defined; f (2) =−2 EXAMPLE 2 Finding the Limit of a Difference 



f (x) = 0; 

x→1 

Find lim(6 − x). 

x→0− x→0 + x→4 

f (0) =−1; f (1) = 2 

h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1 

Solution F(x) = 6 − x is the difference of two functions f (x) = 6 and g(x) = x. 


m sec 

x→2 x→−1 x→1 

lim f (x) = lim 6 = 6 and lim g(x) = lim 

f (−1) = 1; f (2) = 3 

x = 4 

x→4 x→4 (c) Investigate the limit of x→4 the slope ofx→4 the secant line found 

in (a) as h → 0. 

In Problems 41–50, use either a graph or a table to Then, investigate using the Limit of a Difference, we have 


each limit. 

lim(6 − point x) = P lim= |x − 5| 

|x − 5| 

6 (−1, − lim f (−1))? x = 6 − 4 = 2 

x→4 x→4 x→4 

■ 

41. lim 

42. lim 

43. lim 

x→5 + x − 5 

x→5 − x − 5 

 

x→ 12 

2x 


− 

at P = (−1, f (−1)). 

THEOREM Limit of a Product 

PAGE 

44. lim 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. If f and lim 

− 

x→ 23 

 

g are 2xfunctions + 

85 55. for (a) which Investigate lim f (x) lim and cos π limbyg(x) usingboth a table exist, and evaluating the 

x→c x→0 x x→c 

then lim[ f (x) · g(x)] exists and 

function f (x) = cos π x→c 

47. IN WORDS lim |x|−x The limit of 48. the product lim |x|−x of 

x at 

x→2 + x→2 − lim[ f (x) · g(x)] x =− 1 = 

3 3 2 , − 1 lim 

4 , − 1 f (x) 

8 , − 1 · lim 

two functions equals the product of their 

10 , − 1g(x) 


12 ,..., 1 

12 , 1 10 , 1 8 , 1 4 , 1 limits. 

2 . 


x→2 + x→2 − A proof is given in Appendix B. 

51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π by using a table and evaluating the 

: 

x→0 x 

EXAMPLE 3 Finding the Limit 


function off a(x) Product = cos π x at 

and (3, 27). 

Find: 


x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 3 , 1. 

and (x, f (x)), x = 2. 

(a) lim x 2 (b) lim (−4x) 

x→3 x→−5 




graph of f at 2 using the result from (b). Solution (a) F(x) = x 2 is the product of two functions, f (x) = x and g(x) = x. 


(d) On the same set of axes, graph f , the tangent 

Then, 

line 

using 

to the 

the 

graph 

Limit of a Product, we have 

limits? 


lim 

(d) x 2 = lim x · lim x = (3)(3) = 9 

Use technology to graph f . Begin with the x-window 

52. Slope of a Tangent Line For f (x) = x 3 x→3 x→3 x→3 

: 


(b) F(x) =−4x is the product of two functions, f (x) =−4 and g(x) = x. Then, 



using the Limit of a Product, wex→0 have 

and (3, 27). 


(b) Find the slope of the secant line containing the points (2, 8) lim (−4x) = calculator lim (−4) is· set lim to the x = radian (−4)(−5) mode.) = 20 

x→−5 x→−5 x→−5 

■ 

and (x, f (x)), x = 2. 


(c) Create a table to investigate the slope of the tangent 

A corollary 

line to ∗ the 

of the Limit of a Product x→0 Theorem x2 is the special case when f (x) = k 

graph of f at 2 using the result from (b). is a constant function. 

function f (x) = cos π at x =−0.1, −0.01, −0.001, 


x2 of f at the point (2, 8), and the secant line ∗ Afrom corollary (a). is a theorem that follows 

−0.0001, 

directly 

0.0001, 

from a 

0.001, 

previously 

0.01, 

proved 

0.1. 

theorem. 

92 



11/01/17 9:52 am


Section 1.2 • Limits Section of Functions 1.1 • Assess UsingYour Properties Understanding of Limits 89 93 


x→0 x2 COROLLARY Limit of a Constant (c) GraphTimes the function a Function C. 


If g is a function for which(d) lim Use g(x) the exists graphand to investigate if k is any real lim number, C(w) and then lim limC(w). [kg(x)] Do 

x→c w→1− w→1 x→c + 

exists and 


IN WORDS The limit of a constant 

3 . 

w→1 

times a function equals the constant 

(e) lim Use[kg(x)] the graph= tok investigate lim g(x) lim C(w) and 

x→c x→c lim C(w). 

w→12− w→12 

times (c) theCompare limit of the function. results from (a) and (b). What do you conclude 

+ Do these suggest that lim C(w) exists? 

about the limit? Why do you think this happens? You areWhat askedisto your prove this corollary in Problem 103. w→12 


Limit properties often are used in combination. 

C(w). 

w→0 + 



w→13 


− 


x→0 

on the graph. Describe what you see. (Hint: Find: Be sure your 


PAGE 

x − 8 (a) lim 


x→2 2 

(b) How close must x be to 2, so that f (x) isSolution within 0.1(a) 

of the 

limit? 

lim 


limit? 


x→2 


limit? 

NOTE (c) The Howlimit close properties must x beare toalso 2, sotrue 

that f (x) is within 0.01 of the 

(b) We use properties of limits Source: to find Submitted the one-sided by the students limit. of Millikin University. 

for one-sided limit? limits. 

62. The definition of the slope of [ the tangent ][ line to the graph ] of 

lim 

f (x) − f (c) 


x→2 +[4x(2 

− x)] = 4 lim 

y = x→2 f (x) +[x(2 at the − point x)] (c, = 4 lim 

f (c)) is m x lim 

x→2 + tan = x→2 lim +(2 − x) . 


[ 

] 

x→c x − c 


Another = 4 · 2 way limto 2 express − lim this x slope = 4 is · 2 to· define (2 − 2) a new = 0variable 

■ 

x→2 + x→2 + 




63. If f (2) = 6, can you conclude anything NOW about WORK lim f (x)? Problem Explain 13. 


x→2 




To find the limit of 64. piecewise-defined If lim f (x) = 6, can x→2 

functions you conclude at numbers anythingwhere about f the (2)? defining Explain 

letter rates do not apply to letters equation changes requires the your use reasoning. of one-sided limits. 

RECALL weighing The more limit Lthan of a3.5 function ounces. 


y = Source: f (x) asU.S. x approaches Postal Service a Notice 123 

3 − x 

EXAMPLE 5 Finding aout. 

Limit for a Piecewise-defined Function 

number c exists if and only if 

lim 

(a) 

f (x) 

Find 

= 

a 

lim 

function C that models the first-class postage charged, 

in dollars, for f (x) a letter = L. Find lim f (x), if it exists. (a) What straight line and what point? 

x→c− x→2 

x→c + weighing w ounces. Assume w>0. 

{ 



3x + 1 if x < 2 

f (x) = x approaches 3. 


(c) Does 

2x(x 

the graph 

− 1) 

suggest 

if 

that 

x 

lim 

≥ 2 



x→3 

w→2− w→2 + 

Solution Since the rule66. for(a) f changes Use a table at 2, towe investigate need tolim 

find(1 the + x) one-sided 


1/x . limits of f as 

x→0 

w→2 x approaches 2. 



w→0 + For x < 2, we use the left-hand 

(c) What 

limit. 

do 

Also, 

(a) and 

because 

(b) suggest 

x < 

about 

2, f (x) 

lim 

= 

(1 

3x 

+ x) 

+ 1/x 1. 

? 

y 


x→0 

20 

w→3.5 − lim f (x) = lim CAS (d) Find lim(1 x) 1/x . 

x→2 − 

x→2−(3x + 1) = lim 

x→2 x→0 −(3x) 

+ lim 1 = 3 lim x + 1 = 3(2) + 1 = 7 

x→2 − x→2 

60. First-Class Mail 16 

− As of April 2016, the U.S. Postal Service 

charged $0.94 12 postage for first-class large envelope 

For x 

weighing 

≥ 2, we 

up 

use 

to 

the right-hand limit. Also, because x ≥ 2, f (x) = 2x(x − 1). 


8 

or partial ounce up to and including 13 ounces. First-class rates do Challenge Problems 

not apply to large 4 envelopes weighing more than 13 ounces. lim f (x) = lim 

x→2 + Forx→2 Problems +[2x(x 

− 1)] = lim 

67–70, investigate x→2 +(2x) 

· lim 

eachx→2 of the +(x 

− 1) 

(2, 4) 

following limits. 


[ { ] 1 if x is an integer 

4 2 

2 4 x 

= 2 lim f (x) = 


x · lim x − lim 1 = 2 · 2 [2 − 1] = 4 

x→2 + x→2 + x→2 0 + if x is not an integer 


w>0. 

Since lim f (x) = 7 = 67. lim limf (x) f (x) = 4, 68. lim lim f (x) f does (x) not 69. exist. lim ■f (x) 70. lim 

{ f (x) 

x→2 − 

x→2 x→2 + 

x→2x→1/2 x→3 x→0 

(b) What is the domain 3x + 1 of C? if x < 2 

Figure 19 f (x) = 

2x(x − 1) if x ≥ 2 See Figure 19. 

NOW WORK Problem 73 and AP® Practice Problems 1 and 5. 


EXAMPLE 4 

Finding a Limit 




[2x(x + 4)] seven (b) hours limone studies, the closer to 100% the student scores 

x→1 x→2 +[4x(2 

− x)] 




lim“burnout.” 

lim 

Limit of a Product 

[ ][ ] 

[(2x)(x + 4)] = (2x) (x + 4) 

x→1 x→1 x→1 

[ ] [ 

] 

= 2 ·(a) limWrite x · Professor lim x + Smith’s lim 4 claimLimit symbolically of a Constant as a limit. Times a 

(b) x→1 x→1 x→1 Write Professor Smith’s claim 

Function, 

using the 

Limit 

ε-δ definition 

of a Sum 

= (2 · 1) · of (1 limit. + 4) = 10 Use (2) and (1), and simplify. 

AP® Exam Tip 

If you like to use released AP ® Calculus 

multiple-choice questions, be sure to 

review each question carefully, as many 

questions about limits require more 

knowledge than students will have 

at this point in the term. Once they 

complete this chapter, students will be 

able to find limits by direct substitution 

or with some algebraic manipulation 

followed by substitution. 

AP® Exam Tip 

Starting with the 2017 exam, L’Hôpital’s 

Rule is on the Calculus AB exam. 

L’Hôpital’s Rule is a powerful technique 

that can be used to find limits of 

expressions of the form 0 0 or ∞ ∞ . This 

topic will be studied in Section 4.5. 


Finding a Limit for a Piecewise-Defined 

Function 

⎧ x 

Given 

⎪ | + 1| 

fx ( ) = 

x ≤ 0 

⎨ 

2 

⎩⎪ 2− 

2x 

x > 0 

. 

Find 

lim fx ( ) if it exists. 

x→0 

Solution 

Because the rule for f changes at 0, we find 

the one-sided limits of f as x approaches 

0. For values of x near 0 and less than 0, 

f (x) = |x + 1| = x + 1 and 

lim fx ( ) = lim x + 1= 

1 

− 

− 

x→0 x→0 

For values of x near 0 and greater than 

zero, f (x) = 2 – 2x 2 and 

2 

lim fx ( ) = lim 2− 2x 

= lim 2− 2(0) = 2 

+ + + 

x→0 x→0 

x→0 

Because 

lim fx ( ) 

x→0 

lim fx ( ) ≠ lim fx ( ), 

x→0 

− x→ 0 

+ 

does not exist. 



93 


11/01/17 9:52 am




Teaching Tip 

Most of the limit problems in this section 

can be solved using direct substitution. 

Consider teaching this section using direct 

substitution as the go-to technique. When 

substitution does not work, try algebraic 

manipulation, a graphical approach, or a 

tabular approach. It is not necessary to 

teach each of the theorems presented in 

this section one at a time in class because 

students will not be expected to write a 

proof step by step on the AP ® Exam. 

AP® Calc Skill Builder 

for Example 7 

Finding the Limit of a Power 

2 2 

Investigate lim ( x − 2x+ 

1) 

x→a 

Solution 

We can rewrite x 2 − 2x + 1 = (x − 1) 2 , which 

leads to 

lim ( x − 2x+ 1) = lim (( x −1) ) 

2 2 2 2 

x→a x→a 

= lim ( x− 1) = ( a−1) 

x→a 

4 4 

{ 2x 2 if x < 1 

33. f (x) = 

3x 2 at c = 1 

53. Slope of a Tangent 0 if Line t < For c 

The Heaviside function, u f (x) = 1 

− 1 if x > 1 

c (t) = 

, is a step 

2 x2 function − 1: that is used 

1 if t ≥ c 

x 3 if x < −1 

in electrical engineering to(a) model Findathe switch. slope mThe sec ofswitch the secant is off lineif containing t < c, and the it is on 

34. f (x) = 

x 2 at c =−1 

− 1 if x > −1 

if t ≥ c. 



x 2 if x ≤ 0 

35. f (x) = 

at c = 0 

2x + 1 if x > 0 

EXAMPLE 6 Finding a Limit 

⎧ 

h of the −0.5Heaviside −0.1 −0.001 Function 0.001 0.1 0.5 

{ 

⎨ x 2 if x < 1 

m 0sec 

if t < 0 

Find lim u 

36. f (x) = 2 if x = 1 at c = 1 

0 (t), where u 0 (t) = 

t→0 1 if t ≥ 0 

⎩ 

−3x + 2 if x > 1 


ORIGINS Oliver Heaviside 

Solution Since the Heaviside function changes rules at t = 0, we find the one-sided 

as h → 0. 

(1850--1925) was a self-taught 

limits as t approaches 0. 


Applications mathematician and Extensions 

electrical engineer. For t < 0, lim 

point P = (2, f (2))? 

InHe Problems developed 37–40, a branch sketchofamathematics 

u 0(t) = lim 0 = 0 and for t ≥ 0, lim u 0(t) = lim 1 = 1 

graph a function with the given t→0 − t→0− t→0 

(e) On the same set of axes, graph f and + t→0 

the tangent line + to f at 

properties. called operational Answers will calculus vary. in which 

P = (2, f (2)). 

37. 

differential 

lim f (x) 

equations 

= 3; 

are 

lim 

solved 

f (x) 

by 

Since the one-sided limits as t approaches 0 are not equal, lim u 0 (t) does not 

= 3; lim f (x) = 1; 

t→0 

converting x→2 them to algebraic x→3− equations. x→3 + exist. ■ 

Heaviside f (2) = applied 3; f (3) vector = 1calculus to 



electrical engineering and, 

38. lim f (x) = 0; lim perhaps most 

f (x) =−2; lim f (x) =−2; 


significantly, x→−1 he simplified x→2Maxwell's 

− x→2 + 


equations f (−1) toisthe notform defined; used by f (2) electrical =−2 

engineers to this day. In 1902 Heaviside 



claimed 

f (x) = 0; 

x→1 there is a layer x→0 surrounding 

− x→0 + 

Earthf (0) from=−1; which radio f (1) signals = 2 bounce, 

h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1 

allowing the signals to travel around the 2 Find the Limit of a Power and the Limit of a Root 


m sec 

Earth. x→2 Heaviside's claim x→−1 was proved truex→1 Using the Limit of a Product, if lim f (x) exists, then 

in 1923. 

f (−1) 

The 

= 

layer, 

1; 

contained 

f (2) = 3 

in the 

x→c 

[ ] 

(c) Investigate the limit of the slope of the secant line found 2 

ionosphere, is named the Heaviside 

lim 

layer. The function we discuss here is 

[ f x→c (x)]2 = lim[ f (x) 

in 

· 

(a) 

f (x)] 

as h 

= 

→ 

lim 

0. 

f (x) · lim f (x) = lim f (x) 

x→c x→c x→c x→c 


one of his minor contributions to 


each limit. 

Repeated use of this property produces 

point P = 

the 

(−1, 

next 

f (−1))? 

corollary. 

mathematics and electrical engineering. 

|x − 5| 

|x − 5| 

41. lim 

42. lim 

43. lim 

x→5 + x − 5 

x→5 − x − 5 

 

x→ 12 

2x 


COROLLARY − 

Limit of a Power at P = (−1, f (−1)). 

If lim f (x) exists and if n is a positive integer, then 

PAGE 

44. lim 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. lim 

− 

x→ 23 

2x 

+ 

85 55. (a) Investigate lim cos π x→c by using a table and evaluating the 

x→0 

[ x 

function f (x) = cos π ] n 

lim [ f 

47. lim |x|−x 48. lim |x|−x 

x at 

x→c (x)]n = lim f (x) 

x→c 

x→2 + x→2 − 

x =− 1 

3 3 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1 

12 , 1 10 , 1 8 , 1 4 , 1 2 . 


x→2 + x→2 − 

51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π by using a table and evaluating the 

: 

x→0 x 

EXAMPLE 7 Finding the Limit of a Power 


function f (x) = cos π Find: 

x at 

and (3, 27). 


x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 (a) lim x 5 (b) lim(2x − 3) 3 (c) lim x n n a positive 3 , 1. integer 

and (x, f (x)), x = 2. 

x→2 x→1 x→c 



( ) 5 



Solution (a) lim x 5 = lim x = 2 5 = 32 

x→2 x→2 



[ 

] 3 [ 

] 3 

limits? 

of f at the point (2, 12), and the secant line 

(b) 

from 

lim(2x (a). 

− 3) 3 = lim (2x − 3) = lim(2x) − lim 3 = (2 − 3) 3 =−1 

x→1 x→1 x→1 x→1 



[ ] n 

(c) lim x n = lim x = c 

n ■ 


x→c x→c 



x→0 

and (3, 27). 


The result from Example 7(c) is worth remembering since it is used frequently: 



and (x, f (x)), x = 2. 

56. (a) Investigate lim lim cos π by using a table and evaluating the 


x n x→0 

= c n 

x2 x→c 


function f (x) = cos π at x =−0.1, −0.01, −0.001, 

where c is a number and n is a positive integer. 


x2 of f at the point (2, 8), and the secant line from (a). 

−0.0001, 0.0001, 0.001, 0.01, 0.1. NOW WORK Problem 15. 

Science and Society / Science and Society 

94 



11/01/17 9:52 am



(b) Investigate lim cos π by using a table and THEOREM evaluating the Limit of a Root (c) Graph the function C. 


If lim f (x) exists and if n (d) ≥ 2Use is antheinteger, graph tothen 

investigate lim C(w) and lim C(w). Do 

x→c w→1− w→1 + 

x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 these √suggest that √ lim C(w) exists? 

3 . 

n 

lim f (x) = n w→1 lim f (x) 

(e) x→c Use the graph to investigate x→c lim C(w) and lim C(w). 

w→12− w→12 


+ provided f (x) >0 if n is even. Do these suggest that lim C(w) exists? 


w→12 

view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim C(w). 

w→0 + 

(d) Use technology to graph f . Begin with the EXAMPLE x-window 8 Finding the (g) Limit Use theof grapha Root to investigate lim C(w). 

w→13 


− 


x→0 

on the graph. Describe what you see. (Hint: 

Solution 

Be sure your 


PAGE 

x − 8 

3 


lim 

x→2 

x→4 

2 


limit? 


limit? 


x→2 


limit? 


limit? 





and including 1 ounce, plus a flat integers m and n, then 





yweighing more than 3.5 ounces. 

30Source: U.S. Postal Service Notice 123 EXAMPLE 9 


The Limit of a Power 

62. 

and 

The 

the 

definition 

Limit 

of 

of 

the 

a Root 

slope 

are 

of the 

used 

tangent 

together 

line to 

to 

the 

find 

graph 

the limit 

of 

of 

a function with a rational exponent. 

f (x) − f (c) 


. 

x→c x − c 

THEOREM Limit of a Fractional 

Another way 

Power 

to express 

[ f (x)] 

this m/n 

slope is to define a new variable 

If f is a function for which h = limx −f c. (x) Rewrite existsthe and slope if [ off (x)] the tangent m/n is line defined m tan using for positive h and c. 

x→c 


x→2 

your reasoning. [ ] m/n 

64. If lim lim [ f f = 6, can you conclude anything about f (2)? Explain 

x→c (x)]m/n = lim f (x) 

x→2 x→c 



Finding the 3 − x 

out. Limit of a Fractional Power [f(x)] m/n 

27 

(8, 27) 

(a) Find a function C that models the first-class Findpostage lim(x charged, + 1) 3/2 . 

x→8 (a) What straight line and what point? 

20 in dollars, for a letter weighing w ounces. Assume w>0. 



Solution Let f (x) = x + 1. Near 8, x + 1 > 0, so (x + 1) 3/2 is defined. Then 


[ ] 


3/2 

10 

lim 

(d) Use the graph to investigate lim C(w) and lim 

[ C(w). f (c) Does the graph suggest that lim f (x) exists? If so, what is it? 

Do 

x→3 

x→8 (x)]3/2 = lim(x + 1) 3/2 = lim (x + 1) = [8 + 1] 3/2 = 9 3/2 = 27 

x→8 ↑[ 

x→8 

w→2− w→2 + m/n 

66. lim [(a) f ( x)] Use m/n a= 

table limtof ( investigate x)] 

lim(1 + x) 


1/x . 

■ 

x→c 

x→c 

x→0 

5 w→2 8 10 x 



w→0 + See Figure 20. 

Figure 20 f (x) = (x + 1) 3/2 (c) What do (a) and (b) suggest about lim(1 + x) 1/x ? 


x→0 

w→3.5 − CAS (d) Find NOW lim(1 WORK + x) 1/x Problem . 23 and AP® Practice Problem 8. 

x→0 


charged $0.94 postage for first-class large envelope 3 Find weighing theup Limit to of a Polynomial 

and including 1 ounce, plus a flat fee of $0.21Sometimes for each additional lim f (x) can be found by substituting c for x in f (x). For example, 

or partial ounce up to and including 13 ounces. First-class x→c 

rates do Challenge Problems 

not apply to large envelopes weighing more than 13 ounces. Forlim 

Problems (5x 2 ) = 67–70, 5 liminvestigate x 2 = 5 · 2each 2 = of 20the following limits. 

x→2 x→2 


{ 

Since lim x n = c n if n is a positive integer, 1 if x is an integer 

f (x) we = can use the Limit of a Constant Times a 

(a) Find a function C that models the first-class postage x→c charged, 


in dollars, for a large envelope weighingFunction w ounces. to Assume obtain a formula for the limit of a monomial f (x) = ax n . 

w>0. 


x→2 lim x→1/2 x→3 x→0 


x→c (axn ) = ac n 


Find lim 

√ 3 

x 2 + 11. 

x→4 

where a is any number. 




√ √ √ 

x 

2 

+ 

seven 

11 = 

hours 3 lim 

one 

(x 2 studies, 

+ 11) 

the 

= 

closer 3 lim 

to 

x 2 100% 

+ lim 

the 

11 

student scores 

on the ↑ final. x→4 He claims that ↑ studying x→4 significantly x→4 less than seven 

Limit hours of amay Root cause one Limit to be ofunderprepared a Sum for the test, while 


“burnout.” = 3√ 4 2 + 11 = 3√ 27 = 3 

↑ 


lim x 2 = c 2 

■ 

x→c (b) Write Professor Smith’s claim using the ε-δ definition 

NOWofWORK limit. Problem 19 and AP® Practice Problems 6 and 7. 


95 


11/01/17 9:52 am




2x 2 if x < 1 

Since a polynomial 

33. f (x) = 

3x 2 at c = 1 

53. is the Slope sum of aofTangent monomials Line and Forthe f (x) limit = 1 of a 

− 1 if x > 1 

2 x2 −sum 1: is the sum of 

the limits, we have the following result. 


if x < −1 


34. f (x) = 

x 2 at c =−1 


− 1 if x > −1 

THEOREM Limit of a Polynomial Function 


x 2 if x ≤ 0 

If P is a polynomial function, then 

35. f (x) = 

at c = 0 

2x + 1 if x > 0 

⎧ 

h lim−0.5 P(x) = −0.1 P(c) 

IN WORDS To find the limit of a 

−0.001 0.001 0.1 0.5 

⎨ x 2 x→c 

polynomial as x approaches if c, x evaluate < 1 

m sec 

36. the polynomial f (x) = at c. 2 if x = 1 at c = 1 for any number c. 

⎩ 

−3x + 2 if x > 1 


Proof If P is a polynomial function, as h →that 0. is, if 



P(x) = a 

point n x n + a 

P = n−1 x n−1 +···+a 

(2, f (2))? 1 x + a 0 



where n is a nonnegative integer, (e) Onthen 

same set of axes, graph f and the tangent line to f at 


( P = (2, f (2)). 

f (x) = 1; lim P(x) = lim an x n + a 

x→2 x→3− x→3 + n−1 x n−1 ) 

+···+a 1 x + a 0 

x→c x→c 

( 

f (2) = 3; f (3) = 1 


= lim an x n) ( 

+ lim an−1 x n−1) +···+ lim(a 1 x) + lim a 0 

x→c x→c x→c x→c 



x→−1 x→2− x→2 + 

= a n c n + a n−1 c n−1 points +···+a P = (−1, 1 c + f (−1)) a 0 and Q = (−1 Limit + of h, af Monomial (−1 + h)). 


= P(c) 


(b) Use the result from (a) to complete the following table: ■ 

f (x) = 0; 

x→1 x→0− x→0 + 

f (0) =−1; f (1) = 2 

h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1 

40. lim f (x) = 2; lim f (x) = 0; lim EXAMPLE 10 Finding the 

f (x) = 1; 

mLimit sec 

of a Polynomial 

x→2 x→−1 x→1 

Find the limit of each polynomial: 

f (−1) = 1; f (2) = 3 


(a) lim(4x 2 − x + 2) = 4(3) 2 in −(a) 3 + as2 h= →35 

x→3 

0. 



each limit. 

(b) lim (7x 5 + 4x 3 − 2x 2 ) = 7(−1) 5 + 4(−1) 3 − 2(−1) 2 =−13 

x→−1 point P = (−1, f (−1))? 

|x − 5| 

|x − 5| 

41. lim 

42. lim 

43. (c) lim lim 

x→5 + x − 5 

x→5 − x − 5 

(10x 

x→ 12 

2x 6 (e) On the same set of axes, graph f and the tangent line to f 

− 

− 4x 5 − 8x + 5) = 10(0) 6 − 4(0) 5 − 8(0) + 5 = 5 

■ 

x→0 at P = (−1, f (−1)). 

44. lim 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. lim 

− 

x→ 23 

2x 

+ 

 

47. lim |x|−x 48. lim |x|−x 

x→2 + x→2 − 

3 3 


x→2 + x→2 − the following result. 

51. Slope of a Tangent Line For f (x) = 3x 2 : 


and (3, 27). 


and (x, f (x)), x = 2. 

exists and 



IN WORDS 

(d) On the 

The 

same 

limit 

set 

of 

of 

the 

axes, 

quotient 

graph 

of 

f , the tangent line to the graph 

two functions 

of f at 

equals 

the point 

the quotient 

(2, 12), and 

of 

the secant line from (a). 

their limits, provided that the limit of the 

52. denominator Slope of aisTangent not zero. Line For f (x) = x 3 : provided lim g(x) = 0. 

x→c 


NEED TO 

and 

REVIEW? 

(3, 27). 

Rational functions 

are discussed (b) Find the in Section slope of P.2, thep. secant 22. line containing the points (2, 8) 

and (x, f (x)), x = 2. 





PAGE 

85 55. (a) Investigate lim cos π NOW WORK Problem 29. 


x→0 x 


4 Find the Limit of a Quotient 

x =− 1 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1 

12 , 1 10 , 1 8 , 1 4 , 1 2 . 

To find the limit of a rational function, which is the quotient of two polynomials, we use 

THEOREM Limit of a Quotient 

If f and g are functions for which lim 


x→0 x 


[ f (x) 

x =−1, f −(x) 1 

3 , − and 1 

5 , − lim 1 7 , g(x) − 1 

9 ,..., both 1 exist, 

9 , 1 7 , 1 then 

5 , 1 x→c x→c 

3 , 1. lim 

x→c g(x) 


about [ the limit? ] f (x) Why lim do f (x) you think this happens? What is 

x→c 

lim your view about = using a table to draw a conclusion about 

x→c g(x) lim 

limits? 

g(x) 

x→c 




x→0 

COROLLARY Limit of a Rational the Function graph. Describe what you see. (Hint: Be sure your 


If the number c is in the domain of a rational function 

56. (a) Investigate lim cos π R(x) = p(x) 


x→0 x2 q(x) , then 

function lim R(x) f = cos R(c) π at x =−0.1, −0.01, −0.001, (3) 

x→c x2 −0.0001, 0.0001, 0.001, 0.01, 0.1. 

You are asked to prove this corollary in Problem 104. 

] 

96 



11/01/17 9:52 am




x→0 x2 EXAMPLE 11 Finding(c) theGraph Limit theofunction a Rational C. Function 


Find: 


w→1− w→1 + 

x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 3x 3 − 2x + 1 these suggest that 2x + lim 

3 . 

w→1 

4 C(w) exists? 

(a) lim 

(b) lim 

x→1 4x 2 + 5 (e) Use thex→−2 

graph3x to 2 investigate − 1 lim C(w) and lim C(w). 

w→12− w→12 


+ 

about the limit? Why do you think this happens? Solution What (a) isSince your 1 is in the domain 

Do these 

of 

suggest 

the rational 

that lim 

w→12 function 

C(w) exists? 

R(x) = 3x 3 − 2x + 1 

, 

view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim C(w). 4x 2 + 5 

w→0 

lim 

+ 


(g) R(x) Use = R(1) = 3 − 2 + 1 = 2 

x→1 ↑the graph to investigate 4 + 5 lim9 

C(w). 

w→13 


− Use (3) 


x→0 

on the graph. Describe what you see. (Hint: (b) Be Since sure −2 youris in the domain claims of the thatrational a student’s function final exam H(x) score = is 2x a function + 4 of the time t 


(in hours) that the student studies. He claims 3x 2 −that 1 , 

the closer to 

PAGE 

x − 8 



lim 

x→2 2 

on H(x) the = H(−2) = −4 + 4 

x→−2 ↑final. He claims12that − studying 1 = 0 

11 significantly = 0 less than seven 

■ 

(b) How close must x be to 2, so that f (x) is within 0.1 of the hours Use may (3) cause one to be underprepared for the test, while 

limit? 


(c) How close must x be to 2, so that f (x) is within 0.01 of the “burnout.” 


limit? 



x→2 

EXAMPLE 12 Finding(b) theWrite Limit Professor a Quotient Smith’s claim using the ε-δ definition 

√ 


of limit. 

3x 2 

+ 1 

limit? 

Find lim . 

x→4 x − 1 Source: Submitted by the students of Millikin University. 


62. The definition of the slope of the tangent line to graph of 

limit? 

Solution We seek the limit of the quotient of two functions. Since the limit of the 

f (x) − f (c) 


denominator lim(x − 1) = 0, we use the Limit of a Quotient. 

x→4 y = f (x) at the point (c, f (c)) is m tan = lim 

. 

x→c 


x − c 

√ 

√ √ 


3x 2 

+ 1 

lim 3x Another 2 

+ 1way to express lim (3x 2 √ 

this + slope 1) 

√ 

is to define a new variable 

x→4 

x→4 3 · 42 + 1 49 


lim = h = x − c. Rewrite the slope of the tangent line m tan using h and c. 

x→4 x − 1 lim 


63. (x If − f (2) 1) = 

lim 

= 6, can you (x conclude − 1) = 

= = 7 4 − 1 3 3 

↑ x→4 ↑ x→4 anything about lim f (x)? Explain 


Limit of a Quotient 

x→2 

your 

Limit 

reasoning. 

of a Root 

■ 


64. If lim f (x) = 6, can you conclude anything NOW WORK about f (2)? Problem Explain 31. 


x→2 


Based on these examples, your you reasoning. might be tempted 


65. The graph of f (x) = x − 3 to conclude that finding a limit as x 

approaches c is simply a matter of substituting the number is a straight c intoline thewith function. a pointThe punched next 

Source: U.S. Postal Service Notice 123 few examples show that substitution 

3 − x 

out. cannot always be used and other strategies need to 

be employed. 

(a) Find a function C that models the first-class postage charged, (a) What straight line and what point? 

in dollars, for a letter weighing w ounces. Assume The w>0. limit of a rational function can be found using substitution, provided the 



number c being approached is in the domain of the rational function. The next example 



shows a strategy that can be tried when c is not in the domain. 



x→3 

w→2− w→2 + 

66. (a) Use a table to investigate lim(1 + x) 

these suggest that lim C(w) exists? EXAMPLE 13 Finding the Limit of a Rational 1/x . 

x→0 

Function 

w→2 

(e) Use the graph to investigate lim 

x 2 + 5x + 6 (b) Use graphing technology to graph g(x) = (1 + x) 1/x . 

. 

x 2 − 4 (c) What do (a) and (b) suggest about lim 


C(w). 

w→0 + Find lim 


x→−2 

(1 + x) 1/x ? 

x→0 

w→3.5 − Solution Since −2 is not CAS in(d) theFind domain lim(1 of+ the x) 1/x rational . function, substitution cannot be 

60. First-Class Mail As of April 2016, the U.S. used. Postal But Service this does not mean that the x→0 

limit does not exist! Factoring the numerator and 

charged $0.94 postage for first-class large envelope the denominator, weighing up to we find 


or partial ounce up to and including 13 ounces. First-class rates do Challenge x 2 + 5x Problems + 6 (x + 2)(x + 3) 

= 


x 

For Problems 2 − 4 (x + 2)(x − 2) 

67–70, investigate each of the following limits. 

Source: U.S. Postal Service Notice 123 Since x = −2, and we are interested in the limit 

{ 

as 1 x approaches if x is an integer −2, the factor x + 2 

f (x) = 

(a) Find a function C that models the first-class canpostage be divided charged, out. Then 


in dollars, for a large envelope weighing w ounces. Assume x 2 + 5x + 6 (x + 2)(x + 3) 

w>0. 

lim 

67. = lim 

f (x) 68. lim f (x) 69. lim f (x) 70. lim f (x) 

x→−2 x 2 − 4 x→−2 x→2 (x + 2)(x −x→1/2 2) = lim x + 3 

x→−2 x − 2 = −2 + 3 

↑ 

↑ 

x→3 

↑ −2 − 2 =−1 x→0 4 


Factor 

x = −2 

Divide out ( x + 2) 

Use the Limit of a 

Rational Function 


■ 


for Example 11 

Finding the Limit of a Rational Function 

3 

x − 9x 

Find lim 

→ x − 3 . 

x 3 

Solution 

3 

x 9x 

x x 

lim lim ( 2 

− 

− 

= 

9) 

x→3 

x − 3 x→3 

x − 3 

x( x− 3)( x + 3) 

= lim 

x→3 

x − 3 

= lim x( x+ 3) = 3(3+ 3) = 18 

x→3 

To reinforce notational fluency, remind 

students that the limit notation must be 

included in each step of the factorization. 

Also, encourage them to proceed gradually, 

step by step, when factoring expressions. 

For your visual learners, use the graph of 

the function to explain this limit. 

24 

22 

y 

18 

15 

12 

9 

6 

3 


Finding the Limit of a Rational Function 

x − x 

Find lim 9 3/2 7/2 

. 

→ + 5/2 3/2 

x 0 x − 3x 

Solution 

Factor numerator and denominator: 

x − x 

lim 9 

→0 

x − 3x 

x 

3/2 7/2 

= 

lim 

+ 5/2 3/2 + 

x→0 

x 

x 

2 

4 

x 

(9 − x ) 

( x− 

3) 

3/2 2 

3/2 


In this two-page worksheet, graphs are provided 

for 6 limit questions that each require some 

form of algebraic manipulation. An additional 6 

questions cover the properties of limits given a 

table of values. 


This worksheet contains 8 limit questions that can 

be solved using techniques learned in this section. 

Graphs are not provided for answer verification. 

= 

−x 

lim 

x 

+ 

x→0 

( x −9) 

( x− 

3) 

( x− 3)( x+ 

3) 

=−lim + 

x→0 

( x − 3) 

=− lim ( x + 3) =−3 

+ 

x→0 

3/2 2 

3/2 


97 


11/01/17 9:52 am





Finding the Limit of a Quotient 

2 

x − 81 

Find lim . 

x→9 

x − 3 

Solution 

First, we rationalize the denominator 

to eliminate the radical term in the 

denominator: 

2 

x 81 x 

x 

lim lim ( 2 

81) 3 

x 9 

x 9 

x 

−− 3 

= ( x 

−− 3) 

⋅ + 

→ 

→ 

x + 3 

x x 

lim ( 2 

− 81) + 3 

= 

x→9 

x − 9 

x x x 

lim ( − 9)( + 9) + 3 

= 

x→9 

x − 9 

= lim ( x + 9) x + 3 

x→9 

( ) 

( ) 

( ) 

( ) 

( ) 

( ) 

= (9+ 9) 9+ 3 = 108 


for Example 15 

Finding the Limit of an Average Rate of 

Change 

Find the average rate of change of 

fx ( ) = x from x = 1 to x. Use this result 

to find the limit as x approaches 1 of the 

average rate of change of f from 1 to x. 

Solution 

Since f (1) = 1, the average rate 

of change of f from 1 to x is 

∆y 

∆ = fx ( ) − f (1) x −1 

= . The limit of 

x x −1 

x −1 the average rate of change from 1 to x is 

x −1 

x −1 

= 

− − ⋅ x + 1 

lim lim 

x 1 x→ 

x 1 x + 1 

x −1 

= lim 

x→1( x− 1)( x + 1) 

= lim 1 + = 1 

x→1 

x 1 2 

x→1 1 

Explain that the formula for the average 

rate of change is fundamental to differential 

calculus. Observe that the limit of this 

− 

formula, lim fx ( ) fc () 

→ x− 

c 

, gives the 

x c 

instantaneous rate of change, that is, the 

rate of change at a particular moment. 

Teaching Tip 

When you teach the students to find the 

average rate of change of a function 

between two points, choose problems that 

allow you to clearly illustrate the principle 

graphically, as well. 

2x 2 if x < 1 

The Limit of a Quotient 

33. f (x) = 

3x 2 at c = 1 

53. Slope property of a can Tangent onlyLine be usedFor when f (x) the = limit 1 

− 1 if x > 1 

2 x2 −of1: 

the denominator 

of the function is not zero. The next example illustrates a strategy to try if radicals are 

x 3 present. 

(a) Find the slope m 

if x < −1 


34. f (x) = 

x 2 at c =−1 


− 1 if x > −1 

x 2 EXAMPLE 14 Finding(b) theUse Limit the result of a from Quotient (a) to complete the following table: 

if x ≤ 0 

35. f (x) = 

at c = 0 

√ √ 

2x + 1 if x > 0 

x − 5 

⎧ 

Find lim 

h −0.5 −0.1 −0.001 0.001 0.1 0.5 

⎨ x 2 if x < 1 

x→5 x − 5 . 

m sec √ √ 

36. f (x) = 2 if x = 1 at c = 1 

x − 5 

⎩ 

Solution The domain of h(x) = is {x|x ≥ 0, x = 5}. Since the limit of the 

−3x + 2 if x > 1 

(c) Investigate x − 5the limit of the slope of the secant line found in (a) 

denominator is 

as h → 0. 



lim g(x) = lim(x − 5) = 0 

x→5 point P (2, x→5 f (2))? 




we cannot use the Limit of a Quotient 

P = (2, f 

property. 

(2)). 

A different strategy is necessary. We 

37. lim f (x) = 3; lim f (x) = 3; lim f (x) rationalize = 1; the numerator of the quotient. 

x→2 x→3− x→3 + 

√ √ 

f (2) = 3; f (3) = 1 

x − 5 54. Slope of a Tangent Line For f (x) = x 2 − 1: 

= (√ x − √ 5) 

· (√ x + √ 5) 


x − 5 (x − 5) (a) Find ( √ the x + √ slope 5) = x − 5 

m sec 

(x of the − 5)( √ secant x line + √ containing 5) = 1 

√ √ 

↑ x the + 5 

x→−1 x→2− x→2 + 

points P = (−1, f (−1)) and Q = (−1 + x = h, 5f (−1 + h)). 


39. lim f (x) = 4; lim f (x) =−1; lim Do you see why rationalizing (b) the Usenumerator the result from works? (a) to Itcomplete causes the term following x − 5table: 

to appear 

f (x) = 0; 

x→1 x→0− x→0 + in the numerator, and since x = 5, the factor x − 5 can be divided out. Then 

f (0) =−1; f (1) = 2 

√ √ h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1 

x − 5 

1 

lim 1 


x→5 

f (x) = lim1; 

= lim √ √m sec = 

x→2 x→−1 x→1 x→5 x − 5 x→5 x + 5 lim (√ x + √ 5) = 1 

√ √ = 1 

√ 

5 

5 + 5 2 √ 5 = ■ 

NOTE When finding a limit, remember 

10 

f (−1) = 1; f (2) = 3 

↑ x→5 

to include ''lim'' at each step until you 

x→c Use the (c) Limit Investigate of a Quotient the limit of the slope of the secant line found 

let x → c. 

in (a) as h → 0. 



each limit. 

|x − 5| 

41. lim 

x→5 + x − 5 

|x − 5| 

42. lim 

x→5 − x − 5 

43. lim 

x→ 12 

2x 

− 


point P = (−1, f (−1))? 


at P = (−1, f (−1)). 

PAGE 

44. lim 

x→ 12 

2x 45. lim 

+ 

x→ 23 

2x 46. lim 

− 

x→ 23 

2x 

+ 


5 Find the Limit of an Average x→0 Ratexof Change 

The next two examples illustrate function limits that f (x) we = cos encounter π 

47. lim |x|−x 48. lim |x|−x 

x at in Chapter 2. 

NEED TO x→2 + REVIEW? Average rate x→2 − of 

In Section P.1, we defined average x =− 1 rate 

3 3 2 , − 1 of 

4 , − 1 change: 

8 , − 1 If a 

10 , − 1 and b, 

12 ,..., 1 where 

12 , 1 10 , 1 a = 

8 , 1 b, 

4 , 1 are in 

change is discussed in Section P.1, p. 12. the domain of a function y = f (x), the average rate of change of f from a to b2 is . 


x→2 + x→2 − 

51. Slope of a Tangent Line For f (x) = 3x 2 (b) y Investigate f (b) −limf (a) cos π by using a table and evaluating the 

: 

= x→0 x 


function f (x) = cos π a = b 

x b − a 

x at 

and (3, 27). 


x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 EXAMPLE 15 Finding the Limit of an Average Rate of Change 3 , 1. 

and (x, f (x)), x = 2. 

(a) Find the average rate of(c) change Compare of f the (x) results = x 2 from + 3x(a) from and 2(b). toWhat x, x do = 2. you conclude 


graph of f at 2 using the result from (b).(b) Find the limit as x approaches about 2 the of the limit? average Why do rate youofthink change this of happens? f (x) = What x 2 + is 3x 


(d) On the same set of axes, graph f , the tangent line fromto2the tograph 

x. 

limits? 


Solution (a) The average 

(d) 

rate 

Use 

of change 

technology 

of f 

to 

from 

graph 

2 

f 

to 

. Begin 

x is 

with the x-window 



y f (x) − f (2) 

(a) Find the slope of the secant line containing the= points (2, 8) = (x 2 + 3x) lim − f (x) [2 2 using + 3(2)] a graph, what would you conclude? Zoom in 

x→0 = x 2 + 3x − 10 (x + 5)(x − 2) 

= 

and (3, 27). 

x x − 2 

on the 

x − 

graph. 

2 

Describe what you 

x − 

see. 

2 

(Hint: Be sure 

x − 

your 

2 

(b) Find the slope of the secant line containing (b) the The points limit (2, of 8) the average rate calculator of change is set is to the radian mode.) 

and (x, f (x)), x = 2. 


(c) Create a table to investigate the slope of the tangent line to thef (x) − f (2) (x x→0 + 5)(x x− 2 2) 

lim 

= lim 


function f (x) = cos π = lim(x + 5) = 7 

x→2 x − 2 x→2 x − 2 at x x→2 ■ 

=−0.1, −0.01, −0.001, 


x2 of f at the point (2, 8), and the secant line from (a). 

−0.0001, NOW WORK 0.0001, Problem 0.001, 0.01, 63 0.1. and AP® Practice Problem 3. 


MPAC 4: Connecting Multiple Representations 

Draw the following curve and secant line. Ask 

students to use the illustration to justify why the 

average rate of change of a function between two 

points is the slope of the secant line that passes 

through the points. Remind them that slope = 

y2− 

y1. Then have them demonstrate what 

x2− 

x1 

happens to the secant lines through the points as 

the left-hand point gradually approaches the righthand 

point (which is fixed). Ask them to use this 

result to show that the limit of the average value 

between two points is the slope of the tangent line 

at the fixed point. 

f(x) 

3 

2 

1 

2 4 6 8 10 x 

Use the tangent line applet on the Mathscoop Web 

site to illustrate this process. 

98 



11/01/17 9:53 am

Sullivan AP˙Sullivan˙Chapter01 October September 8, 2016 20, 2016 17:414:45 

Section 1.1 1.2 • Assess Your Understanding 89 99 




In Section P.1, we defined the (d) difference Use the graph quotient to investigate of a function lim C(w) f at xand as lim C(w). Do 

w→1− w→1 + 

f 

x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 these (x + suggest h) − f that (x) lim 

3 . 

w→1 

h C(w) = 0 exists? 

h 

(e) Use the graph to investigate lim C(w) and lim C(w). 

w→12− CALC 

w→12 


+ EXAMPLE 16 Finding theDo Limit theseof suggest a Difference that lim C(w) Quotient exists? 

about the limit? Why do you think this CLIPhappens? What is your 

w→12 


(a) For f (x) = 2x 2 f (x C(w). + h) − f (x) 

− 3x + 1, find the difference quotient w→0 + , h = 0. 


(g) Use the graph to investigate lim C(w). h 

w→13 

[−2π, 2π] and the y-window [−1, 1]. If 

(b) 

you were 

Find 

finding 

the limit as h approaches 0 of the difference quotient − of f (x) = 2x 2 − 3x + 1. 


x→0 Solution (a) To find the difference quotient of f, we begin with f (x + h). 



calculator is set to the radian mode.) f (x + h) = 2(x + h) 2 (in− hours) 3(x + that h) the + 1 student = 2(xstudies. 2 + 2xh He+ claims h 2 ) −that 3x the − 3h closer + 1to 

PAGE 

85 57. (a) Use a table to investigate lim 

2 

. 


limit? 

Now 


limit? 


x→2 

(b) How close must x be to 2, so that f (x) is within f (x + 0.1h) of− thef (x) 

limit? 

h 


x→2 

x − 8 

limit? 







Summary or partial ounce up to and 


Twoletter Basic rates Limits do not apply to letters 


• lim A = A, where A is a constant. 

x→c Source: U.S. Postal Service Notice 123 

• lim x = c 

x→c 



Properties (b) What ofisLimits 

the domain of C? 

If f and (c) gGraph are functions the function for which C. lim f (x) and lim g(x) both exist, 

x→c x→c 


and k is a constant, then 

w→2− w→2 + 

• Limit 

these 

of a 

suggest 

Sum or 

that 

a Difference: 

lim C(w) exists? 

w→2 

lim (e) [ Use f (x) the ± g(x)] graph = to investigate lim f (x) ± lim 

g(x) C(w). 

x→c x→c w→0 x→c + 

• Limit (f) Use of the a Product: graph tolim 

investigate [ f (x) · g(x)] lim = C(w). lim f (x) · lim g(x) 

x→c w→3.5 − x→c x→c 

• Limit of a Constant Times a Function: lim[kg(x)] = k lim g(x) 

60. First-Class Mail As of April 2016, the x→c U.S. Postal Service x→c 



or partial ounce up to and including 13 ounces. First-class rates do 


1.2Source: Assess 

U.S. 

Your 

Postal Service 

Understanding 

Notice 123 


Concepts and Vocabulary 


1. (a) lim w>0. (−3) = ; (b) lim π = 

x→4 x→0 


2. If lim f (x) = 3, then lim[ f (x)] 5 = . 

x→c x→c 

3. If lim x→c 

f (x) = 64, then lim x→c 

3 √ f (x) = . 

The limit of the difference quotient 

tends to appear as a topic for a 

multiple-choice question on the exam. 

In the next chapter, students learn 

to recognize this as the definition of 

the derivative. Once this connection 

is made, the limit of the difference 

quotient can be found by simply taking 

the derivative of the function. This will 

be a great time saver. 


AP® Exam Tip 

6 Find the Limit of a Difference Quotient 


= 2x 2 + 4xh on+ the2h final. 2 − He 3x claims − 3h + that 1 studying significantly less than seven 




f (x +h)− f (x) = (2x 2 +4xh+2h 2 −3x −3h +1)−(2x 2 −3x +1) = 4xh+2h 2 −3h 


Then, the difference quotient is 


of limit. 

= 4xh + 2h2 − 3h h(4x + 2h − 3) 

= = 4x + 2h − 3, h = 0 

h 

h 



f (x + h) − f (x) 

(b) lim 

= lim(4x + 2h − 3) = 4x + 0 − 3 = 4x −f (x) 3 − f (c) ■ 

h→0 h y = h→0 f (x) at the point (c, f (c)) is m tan = lim 

. 

x→c x − c 

Another way to express this slope is toNOW defineWORK a new variable Problem 71. 



x→2 



x→2 [ ] n 

• Limit your reasoning. of a Power: lim 

65. The graph of f (x) = x [ − f (x)] n 3 = lim f (x) x→c x→c 

where n ≥ 2 is an integer is a straight line with a point punched 

3 − x 

• 

out. 

√ √ 

n 

Limit of a Root: lim f (x) = n lim f (x) 

x→c x→c 

provided (a) Whatf straight (x) >0line if nand ≥ 2what is even point? 

(b) Use the graph of f to investigate 

[ 

the one-sided 

] m/n limits of f as 

• Limit of [f(x)] 

x approaches m/n : lim[ f (x)] 

3. 

m/n = lim f (x) x→c x→c 

provided (c) Does[ the f (x)] graph m/n is suggest defined that forlim 

positive f (x) integers exists? If mso, andwhat n is it? 

[ x→3 ] f (x) 

lim 

66. • f (x) 

x→c 

Limit (a) Use of a table Quotient: to investigate lim lim(1 = + x) 1/x . 

x→c g(x) x→0 lim g(x) x→c 

provided 

(b) Use graphing 

lim g(x) 

technology 

= 0 

to graph g(x) = (1 + x) 1/x . 

(c) What x→c do (a) and (b) suggest about lim(1 + x) 1/x ? 

• Limit of a Polynomial Function: lim P(x) x→0 = P(c) 


x→c 

• Limit of a x→0 Rational Function: lim R(x) = R(c) 

x→c 

if c is in the domain of R 




f (x) = 

4. (a) lim x = 0 ; (b) if xlim 

is not an integer 

x→−1 x→e 

67. 5. (a) lim f (x) 68. lim f (x) 69. lim f (x) x→2 lim (x − 2) = ; (b) lim (3 + x) = 

x→0 

x→1/2 

x→1/2 

x→3 

70. lim x→0 

6. (a) lim (−3x) = x→2 

; (b) lim (3x) = 

x→0 

7. True or False If p is a polynomial function, 

then lim p(x) = p(5). 

x→5 


Finding the Limit of a Difference 

Quotient 

For fx ( ) = x + 1 find the limit of the 

difference quotient as h approaches 0 + . 

Solution 

fx+ h −f x 

lim ( ) ( ) 

+ 

→0 

h 

h 

= 

= 

= 

= 

= 

lim 

+ 

h→0 

lim 

+ 

h→0 

lim 

+ 

h→0 

lim 

+ 

h→0 

lim 

+ 

h→0 

( x+ h+ 1) − ( x + 1) 

h 

x+ h− 

x 

h 

( x+ h− 

x ) ( x+ h+ 

x ) 

⋅ 

h 

( x+ h+ 

x ) 

x+ h−x 

h( x+ h+ 

x) 

h 

h( x+ h+ 

x) 

1 

= lim 

+ 

h→0 

x+ h+ 

x 

1 

= 

x+ 0 + x 

1 

= 

2 x 



AB: 35, 37, 39, 41, 43, 47, 48, 51–65 

odd, 67, 71, 73–79 odd, AP ® Practice 

Problems 

BC: 10, 19, 37, 39, 43, 59, 75, and all AP ® 

Practice Problems (especially 5) 



Problems 


Problems 

1. (a) −3 

(b) π 

2. 243 

3. 4 

4. (a) −1 

(b) e 

5. (a) −2 

(b) 7 2 

6. (a) −6 

(b) 0 

7. True 


99 


11/01/17 9:53 am




8. 2 

9. False 

10. True 

11. 14 

12. –3 

13. 0 

14. 18 

15. 1 

16. 1 

17. 6 

18. 1 2 

19. 11 

20. 10 

21. 2 78 

22. 0 

23. 7+ 

3 

24. 2 

25. 0 

26. 0 

27. 5 

28. 1 

29. − 31 

8 

30. –3 

31. 10 

32. 14 3 

33. 13 4 

34. 1 5 

35. 4 

8. If the domain of a rational function R is {x | x = 0}, 

then lim R(x) = R( ). 

x→2 

9. True or False Properties of limits cannot be used for one-sided 

limits. 

(x + 1)(x + 2) 

10. True or False If f (x) = and g(x) = x + 2, 

x + 1 

then lim f (x) = lim g(x). 

x→−1 x→−1 

Skill Building 

In Problems 11–44, find each limit using properties of limits. 

11. lim [2(x + 4)] 12. lim [3(x + 1)] 

x→3 x→−2 

PAGE 

93 13. lim [x(3x−1)(x + 2)] 

x→−2 

14. lim [x(x − 1)(x + 10)] 

x→−1 

PAGE 

94 15. lim (3t − 2) 3 

t→1 

16. lim (−3x + 1) 2 

x→0 

17. lim (3 √ ( ) 1 

x) 18. lim 

3√ x x→4 x→8 4 

√ 

PAGE 

95 19. lim 5x − 4 

x→3 

20. lim 

t→2 

√ 

3t + 4 

21. lim t→2 

[t √ (5t + 3)(t + 4)] 22. lim 

t→−1 [t 3√ (t + 1)(2t − 1)] 

PAGE 

95 23. lim ( √ x + x + 4) 1/2 

x→3 

24. lim t→2 

(t √ 2t + 4) 1/3 

25. lim 

t→−1 [4t(t + 1)]2/3 26. lim x→0 

(x 2 − 2x) 3/5 

27. lim t→1 

(3t 2 − 2t + 4) 28. lim x→0 

(−3x 4 + 2x + 1) 

PAGE 

96 29. lim (2x 4 − 8x 3 + 4x − 5) 30. lim (27x 3 + 9x + 1) 

x→ 1 2 

x→− 1 3 

PAGE 

97 31. 

x 2 + 4 

lim √ x→4 x 

32. 

x 2 + 5 

lim √ x→3 3x 

PAGE 

97 33. 

2x 3 + 5x 

lim 

x→−2 3x − 2 

PAGE 

97 35. 

x 2 − 4 

lim x→2 x − 2 

x 3 − x 

37. lim 

x→−1 x + 1 

39. lim 

x→−8 

( 2x 

x + 8 + 16 

x + 8 

√ √ 

PAGE 

x − 2 

98 41. lim x→2 x − 2 

43. lim x→4 

√ 

x + 5 − 3 

(x − 4)(x + 1) 

) 

34. lim x→1 

2x 4 − 1 

3x 3 + 2 

x + 2 

36. lim 

x→−2 x 2 − 4 

x 3 + x 2 

38. lim 

x→−1 x 2 − 1 

40. lim x→2 

( 3x 

x − 2 − 6 

x − 2 

42. lim x→3 

√ x − 

√ 

3 

x − 3 

44. lim x→3 

√ 

x + 1 − 2 

x(x − 3) 

In Problems 45–50, find each one-sided limit using properties of limits. 

45. lim 

x→3 −(x2 − 4) 46. lim 

x→2 +(3x2 + x) 

x 2 − 9 

47. lim 

x→3 − x − 3 

x 2 − 9 

48. lim 

x→3 + x − 3 

√ 

√ 

49. lim 

x→3 −( 9 − x 2 + x) 2 50. lim 

x→2 +(2 x 2 − 4 + 3x) 

) 

In Problems 51–58, use the information below to find each limit. 

lim f (x) = 5 

x→c 

lim g(x) = 2 

x→c 

lim h(x) = 0 

x→c 

51. lim x→c 

[ f (x) − 3g(x)] 52. lim x→c 

[5 f (x)] 

53. lim [g(x)] 3 f (x) 

54. lim 

x→c x→c g(x) − h(x) 

55. lim x→c 

h(x) 

g(x) 

[ 1 

57. lim x→c g(x) 

56. lim x→c 

[4 f (x) · g(x)] 

] 2 

58. lim x→c 

3 √ 5g(x) − 3 

In Problems 59 and 60, use the graphs of the functions and properties 

of limits to find each limit, if it exists. If the limit does not exist, write, 

“the limit does not exist,” and explain why. 

59. (a) lim [ f (x) + g(x)] 

x→4 

y 

y h(x) 

10 y f (x) 

(b) lim { f (x) [g(x) − h(x)]} 

x→4 8 

(c) lim[ f (x) · g(x)] 

x→4 6 

(4, 6) 

(d) lim[2h(x)] 

x→4 

g(x) 

2 

(e) lim x→4 f (x) 

(4, 0) 

(f) 

8 x 

h(x) 

2 

lim 

y g(x) 

x→4 

(3, 2) 

f (x) 

60. (a) lim x→3 

{2 [ f (x) + h(x)]} 

(b) 

lim + h(x)] 

x→3−[g(x) (c) lim x→3 

3 √ h(x) 

(d) lim x→3 

f (x) 

h(x) 

(e) lim x→3 

[h(x)] 3 

(f) 

lim[ f (x) − 2h(x)] 3/2 

x→3 

y 

6 

4 

2 

2 

y f (x) 

(3, 2) 

y g(x) 

3 

(3, 6) 

y h(x) 

In Problems 61–66, for each function f, find the limit as x approaches c 

of the average rate of change of f from c to x. That is, find 

f (x) − f (c) 

lim 

x→c x − c 

61. f (x) = 3x 2 , c = 1 62. f (x) = 8x 3 , c = 2 

PAGE 

98 63. f (x) =−2x 2 + 4, c = 1 64. f (x) = 20 − 0.8x 2 , c = 3 

65. f (x) = √ x, c = 1 66. f (x) = √ 2x, c = 5 

In Problems 67–72, find the limit of the difference quotient for each 

f (x + h) − f (x) 

function f . That is, find lim 

. 

h→0 h 

67. f (x) = 4x − 3 68. f (x) = 3x + 5 

69. f (x) = 3x 2 + 4x + 1 70. f (x) = 2x 2 + x 

PAGE 

99 71. f (x) = 2 x 

72. f (x) = 3 x 2 

x 

36. − 1 4 

37. 2 

38. − 1 2 

39. 2 

40. 3 

2 

41. 

4 

3 

42. 

6 

1 

43. 

30 

1 

44. 

12 

45. 5 

46. 14 

47. 6 

48. 6 

49. 9 

50. 6 

51. –1 

52. 25 

53. 8 

54. 5 2 

55. 0 

56. 40 

57. 1 4 

3 

58. 7 

59. (a) 6 

(b) –16 

(c) –16 

(d) 0 

(e) − 1 4 

(f) 0 

60. (a) –4 

(b) 4 

3 

(c) − 2 

(d) 0 

(e) –8 

(f) 8 

61. 6 

62. 96 

63. –4 

64. –4.8 

65. 1 2 

1 

66. 

10 

67. 4 

68. 3 

69. 6x + 4 

70. 4x + 1 

2 

71. − 

x 

2 

6 

72. − 

x 

3 

100 



11/01/17 9:53 am



In Problems 73–80, find lim f (x) and lim f (x) for the 

x→c− x→c + 

given number c. Based on the results, determine whether lim f (x) 

x→c 

exists. 

2x − 3 if x ≤ 1 

PAGE 

93 73. f (x) = 

3 − x if x > 1 

at c = 1 

5x + 2 if x < 2 

74. f (x) = 

1 + 3x if x ≥ 2 

at c = 2 

⎧ 

⎨ 3x − 1 if x < 1 

75. f (x) = 4 if x = 1 at c = 1 

⎩ 

2x if x > 1 

⎧ 

⎨ 3x − 1 if x < 1 

76. f (x) = 2 if x = 1 at c = 1 

⎩ 

2x if x > 1 

x − 1 if x < 1 

77. f (x) = √ 

x − 1 if x > 1 

at c = 1 

 

78. f (x) = 

9 − x 

 

2 if 0 < x < 3 

at c = 3 

x 2 − 9 if x > 3 

79. 

⎧ 

⎨ x 2 − 9 

if x = 3 

f (x) = x − 3 

⎩ 

6 if x = 3 

at c = 3 

⎧ 

⎨ x − 2 

if x = 2 

80. f (x) = x 2 − 4 

at c = 2 

⎩ 

1 if x = 2 


Heaviside Functions In Problems 81 and 82, find the limit, if it 

exists, of the given Heaviside function at c. 

0 if t < 1 

PAGE 

94 81. u 1(t) = 

1 if t ≥ 1 

at c = 1 

0 if t < 3 

82. u 3(t) = 

1 if t ≥ 3 

at c = 3 

In Problems 83–92, find each limit. 

(x + h) 2 − x 2 

83. lim h→0 h 

1 

85. lim 

x + h − 1 x 

h→0 h 

1 1 

87. lim x→0 x 4 + x − 1 

4 

√ √ x + h − x 

84. lim h→0 h 

1 

(x + h) 

86. lim 

3 − 1 x 3 

h→0 

h 

2 

88. lim 

x→−1 x + 1 

x − 7 

x − 2 

89. lim √ 90. lim √ x→7 x + 2 − 3 x→2 x + 2 − 2 

91. lim x→1 

x 3 − 3x 2 + 3x − 1 

x 2 − 2x + 1 

1 

3 − 1 

x + 4 

x 3 + 7x 2 + 15x + 9 

92. lim 

x→−3 x 2 + 6x + 9 

93. Cost of Water The Jericho Water District determines quarterly 

water costs, in dollars, using the following rate schedule: 

Water used 

(in thousands of gallons) Cost 

0 ≤ x ≤ 10 $9.00 

10 < x ≤ 30 $9.00 + 0.95 for each thousand 

gallons in excess of 10,000 gallons 



x > 100 $143.50 + 2.20 for each thousand 


Source: Jericho Water District, Syosset, NY. 

(a) Find a function C that models the quarterly cost, in dollars, of 

using x thousand gallons of water. 

(b) What is the domain of the function C? 

(c) Find each of the following limits. If the limit does not exist, 

explain why. 

lim C(x) lim C(x) 

x→5 x→10 

(d) What is lim C(x)? 

x→0 + 

(e) Graph the function C. 

lim C(x) 

x→30 

lim C(x) 

x→100 

94. Cost of Electricity In June 2016, Florida Power and Light had 

the following monthly rate schedule for electric usage in 

single-family residences: 

Monthly customer charge $7.87 

Fuel charge 

≤ 1000 kWH $0.02173 per kWH 

> 1000 kWH $21.73 + 0.03173 for each kWH 

in excess of 1000 

Source: Florida Power and Light, Miami, FL. 

(a) Find a function C that models the monthly cost, in dollars, of 

using x kWH of electricity. 

(b) What is the domain of the function C? 

(c) Find lim C(x), if it exists. If the limit does not exist, 

x→1000 

explain why. 

(d) What is lim C(x)? 

x→0 + 

(e) Graph the function C. 

95. Low-Temperature Physics In thermodynamics, the average 

molecular kinetic energy (energy of motion) of a gas having 

molecules of mass m is directly proportional to its temperature T 

on the absolute (or Kelvin) scale. This can be expressed as 

1 

2 mv2 = 3 kT, where v = v(T ) is the speed of a typical molecule 

2 

at time t and k is a constant, known as the Boltzmann constant. 

(a) What limit does the molecular speed v approach as the gas 

temperature T approaches absolute zero (0 K or −273 ◦ C 

or −469 ◦ F)? 

(b) What does this limit suggest about the behavior of a gas as 

its temperature approaches absolute zero? 

88. 2 9 

89. 6 

90. 4 

91. 0 

92. –2 

93. (a) 

⎧ 

⎪ 

⎪ 

Cx ( ) = ⎨ 

⎪ 

⎪ 

⎩ 

9.00 0≤x 

≤10 

9.00 + 0.95( x− 10) 10 < x ≤30 

28.00 + 1.65( x− 30) 30 < x ≤100 

143.50 + 2.20( x− 100) x > 100 

(b) { xx≥ 

0} 

(c) 

lim Cx ( ) = 9.00, lim Cx ( ) = 9.00, 

x→5 

lim Cx ( ) = 28.00, 

x→30 

(d) 9.00 

(e) 

94. (a) 

Quarterly water cost 

(dollars) 

y 

250 

200 

150 

100 

50 

x→10 

lim Cx ( ) = 143.50 

x→100 

50 100 150 x 

Water usage (thousands of gallons) 

⎪ 

x 

( ) = ⎨ 

( x ) 

C x 

⎧ 7.87 + 0.02173 if 0 ≤x 

≤ 1000 

+ − > 

⎩ 

⎪ 

29.60 0.3173 1000 if x 1000 

(b) { xx≥ 

0} 

(c) lim Cx ( ) = 8.9565; 

− 

x→50 

lim Cx ( ) = 8.9565; 

+ 

x→ 

50 

lim Cx ( ) = 8.9565 

x→50 

(d) lim Cx ( ) = 7.87 

x→ 0 

+ 

73. lim fx ( ) =−1, 

lim fx ( ) = 2, 

x→1 

− 

not exist. 

74. lim fx ( ) = 12, 

x→2 

− 

not exist. 

75. lim fx ( ) = 2, 

x→1 

− 

76. lim fx ( ) = 2, 

x→1 

− 

77. lim fx ( ) = 0, 

x→1 

− 

78. lim fx ( ) = 0, 

x→3 

− 

x→ 1 

+ 

lim fx ( ) = 7, 

x→ 2 

+ 

lim fx ( ) does 

x→1 

lim fx ( ) does 

x→ 2 

lim fx ( ) = 2, lim fx ( ) = 2 

x→ 1 

+ 

x→1 

lim fx ( ) = 2, lim fx ( ) = 2 

x→ 1 

+ 

x→1 

lim fx ( ) = 0, lim fx ( ) = 0 

x→ 1 

+ 

x→1 

lim fx ( ) = 0, lim fx ( ) = 0 

x→ 3 

+ 

x→3 

79. lim fx ( ) = 6, 

x→3 

− 

lim fx ( ) = 6, 

x→ 3 

+ 

1 

80. lim fx ( ) = 

4 , lim fx ( ) 

x→2 

− 

x→ 2 

+ 



83. 2x 

1 

84. 

2 x 

1 

85. − 

x 

2 

86. − 

x 

3 

4 

87. − 1 

16 

= 

1 

4 , 

lim fx ( ) = 6 

x→3 

lim fx ( ) = 

x→2 

1 

4 

(e) 

C 

50 

40 

30 

20 

10 

95. (a) 0 

1 2 3 4 5 6 7 8 

(b) As the temperature of a gas 

approaches zero, the molecules in 

the gas stop moving. 

x 


101 


11/01/17 9:53 am




96. (a) 3 

(b) –1 


97. lim x = 0 since lim x = 0 and 

x→0 

x→0 

− 

lim = 0. 

x 

x→ 0 

+ 

98. lim x = lim 

2 

x = 

2 

lim x = 0 

x→0 

x→0 

x→0 

= 0 


⎧⎪ 

1, if x < 2 

f( x)= 

⎨ 

⎩⎪ 0, if x ≥ 2 , 

⎧⎪ 

1, if x < 2 

gx ( )= ⎨ 

⎩⎪ 0, if x ≥ 2 


⎧⎪ 

fx ( ) = ⎨ 

⎩⎪ 

⎧⎪ 

gx ( ) = ⎨ 

⎩⎪ 

0, if x < 2 

1, if x ≥ 2 , 

1, if x < 2 

0, if x ≥ 2 

101. Answers will vary. Sample answer 

⎧⎪ 

fx ( ) = ⎨ 

⎩⎪ 

⎧⎪ 

gx ( ) = ⎨ 

⎩⎪ 

2, if x < 1 

−1, if x ≥1 , 

1, if x < 1 

−1, if x ≥1 


⎧⎪ 

1, if x < 0 

f( x)= 

⎨ 

⎩⎪ −1, if x ≥0 

103. See TSM. 

104. See TSM. 

− 

105. na n 1 

106. If n is an even positive integer, then 

limit does not exist. If n is an odd 

positive integer, then the limit equals 

− 

na n 1 . 

107. m n 

108. 1 3 

a+ 

b 

109. 

2 

110. 

a1+ a2+ + 

a 

2 

111. 0 

n 

2x 2 if x < 3x 1 + 5 if x ≤ 2 

96. 33. For f (x) the = function 

3x 2 f (x) = at c = 1 

− 1 if x > 13 1 − x if x > 2 , find 

103. 

x 3 if x < −1 

34. f (x) = f (2 

x 2 + h) − f (2) at c =−1 

(a) lim − 1 if x > −1 

h→0 − h 

x 2 if x ≤ 0 

35. f (x) = 

f (2 + h) − f (2) 

(b) lim 

at c = 0 

h→0 + 2x + 1 h if x > 0 

⎧ 

⎨ xf 2 (2 + h) if−x < f (2) 1 

(c) Does lim 

exist? 

36. f (x) = h→0 2 hif x = 1 at c = 1 

⎩ 

−3x + 2 if x > 1 if x ≥ 0 

97. Use the fact that |x| = 

to show that lim |x| =0. 

−x if x < 0 x→0 

Applications 98. Use the fact and that Extensions |x| = √ x 2 to show that lim |x| =0. 

x→0 


99. Find functions f and g for which lim [ f (x) + g(x)] may exist 


x→c 

37. even lim f though (x) = 3; lim f lim (x) and f (x) lim= g(x) 3; do lim notf exist. (x) = 1; 

x→2 x→c x→3− x→c x→3 + 

100. f Find (2) = functions 3; f (3) f and = 1g for which lim [ f (x)g(x)] may exist even 

x→c 

38. though lim f (x) lim= f 0; (x) and lim limf g(x) =−2; do not exist. lim f (x) =−2; 

x→−1 x→c x→2 x→c − x→2 + 

 

f (−1) is not defined; f (2) =−2 f (x) 

101. Find functions f and g for which lim 


may exist even 

x→c g(x) f (x) = 0; 

x→1 x→0− x→0 + 

though lim f (x) and lim g(x) do not exist. 

f (0) =−1; x→c f (1) = x→c 2 

102. 

40. lim 

Find 

f 

a 

(x) 

function 

= 2; 

f 

lim 

for which 

f (x) = 

lim 

0; 

| f (x)| lim may 

f (x) 

exist 

= 1; 

even though 

x→c x→2 lim f (x) does not x→−1 exist. 

x→1 

f x→c (−1) = 1; f (2) = 3 


each limit. 

|x − 5| 

|x − 5| 

41. AP® lim Practice Problems 42. lim 

43. lim 

x→5 + x − 5 

x→5 − x − 5 

 

x→ 12 

2x 

− 

PAGE 

PAGE 

93 1. Consider the piecewise-defined function f given by 

PAGE 

44. lim 

x→ 12 

2x ⎧45. lim 

+ 

x→ 23 

2x 46. lim 

− 

x→ 23 

2x 

⎨ −x − 2 if x < −1 

+ 

f (x) = x 

2 if −1 ≤ x < 2 

⎩ 

47. lim |x|−x 48. −4x + lim 12 |x|−x if x ≥ 2 

x→2 + x→2 − 

Investigate the limits below and decide which limit does NOT 

3 3 


x→2 + x→2 − PAGE 

exist. 

(A) lim f (x) (B) lim 

51. Slope of a Tangent Line For f (x) f (x) 

x→−1 + x→2 − = 3x 2 : 

(a) (C) Find limthe f (x) slope of the (D) secant lim line f (x) 

x→2 x→−1 containing the points (2, 12) 

PAGE 

and (3, 27). 

PAGE 

(5 − t) 2 

97 2. (b) limFind the slope of the secant line containing the points (2, 12) 

t→5 

and 

t − 

(x, 

5 = 

f (x)), x = 2. 

(c) (A) Create −5 a table (B) 0 to investigate (C) 1 the(D) slope 5 of the tangent line to the PAGE 


PAGE 

f (x) − f (2) 

98 3. Find lim 

for the function f (x) = 3x 

(d) On the same set of axes, graph f , the tangent line 3 − 4. 

x→2 x − 2 

to the graph 


(A) 0 (B) 12 (C) 24 (D) 36 


PAGE 

x − s 

98 4. lim √ √ = 

(a) x→s 

Find x − 

the slope s 

of the secant line containing the points (2, 8) 

(A) and 2s (3, 27). (B) 2 √ √ 

s (C) 2s (D) s 


and (x, f (x)), x = 2. 





Answers to AP® Practice Problems 

1. D 

2. B 

3. D 

4. B 

5. C 

6. C 

7. C 

8. B 

Prove that if g is a function for which lim 

53. Slope of a Tangent Line For f (x) = 1 g(x) exists and 

x→c 

2 x2 − 1: 

if k is any real number, then lim[kg(x)] exists and 

(a) Find the slope m sec of the x→c 

secant line containing the 

lim[kg(x)] = k lim 

points P = (2, f g(x). 

x→c x→c (2)) and Q = (2 + h, f (2 + h)). 

104. Prove that if the number c is in the domain of a rational 


function R(x) = p(x) , then lim R(x) = R(c). 

q(x) x→c 

h −0.5 −0.1 −0.001 0.001 0.1 0.5 

m sec 


(c) Investigate x n − the a n 

105. Find lim , limit n a positive of the slope integer. of the secant line found in (a) 

as x→a h →x 0. − a 

(d) What isx the n + slope a n of the tangent line to the graph of f at the 

106. Find lim , n a positive integer. 

point x→−aP = x + (2, a f (2))? 

(e) On thex m same − 1set of axes, graph f and the tangent line to f at 

107. Find 

P 

lim x→1 = (2, x n f (2)). 

, m, n positive integers. 

√ − 1 

3 

1 + x − 1 

108. 54. Slope Find lim of a Tangent Line . For f (x) = x 2 − 1: 

x→0 x 

 

(a) Find the (1 

slope + ax)(1 

m sec of + 

the bx) 

secant − 1 

line containing the 

109. Findpoints lim P = (−1, f (−1)) and Q . = (−1 + h, f (−1 + h)). 

x→0 x 


(1 + a1x)(1 + a 2x) ···(1 + a nx) − 1 

110. Find lim 

. 

x→0 x 

h −0.1 

f (h) − 

−0.01 

f (0) 

−0.001 −0.0001 0.0001 0.001 0.01 0.1 

111. Findm lim h→0 sec 

if f (x) = x|x|. 

h 


in (a) as h → 0. 


point P = (−1, f (−1))? 


at P = (−1, f (−1)). 

ax 

93 5. For g(x) = 

2 − 5 if x < 2 

ax + b if x 

85 55. (a) Investigate lim cos π > 2 , 


find values for a and x→0 b so that x 

function f (x) = cos π lim g(x) = 7. 

x→2 

(A) a = 1, b x at 

x =− 1 = 5 

2 , − 1 (B) 

4 , − 1 a 

8 , − 1 = 2, b 

10 , − 1 = 3 

(C) a = 3, b = 1 (D) a = 6, b =−5 

12 ,..., 1 

12 , 1 10 , 1 8 , 1 4 , 1 2 . 

 

95 6. lim 

x→4 

(b) +(5 x 2 − 16 + 3x) = 

Investigate lim cos π by using a table and evaluating the 

(A) −12 (B) x→0 

0 (C) x 

function f (x) = cos π 12 (D) The limit does not exist. 

 

[ f (x)] x at 

2 

x =−1, − 1 − 8x 

3 , − 1 + 3 

95 7. If lim 

5 , − 1 = 

7 , − 9, 1 then 

9 ,..., lim 1 9 , f 1 (x) 

7 , 1 = 

5 , 1 x→2 x + 1 

x→2 

√ √ 3 , 1. 

(A) 22 (B) 2 10 (C) 16 (D) 256 


95 8. lim[x −1/2 about (5xthe −limit? 7) 1/3 ] Why = 

x→3 

do you think this happens? What is 


(A) 3 −1/2 2 

8 

limits? (B) 

3 1/2 (C) 

3 1/2 (D) 6 −1/2 




x→0 




x→0 x2 function f (x) = cos π at x =−0.1, −0.01, −0.001, 

x2 −0.0001, 0.0001, 0.001, 0.01, 0.1. 

102 



11/01/17 9:53 am


Section 1.3 • Continuity 103 

y 

c 

y f (x) 

(c, f (c)) 

x 

y 

1.3 Continuity 

c 

(c, f (c)) 

OBJECTIVES When you finish this section, you should be able to: 

1 Determine whether a function is continuous at a number (p. 103) 

2 Determine intervals on which a function is continuous (p. 106) 

3 Use properties of continuity (p. 108) 

4 Use the Intermediate Value Theorem (p. 110) 

Sometimes lim f (x) equals f (c) and sometimes it does not. In fact, f (c) may not even 

x→c 

be defined and yet lim f (x) may exist. In this section, we investigate the relationship 

x→c 

between lim f (x) and f (c). Figure 21 shows some possibilities. 

x→c 

y f (x) 

x 

y 

c 

y f (x) 

(b) lim f(x) lim f(x) f (c) (c) lim f (x) lim f(x) 

(a) lim f(x) lim f(x) f (c) 

x→c x→c x→c x→c x→c x→c 

f(c) is not defined. 

Figure 21 

x 

y 

c 

(c, f (c)) 

y f (x) 

(d) lim f (x) lim f (x) 

x→c x→c 

f (c) is defined. 

x 

y 

c 

y f (x) 

(e) lim f (x) lim f (x) 

x→c x→c 

f (c) is not defined. 

Of these five graphs, the “nicest” one is Figure 21(a). There, lim 

x→c 

f (x) exists and is 

equal to f (c). Functions that have this property are said to be continuous at the number c. 

This agrees with the intuitive notion that a function is continuous if its graph can be drawn 

without lifting the pencil. The functions in Figures 21(b)–(e) are not continuous at c, since 

each has a break in the graph at c. This leads to the definition of continuity at a number. 

DEFINITION Continuity at a Number 

A function f is continuous at a number c if the following three conditions are met: 

• f (c) is defined (that is, c is in the domain of f ) 

• lim f (x) exists 

x→c 

• lim f (x) = f (c) 

x→c 

If any one of these three conditions is not satisfied, then the function is discontinuous 

at c. 

NOW WORK AP® Practice Problems 1 and 2. 

1 Determine Whether a Function Is Continuous at a Number 

EXAMPLE 1 

Teaching Tip 

Another advantage to the left–right–center 

definition of continuity is that it will help students to 

identify the type of discontinuity a function exhibits. 

If lim fx ( ) ≠ lim fx ( ), then the function jumps 

− + 

x→c x→c 

from one value to another and therefore is 

classified as a jump discontinuity. 

If lim fx ( ) = lim fx ( ), then the function does not 

− + 

x→c x→c 

jump. That doesn’t mean it is continuous, though. 

We still have to determine if the left- and righthand 

limits equal fc (). If they do not, there is a tiny 

hole. This is called a removable discontinuity. 

Determining Whether a Function Is Continuous 

at a Number 

(a) Determine whether f (x) = 3x 2 − 5x + 4 is continuous at 1. 

(b) Determine whether g(x) = x 2 + 9 

is continuous at 2. 

x 2 − 4 

Solution (a) We begin by checking the conditions for continuity. First, 1 is in the 

domain of f and f (1) = 2. Second, lim f (x) = lim(3x 2 − 5x + 4) = 2, so lim f (x) 

x→1 x→1 x→1 

exists. Third, lim f (x) = f (1). Since the three conditions are met, f is continuous at 1. 

x→1 

(b) Since 2 is not in the domain of g, the function g is discontinuous at 2. ■ 

x 

TRM Alternate Examples 

Section 1.3 





Section 1.3 




Teaching Tip 

The definition of continuity can also be 

stated this way: 

A function is continuous at a number c if 

lim fx ( ) = lim fx ( ) = fc (). 

− + 

x→c x→c 

Students may remember this set of 3 

checks because they examine the limit 

from the left, the limit from the right, and 

then the center, the value at c. 

The three checks can become three 

2 

questions. For instance, for lim x 

x→3 

1. What is the limit of f (x) as x 

approaches 3 from the right? 

2. What is the limit of f (x) as x 

approaches 3 from the left? 

3. What is the value of f(x) at x = 3? 

Since these three values are 9, then f(x) = 

x 2 is continuous at x = 3. 

In contrast, consider the following function 

at c = 3: 

⎛ 

⎜ 

fx ( ) = ⎜ 

⎜ 

⎝ 

2 

x , x < 3 

4, x = 3 

x+ 6, x > 3 

1. lim x 2 as x approaches 3 from the 

left = 9 

2. lim x + 6 as x approaches 3 from the 

right = 9 

3. f (3) = 4 

Since these three values are not the same, 

then f is not continuous. 

Teaching Tip 

Remind the students that if 

lim fx ( ) = lim fx ( ) then lim fx ( ) exists. 

− + 

x→c x→c 

x→c 

If it helps their understanding, consider 

using this phrase: If the limit of a function 

from the left equals the limit of the 

function from the right, then the limit in 

general exists. 

Section 1.3 • Continuity 

103 


11/01/17 9:53 am





for example 2 

Determining Whether a Function Is 

Continuous at a Number 

3 

x −25x 

Determine whether f( x) 

= is 

2 

continuous at x = 5. x −5x 

Solution 

We first check the conditions for continuity. 

To determine if 5 is in the domain of f, 

compute 

3 

5 −25(5) 

f (5) = 

2 

5 −5(5) 

This shows that f (5) is not defined, 

because both numerator and denominator 

are 0. Since 5 is not in the domain of f, 

then f is not continuous at 5. 


for example 2 


Continuous at a Number 

4 

( x −1) 

Use the graph of f( x) 

= to 

1− 

x 

determine whether f is continuous at 

x = 1. If f is discontinuous, what type of 

discontinuity does the function have? 

y 

y 

4 

2 

x 

f(x) 2 2 

x 2 4 

4 2 2 4 

2 

(0, 

4 ) 

2 

4 

Figure 23 f is continuous at 0; 

f is discontinuous at −2 and 2. 

x 

Figure 22 shows the graphs of f and g from Example 1. Notice that f is continuous 

at 1, and its graph is drawn without lifting the pencil. But the function g is discontinuous 

at 2, and to draw its graph, you must lift your pencil at x = 2. 

10 

f (x) 3x 2 5x 4 

4 

Figure 22 

2 

y 

20 

15 

5 

(1, 2) 

2 

4 

x 

(a) f is continuous at 1. (b) g is discontinuous at 2. 

4 

2 

y 

10 

5 

5 

2 

g(x) x2 9 

x 2 4 



EXAMPLE 2 at a Number 

 

x 

2 

+ 2 

Determine if f (x) = is continuous at the numbers −2, 0, and 2. 

x 2 − 4 

Solution The domain of f is {x|x = −2, x = 2}. Since f is not defined at −2 and 2, 

the function f is not continuous at −2 and at 2. The number 0 is in the domain of f. 

√ 

2 

That is, f is defined at 0, and f (0) =− 

4 . Also, 

 

x 

2 

+ 2 

lim f (x) = lim 

x→0 x→0 x 2 − 4 

 

lim x 2 

+ 2 lim 

= x→0 

lim (x 2 − 4) = lim 

x→0 

x→0 

(x 2 + 2) 

x→0 x 2 − lim 

x→0 

4 

√ √ 

0 + 2 2 

= 

0 − 4 =− 4 = f (0) 

The three conditions of continuity at a number are met. So, the function f is continuous 

at 0. ■ 

Figure 23 shows the graph of f. 


4 

x 

4 

24 

2 

22 

22 

24 

2 

4 

x 

CALC 

CLIP 

EXAMPLE 3 

Determining Whether a Piecewise-Defined Function 

Is Continuous 

Determine whether the function 

⎧ 

x 2 − 9 

if x < 3 

⎪⎨ x − 3 

f (x) = 

9 if x = 3 

⎪⎩ 

x 2 − 3 if x > 3 

is continuous at 3. 

Solution 

The graph of f reveals a hole at 

x = 1 because f (1) is undefined. We can 

conclude that f is discontinuous at x = 1. 

Since lim − fx ( ) = lim + fx ( ) =−4, 

x→1 x→1 

the function has a removable discontinuity 

at x = 1. If we define f (1) = − 4, then f is a 

continuous function. 


This worksheet contains 7 functions along 

with their graphs. Students are asked to 

identify the point(s) of discontinuity as 

well as to state why the listed points are 

discontinuous. 

Calculator Tip 

Some students may be tempted to graph functions 

on their calculator to make decisions about 

continuity. Most students, though, use a default 

calculator setting in which holes cannot be seen 

on the graph that is made on the calculator screen. 

This might lead the students to (falsely) believe 

a function is continuous when it is not. Students 

mainly should determine points of discontinuity 

analytically and then use a graph to support their 

conclusion. 

104 


TE_Sullivan_Chapter01_PART I.indd 1 

11/01/17 9:57 am



2 

y 

20 

10 

(3, 9) 

2 4 

x 2 9 

if x 3 

x 3 

f (x) 

9 if x 3 

x 2 3 if x 3 

Figure 24 f is discontinuous at 3. 

IN WORDS Visually, the graph of a 

function f with a removable 

discontinuity has a hole at (c, lim x→c 

f (x)). 

The discontinuity is removable because 

we can fill in the hole by appropriately 

defining f . The function will then be 

continuous at x = c. 

NEED TO REVIEW? The floor function 

is discussed in Section P.2, p. 19. 

y 

2 

2 

2 

Figure 25 f (x) =x 

2 

4 

x 

x 

Solution Since f (3) = 9, the function f is defined at 3. To check the second condition, 

we investigate the one-sided limits. 

Teaching Tip 

x 2 − 9 


x→3 − 

x→3 − x − 3 = lim (x − 3)(x + 3) 

= lim 

x→3 − x − 3 x→3−(x + 3) = 6 

↑ 

Divide out x − 3 


x→3 + 

x→3 +(x 2 − 3) = 9 − 3 = 6 

3 

Since lim f (x) = lim f (x), then lim f (x) exists. But, lim f (x) = 6 and f (3) = 9, 

x −25x 

x→3 − 

x→3 + 

x→3 x→3 fx ( ) = 

2 

so the third condition of continuity is not satisfied. The function f is discontinuous 

x −5x 

at 3. ■ 

Figure 24 shows the graph of f . 

NOW WORK Problem 25 and AP® Practice Problems 5, 8, and 11. 

fx 

The discontinuity at c = 3 in Example 3 is called a removable discontinuity because 

we can redefine f at the number c to equal lim f (x) and make f continuous at c. So, in 

x→c fx 

Example 3, if f (3) is redefined to be 6, then f would be continuous at 3. 

DEFINITION Removable Discontinuity 

Let f be a function that is defined everywhere in an open interval containing c, except 

fx 

possibly at c. The number c is called a removable discontinuity of f if the function 

is discontinuous at c but lim f (x) exists. The discontinuity is removed by defining (or 

x→c 

redefining) the value of f at c to be lim f (x). 

x→c fx 

NOW WORK Problems 13 and 35 and AP® Practice Problem 4. 


EXAMPLE 4 at a Number 

Determine whether the floor function f (x) =x is continuous at 1. 

Solution The floor function f (x) =x =the greatest integer ≤ x. The floor function f 

is defined at 1 and f (1) = 1. But 


x→1 − 

x→1−x =0 and lim f (x) = lim 

x→1 + 

x→1 +x 

=1 

So, limx does not exist. Since limx does not exist, f is discontinuous at 1. ■ 

x→1 x→1 

Figure 25 illustrates that the floor function is discontinuous at each integer. Also, 

none of the discontinuities of the floor function is removable. Since at each integer the 

value of the floor function “jumps” to the next integer, without taking on any intermediate 

values, the discontinuity at integer values is called a jump discontinuity. 

evaluate fc 


We have defined what it means for a function f to be continuous at a number. Now 

we define one-sided continuity at a number. 

Teaching Tip 

DEFINITION One-Sided Continuity at a Number 

Let f be a function defined on the interval (a, c]. Then f is continuous from the left 

at the number c if 

lim f (x) = f (c) 

x→c − 

Let f be a function defined on the interval [c, b). Then f is continuous from the right 

at the number c if 

lim f (x) = f (c) 

x→c + 


This worksheet contains 5 piecewise functions. 

The students are asked to determine if the 

function is continuous at various values of each 

function. 

Continuity can also be taught using 

graphical and analytic methods. Students 

may be surprised to see that the function 

looks like a line. 

Notice the algebraic simplification: 

3 

x −25x 

( ) = 

2 

x −5x 

2 

xx ( −25) 

( ) = 

xx ( −5) 

( x− 5)( x+ 

5) 

( ) = 

( x −5) 

( ) = x + 5 

The x’s and the x – 5’s were removed from 

the equation. The values, 0 and 5, were 

not in the domain of the original function. 

Since both were removed through algebraic 

simplification, they are both removable 

discontinuities. This function has no other 

points of discontinuity. 

Common Error 

Remind the students that when they 

(), they must substitute the 

value c into the original function, not the 

simplified version of the original function. 

If you present the left–right–center check 

for continuity at a point, consider having 

the students evaluate f() c first. If f() 

c 

is undefined, the function cannot be 

continuous at that point. 

Teaching Tip 

If a student is given a function and asked to 

identify all points of discontinuity, have the 

student begin by identifying the domain of 

the function. Points not in the domain of the 

function are points of discontinuity. 

AP® Exam Tip 

Although questions about continuity 

seldom appear directly on the multiplechoice 

portion of the exam, it is 

common to see continuity as part of a 

multiple-choice question. For example, 

a principle of continuity may appear as 

one of the answer choices. Also, in a 

problem about continuity, it is common 

to be given a piecewise function. 



11/01/17 9:57 am





for example 5 


Continuous on a Closed Interval 

⎧ 

⎪ 

fx ( ) = ⎨ 

⎪ 

⎩ 

2 

x −3x−4 

x − 4 

k 

x ≠ 4 

x = 4 

Determine the value of k that makes the 

function f defined above continuous at 

x = 4. 

Solution 

The function will be continuous at x = 4 if 

lim fx ( ) = lim fx ( ) = f (4). 

− + 

x→4 x→4 

In Example 4, we showed that the floor function f (x) =x is discontinuous at 

x = 1. But since 

f (1) =1 =1 and lim f (x) =x =1 

x→1 + 

the floor function is continuous from the right at 1. In fact, the floor function is 

discontinuous at each integer n, but it is continuous from the right at every integer n. 

(Do you see why?) 

2 Determine Intervals on Which a Function Is Continuous 

So far, we have considered only continuity at a number c. Now, we use one-sided 

continuity to define continuity on an interval. 

DEFINITION Continuity on an Interval 

• A function f is continuous on an open interval (a, b) if f is continuous at every 

number in (a, b). 

• A function f is continuous on an interval [a, b) if f is continuous on the open 

interval (a, b) and continuous from the right at the number a. 

• A function f is continuous on an interval (a, b] if f is continuous on the open 

interval (a, b) and continuous from the left at the number b. 

• A function f is continuous on a closed interval [a, b] if f is continuous on the 

open interval (a, b), continuous from the right at a, and continuous from the left 

at b. 

Figure 26 gives examples of graphs over different types of intervals. 

x 

lim fx ( ) = lim 

− 

− 

x→4 x→4 

−3x−4 

x −4 

( x+ 1)( x−4) 

= lim 

− 

x→4 

( x −4) 

= lim ( x + 1) 

− 

x→4 

2 

y 

f(a) 

y f (x) 

a 

b x 

lim f(x) f (a) 

x→a 

(a) f is continuous on [a, b). 

y 

f(a) 

y f (x) 

a 

b x 

lim f(x) f (a) 

x→a 

(b) f is continuous on (a, b). 

y 

f(b) 

y f (x) 

a 

b x 

lim f(x) f (b) 

x→b 

(c) f is continuous on (a, b]. 

y 

y f (x) 

f(b) 

a 

b x 

lim f(x) f (b) 

x→b 

(d) f is continuous on (a, b). 

y 

f(b) 

y f (x) 

f(a) 

a 

b x 

lim f (x) f (a) 

x→a 

lim f (x) f (b) 

x→b 

(e) f is continuous on [a, b]. 

= 5 

Figure 26 

x 

lim fx ( ) = lim 

+ + 

x→4 x→4 

−3x−4 

x −4 

( x+ 1)( x−4) 

= lim 

+ 

x→4 

( x −4) 

= lim ( x + 1) 

+ 

x→4 

= 5 

If we define f (4) = 5, the function f will be 

continuous at x = 4. 

2 

For example, the graph of the floor function f (x) =x in Figure 25 illustrates 

that f is continuous on every interval [n, n + 1), n an integer. In each interval, f is 

continuous from the right at the left endpoint n and is continuous at every number in the 

open interval (n, n + 1). 

EXAMPLE 5 


on a Closed Interval 

Is the function f (x) = √ 4 − x 2 continuous on the closed interval [−2, 2]? 

Solution The domain of f is {x|−2 ≤ x ≤ 2}. So, f is defined for every number in 

the closed interval [−2, 2]. 

For any number c in the open interval (−2, 2), 

lim 

x→c 

f (x) = lim 

x→c 

So, f is continuous on the open interval (−2, 2). 

√ 

4 − x 

2 

= √ lim 

x→c 

(4 − x 2 ) = √ 4 − c 2 = f (c) 

AP® Exam Tip 

The topic of continuity on an interval 

is generally not directly tested on the 

exam. The concept, however, is still 

important because many important 

theorems that do appear on the exam 

are applicable only for functions that 

are continuous on a closed interval. 

106 



11/01/17 9:58 am



22 21 

y 

2 

1 

Figure 27 f (x) = √ 4 − x 2 , −2 ≤ x ≤ 2 

y 

4 

2 

(0, 0) 

(1, 0) 

1 

2 4 

2 

x 

x 

f (x) x 2 (x 1) 

Figure 28 f is discontinuous at 0; 

f is continuous on [1, ∞). 

To determine whether f is continuous on [−2, 2], we investigate the limit from the 

right at −2 and the limit from the left at 2. Then, 

√ 


x→−2 + 

4 − x 

2 

= 0 = f (−2) 

x→−2 + 

So, f is continuous from the right at −2. Similarly, 

lim 

x→2 − 

√ 

f (x) = lim 4 − x 2 

= 0 = f (2) 

x→2 − 

So, f is continuous from the left at 2. We conclude that f is continuous on the closed 

interval [−2, 2]. ■ 

Figure 27 shows the graph of f . 

DEFINITION Continuity on a Domain 


A function f is continuous on its domain if it is continuous at every number c in its 

domain. 

EXAMPLE 6 

√ 

Determining Whether f(x) = x 2 (x − 1) Is Continuous 

on Its Domain 

Determine if the function f (x) = √ x 2 (x − 1) is continuous on its domain. 

Solution The domain of f (x) = √ x 2 (x − 1) is {x|x = 0} ∪{x|x ≥ 1}. We need to 

determine whether f is continuous at the number 0 and whether f is continuous on the 

interval [1, ∞). 

At the number 0, there is an open interval containing 0 that contains no other number 

( ) 

in the domain of f . [For example, use the interval − 1 2 , 1 2 

.] This means lim f (x) 

x→0 

does not exist. So, f is discontinuous at 0. 

and 

For all numbers c in the open interval (1, ∞) we have 

f (c) = √ c 2 (c − 1) 

√ 

lim x 2 

(x − 1) = √ lim[x 2 (x − 1)] = √ c 2 (c − 1) = f (c) 

x→c 

x→c 

So, f is continuous on the open interval (1, ∞). 

Now, at the number 1, 

So, f is continuous from the right at 1. 

f (1) = 0 and lim 

x→1 + √ 

x 

2 

(x − 1) = 0 

The function f (x) = √ x 2 (x − 1) is continuous on the interval [1, ∞), but it is 

discontinuous at 0. So, f is not continuous on its domain. ■ 

Figure 28 shows the graph of f. The discontinuity at 0 is subtle. It is neither a 

removable discontinuity nor a jump discontinuity. 

When listing the properties of a function in Chapter P, we included the function’s 

domain, its symmetry, and its zeros. Now we add continuity to the list by asking, “Where 

is the function continuous?” We answer this question here for two important classes of 

functions: polynomial functions and rational functions. 

THEOREM 

• A polynomial function is continuous on its domain, all real numbers. 

• A rational function is continuous on its domain. 


MathIsFun: The Math Is Fun Web site 

has a very clear interactive explanation 

that can help the struggling student to 

grasp continuity. A link to this resource is 

available on the Chapter 1 Additional 

Resources document, available for 

download. 


for Example 7 

Identifying Where Functions Are 

Continuous 

⎧ 

⎪ 

⎪ 

⎪ 

fx ( ) = ⎨ 

⎪ 

⎪ 

⎪ 

⎩ 

| x| x ≤ 1 

1 

1< x ≤3 

x + 1 

2 3< x < 4 

x 

x ≥ 4 

Let f be the function defined here. For what 

values of x is f not continuous? 

Solution 

We have to check each point where one 

piece of the function stops and the next 

one starts to determine if they start and 

stop at the same place. Find the one-sided 

limits at x = 1, 3, and 4. 

lim fx ( ) = |1| = 1 

x→1 

− 1 1 

lim fx ( ) = = 

x→ 1 

+ x + 1 2 

Since the one-sided limits are not equal, f is 

not continuous at x = 1. 

1 1 

lim fx ( ) = = 

x→3 

− x + 1 4 

lim fx ( ) = 2 

x→ 3 

+ 

Since the one-sided limits are not equal, f is 

not continuous at x = 3. 

lim fx ( ) = 2 

x→4 

− 

lim fx ( ) = 4 = 2 

→ + f (4) = 4 = 2 

x 4 

f is continuous at x = 4. 



11/01/17 9:58 am




Teaching Tip 

Consider teaching the overall principles 

without relying on rote learning of 

the theorems presented, such as the 

continuity of a sum, difference, product, 

and quotient presented on this page. 

Students may balk at the number of 

theorems presented, when in reality, they 

only have to understand the overarching 

principle of how to determine if a function 

is continuous at a point and how to identify 

points of discontinuity algebraically. 

Proof If P is a polynomial function, its domain is the set of real numbers. For a 

polynomial function, 

lim P(x) = P(c) 

x→c 

for any number c. That is, a polynomial function is continuous at every real number. 

If R(x) = p(x) is a rational function, then p(x) and q(x) are polynomials and the 

q(x) 

domain of R is {x|q(x) = 0}. The Limit of a Rational Function (p. 96) states that for 

all c in the domain of a rational function, 

lim R(x) = R(c) 

x→c 

So a rational function is continuous at every number in its domain. ■ 

To summarize: 

• If a function is continuous on an interval, its graph has no holes or gaps on that 

interval. 

• If a function is continuous on its domain, it will be continuous at every number 

in its domain; its graph may have holes or gaps at numbers that are not in the 

domain. 

For example, the function R(x) = x 2 − 2x + 1 

is continuous on its domain 

x − 1 

{x|x = 1} even though the graph has a hole at (1, 0), as shown in Figure 29. The 

function f (x) = 1 is continuous on its domain {x|x = 0}, as shown in Figure 30. 

x 

Notice the behavior of the graph as x goes from negative numbers to positive numbers. 

y 

y 

2 

4 

2 

f (x) 1 x 

2 

2 

4 

x 

4 

2 

2 

4 

x 

2 

R(x) x2 2x 1 

x 1 

2 

4 

Figure 29 The graph of R has a hole at (1, 0). 

Figure 30 f is not defined at 0. 

3 Use Properties of Continuity 

So far we have shown that polynomial and rational functions are continuous on their 

domains. From these functions, we can build other continuous functions. 

THEOREM Continuity of a Sum, Difference, Product, and Quotient 

If the functions f and g are continuous at a number c, and if k is a real number, then 

the functions f + g, f − g, f · g, and kf are also continuous at c. If g(c) = 0, the 

function f is continuous at c. 

g 

108 



11/01/17 9:58 am



NEED TO REVIEW? Composite 

functions are discussed in Section P.3, 

pp. 27--30. 

The proofs of these properties are based on properties of limits. For example, the 

proof of the continuity of f + g is based on the Limit of a Sum property. That is, if 

lim f (x) and lim g(x) exist, then lim[ f (x) + g(x)] = lim f (x) + lim g(x). 

x→c x→c x→c x→c x→c 

EXAMPLE 7 

Identifying Where Functions Are Continuous 

Determine where each function is continuous: 

(a) F(x) = x 2 + 5 − 

x 

x 2 + 4 

(b) G(x) = x 3 + 2x + x 2 

x 2 − 1 

Solution First we determine the domain of each function. 

(a) F is the difference of the two functions f (x) = x 2 + 5 and g(x) = 

x 

x 2 + 4 , each 

of whose domain is the set of all real numbers. So, the domain of F is the set of all real 

numbers. Since f and g are continuous on their domains, the difference function F is 

continuous on its domain. 

(b) G is the sum of the two functions f (x) = x 3 + 2x, whose domain is the set of all 

real numbers, and g(x) = x 2 

, whose domain is {x|x = −1, x = 1} . Since f and g 

x 2 − 1 

are continuous on their domains, G is continuous on its domain, {x|x = −1, x = 1} . ■ 


The continuity of a composite function depends on the continuity of its components. 

THEOREM Continuity of a Composite Function 

If a function g is continuous at c and a function f is continuous at g(c), then the 

composite function ( f ◦ g)(x) = f (g(x)) is continuous at c. That is, 

lim( f ◦ g)(x) = lim 

x→c x→c 

f (g(x)) = f [lim g(x)] = f (g(c)) 

x→c 

EXAMPLE 8 Identifying Where Functions Are Continuous 

Determine where each function is continuous: 

(a) F(x) = √ x 2 + 4 (b) G(x) = √ x 2 − 1 (c) H(x) = x 2 − 1 

x 2 − 4 + √ x − 1 

Solution (a) F = f ◦ g is the composite of f (x) = √ x and g(x) = x 2 + 4. f is 

continuous for x ≥ 0 and g is continuous for all real numbers. The domain of F is all 

real numbers and F = ( f ◦ g)(x) = √ x 2 + 4 is continuous for all real numbers. That 

is, F is continuous on its domain. 

(b) G is the composite of f (x) = √ x and g(x) = x 2 − 1. f is continuous for x ≥ 0 

and g is continuous for all real numbers. The domain of G is {x|x ≥ 1}∪{x|x ≤−1} 

and G = ( f ◦ g)(x) = √ x 2 − 1 is continuous on its domain. 

(c) H is the sum of f (x) = x 2 − 1 

x 2 − 4 and the function g(x) = √ x − 1. The domain 

of f is {x|x = −2, x = 2}; f is continuous on its domain. The domain of g is 

x ≥ 1. The domain of H is {x|1 ≤ x < 2} ∪{x|x > 2}; H is continuous on its 

domain. ■ 


Teaching Tip 

Students should have a strong knowledge 

of what all of the basic functions (as shown 

in Section P.2) look like. The following is 

a suggested list of functions the students 

should be familiar with. Students should 

also be comfortable with transformations of 

these functions. 

y = x y = sinx y = e 

2 

y = x y = cos x y = e 

−x 

3 

y = x y = tanx y = lnx 

1 

y = | x| 

y = x y = 

x 


Identifying Where Functions Are 

Continuous 

Which of the following functions are 

continuous at x = 0? 

1 

(a) fx ( ) = 

x 

(b) g(x) = tan x 

Solution 

(a) Since x = 0 is not in the domain of f, 

1 

fx ( ) = is discontinuous at x = 0. 

x 

(b) The function gx ( ) = tan x is defined 

and continuous on − π < x < π , so 

2 2 

gx ( ) = tanx 

is continuous at x = 0. 

x 

NEED TO REVIEW? Inverse functions 

are discussed in Section P.4, pp. 37--40. 

Recall that for any function f that is one-to-one over its domain, its inverse f −1 

is also a function, and the graphs of f and f −1 are symmetric with respect to the line 

y = x. It is intuitive that if f is continuous, then so is f −1 . See Figure 31 on page 110. 

The following theorem, whose proof is given in Appendix B, confirms this. 



11/01/17 9:58 am




AP® Exam Tip 

y 

4 

y f 1 (x) 

y x 

The Intermediate Value Theorem 

sometimes appears on the multiplechoice 

portion of the exam. The 

Intermediate Value Theorem also 

appears sometimes as a portion of a 

free-response question. Even if it only 

appears sporadically on the exams, it 

is important that students learn it. For 

example, see the 2007 AP ® Calculus 

AB Exam, Question 3. 

2 

y f (x) 

4 2 2 4 

Figure 31 

THEOREM Continuity of an Inverse Function 

If f is a one-to-one function that is continuous on its domain, then its inverse 

function f −1 is also continuous on its domain. 

2 

4 

x 

AP® Exam Tip 

The three theorems that students 

must cite and use on the exam as a 

part of an answer or justification for 

an answer are the IVT (Intermediate 

Value Theorem), the EVT (Extreme 

Value Theorem, which will be learned 

in Section 4.2), and the MVT (Mean 

Value Theorem, which will be learned 

in Section 4.3). It is a good idea to have 

students practice them throughout the 

year. 

5000 m 

3765.6 m 

2000 m 

4 Use the Intermediate Value Theorem 

Functions that are continuous on a closed interval have many important properties. One 

of them is stated in the Intermediate Value Theorem. The proof of the Intermediate Value 

Theorem may be found in most books on advanced calculus. 

THEOREM The Intermediate Value Theorem 

Let f be a function that is continuous on a closed interval [a, b] and f (a) = f (b). 

If N is any number between f (a) and f (b), then there is at least one number c in the 

open interval (a, b) for which f (c) = N. 

To get a better idea of this result, suppose you climb a mountain, starting at an 

elevation of 2000 meters and ending at an elevation of 5000 meters. No matter how 

many ups and downs you take as you climb, at some time your altitude must be 3765.6 

meters, or any other number between 2000 and 5000. 

In other words, a function f that is continuous on a closed interval [a, b] must take 

on all values between f (a) and f (b). Figure 32 illustrates this. Figure 33 shows why 

the continuity of the function is crucial. Notice in Figure 33 that there is a hole in the 

graph of f at the point (c, N). Because of the discontinuity at c, there is no number c in 

the open interval (a, b) for which f (c) = N. 

y 

y 

Teaching Tip 

f(b) 

f(b) 

A picture is worth 1,000 words! The 

Intermediate Value Theorem is very much 

a commonsense theorem that can easily 

be conveyed using Figures 32 and 33. 

N 

f(a) 

a 

f (c) N 

y f (x) 

c 

b 

x 

N 

f(a) 

a 

y f (x) 

c 

b 

x 


MPAC 1: Reasoning with Definitions and 

Theorems 

Justifying answers will now include 

confirming that the hypotheses of theorems 

are met before stating the conclusions. The 

Web site AP ® Central, under the heading 

Teaching and Assessing AP ® Calculus, 

offers resources and short videos by AP ® 

Calculus teachers demonstrating how they 

prepare students. The module on continuity 

and differentiability is particularly good 

and can be accessed through a calculus 

teacher’s AP ® Audit account. 

Figure 32 f takes on every value 

between f (a) and f (b). 

Figure 33 A discontinuity at c results in no 

number c in (a, b) for which f (c) = N. 

The Intermediate Value Theorem is an existence theorem. It states that there is at 

least one number c for which f (c) = N, but it does not tell us how to find c. However, we 

can use the Intermediate Value Theorem to locate an open interval (a, b) that contains c. 

An immediate application of the Intermediate Value Theorem involves locating the 

zeros of a function. Suppose a function f is continuous on the closed interval [a, b] and 

f (a) and f (b) have opposite signs. Then by the Intermediate Value Theorem, there is 


This worksheet contains an explanation of 

the Intermediate Value Theorem and one 

problem based upon a portion of a former 

free-response exam question. 

110 



11/01/17 9:58 am



f 0, 3g 3 f22, 10g 

Figure 34 f (x) = x 3 + x 2 − x − 2 

at least one number c between a and b for which f (c) = 0. That is, f has at least one 

zero between a and b. 

EXAMPLE 9 

Using the Intermediate Value Theorem 

Use the Intermediate Value Theorem to show that 

f (x) = x 3 + x 2 − x − 2 

has a zero between 1 and 2. 

Solution Since f is a polynomial, it is continuous on the closed interval [1, 2]. Because 

f (1) =−1 and f (2) = 8 have opposite signs, the Intermediate Value Theorem states 

that f (c) = 0 for at least one number c in the interval (1, 2). That is, f has at least one 

zero between 1 and 2. Figure 34 shows the graph of f on a graphing utility. ■ 

NOW WORK Problem 59 and AP® Practice Problems 6, 7, 9, and 10. 

The Intermediate Value Theorem can be used to approximate a zero in the interval 

(a, b) by dividing the interval [a, b] into smaller subintervals. There are two popular 

methods of subdividing the interval [a, b]. 

The bisection method bisects [a, b], that is, divides [a, b] into two equal subintervals 

( ) b − a 

and compares the sign of f to the signs of the previously computed values 

2 

f (a) and f (b). The subinterval whose endpoints have opposite signs is then bisected, 

and the process is repeated. 

The second method divides [a, b] into 10 subintervals of equal length and compares 

the signs of f evaluated at each of the 11 endpoints. The subinterval whose endpoints 

have opposite signs is then divided into 10 subintervals of equal length and the process 

is repeated. 

We choose to use the second method because it lends itself well to the table feature 

of a graphing utility. You are asked to use the bisection method in Problems 107–114. 

EXAMPLE 10 

Using the Intermediate Value Theorem to Approximate 

a Real Zero of a Function 

The function f (x) = x 3 +x 2 −x −2 has a zero in the interval (1, 2). Use the Intermediate 

Value Theorem to approximate the zero correct to three decimal places. 

Solution Using the TABLE feature on a graphing utility, we subdivide the interval [1, 2] 

into 10 subintervals, each of length 0.1. Then we find the subinterval whose endpoints 

have opposite signs, or the endpoint whose value equals 0 (in which case, the exact 

zero is found). From Figure 35, since f (1.2) =−0.032 and f (1.3) = 0.587, by the 

Intermediate Value Theorem, a zero lies in the interval (1.2, 1.3). Correct to one decimal 

place, the zero is 1.2. 

Repeat the process by subdividing the interval [1.2, 1.3] into 10 subintervals, each 

of length 0.01. See Figure 36. We conclude that the zero is in the interval (1.20, 1.21), 

so correct to two decimal places, the zero is 1.20. 

Figure 35 Figure 36 

AP® Exam Tip 

Although problems like Example 10 have 

not appeared on previous AP ® Exams, 

this example can help the student better 

understand how to find the zeros of a function 

using technology. A much faster method using 

a TI-84 Plus calculator is pressing (2nd Trace 

then Zero), which will find the zero graphically. 

This is one of the four graphing calculator skills 

that students have to know to be successful on 

the AP ® Exam. 


for Example 9 

Using the Intermediate Value Theorem 

Let f be a continuous function on a 

closed interval [−5, 5]. Also suppose that 

f ( − 5) =− 3and f (5) = 3. Which of the 

following statements must be true? 

I. f() c= 0 for at least 1 c between −5 and 5. 

II. f (0) = 0 

III. f() c= 0 for at least 1 c between −3 and 3. 

Solution 

I. Since f ( − 5) < 0 and f (5) > 0according 

to the IVT, there is at least 1 number c in 

the open interval (−5, 5) for which fc () = 0. 

This answer is true. 

II. The IVT only tells us that f (x) = 0 for 

some value of x in (−5, 5), but we do not 

know where in the interval the function 

crosses the x-axis. So we do not know if 

f (0) = 0. This answer is not true. 

III. Also, we do not know if the function 

crosses the x-axis between −3 and 3. It 

might cross between −5 and −4. We do 

not know! This answer is not true. 



Theorems 

To use the IVT, the function must be 

continuous on a closed interval. Ask 

students to explain why the function must 

be continuous. Tell students that if you 

draw a curve with a pencil and do not 

lift the pencil from the page, the result is 

a continuous function. Ask students to 

explain why this pencil drawing shows the 

conclusion of the IVT. 

Section 1.3 • Continuity 

111 


11/01/17 9:58 am






AB: 13–18, 19–35 odd, 59–63 odd, AP ® 

Practice Problems 

BC: 13–18, 24, 25, 29, 51–56, 79, and all 

AP ® Practice Problems 



Problems 


Problems 

1. True. 

2. False. 

3. fc () is defined, lim fx ( ) exists, 

x→c 

lim fx ( ) = fc () 

x→c 

4. True. 

5. False. 

6. False. 

7. False. 

8. True. 

9. Discontinuous. 

10. Continuous. 

11. True. 

12. False. 

13. (a) Discontinuous at c =− 3. 

(b) lim ( ) ≠ f ( −3) 

fx 

x→−3 

(c) Removable. 

(d) f ( − 3) =−2 

14. (a) Continuous at c = 0. 

15. (a) Discontinuous at c = 2. 

(b) lim fx ( ) does not exist. 

x→2 

(c) Not removable. 


(b) lim fx ( ) does not exist. 

x→3 


17. (a) Continuous at c = 4. 


(b) f not defined at c = 5, and 

lim fx ( ) ≠ lim fx ( ) so lim fx ( ) does 

− + 

x→5 x→5 

x→5 

not exist. 


19. Continuous at c =− 1. 

20. Continuous at c = 5. 

21. Continuous at c =− 2. 

22. Discontinuous at c = 2. 


Figure 37 

1.3 Assess Your Understanding 

Now subdivide the interval [1.20, 1.21] into 10 subintervals, each of length 0.001. 

See Figure 37. 

We conclude that the zero of the function f is 1.205, correct to three decimal 

places. ■ 

Notice that a benefit of the method used in Example 10 is that each additional 

iteration results in one additional decimal place of accuracy for the approximation. 


PAGE 

105 13. c =−3 14. c = 0 

1. True or False A polynomial function is continuous at every 15. c = 2 16. c = 3 

real number. 

2. True or False Piecewise-defined functions are never continuous 

at numbers where the function changes equations. 

3. The three conditions necessary for a function f to be continuous 

17. c = 4 18. c = 5 

at a number c are , , and . 

4. True or False If f is continuous at 0, then g(x) = 1 f (x) is 

4 

continuous at 0. 

5. True or False If f is a function defined everywhere in an open 

interval containing c, except possibly at c, then the number c is 

called a removable discontinuity of f if the function f is not 

continuous at c. 

6. True or False If a function f is discontinuous at a number c, 

then lim f (x) does not exist. 

x→c 

7. True or False If a function f is continuous on an open interval 

(a, b), then it is continuous on the closed interval [a, b]. 

8. True or False If a function f is continuous on the closed interval 

[a, b], then f is continuous on the open interval (a, b). 

In Problems 9 and 10, explain whether each function is continuous or 

discontinuous on its domain. 

9. The velocity of a ball thrown up into the air as a function of 

time, if the ball lands 5 seconds after it is thrown and stops. 

10. The temperature of an oven used to bake a potato as a function 

of time. 

11. True or False If a function f is continuous on a closed interval 

[a, b], then the Intermediate Value Theorem guarantees that the 

function takes on every value between f (a) and f (b). 

12. True or False If a function f is continuous on a closed interval 

[a, b] and f (a) = f (b), but both f (a) >0 and f (b) >0, then 

according to the Intermediate Value Theorem, f does not have a 

zero on the open interval (a, b). 


In Problems 13–18, use the graph of y = f (x) (top right). 

(a) Determine if f is continuous at c. 

(b) If f is discontinuous at c, state which condition(s) of the definition 

of continuity is (are) not satisfied. 

(c) If f is discontinuous at c, determine if the discontinuity is 

removable. 

(d) If the discontinuity is removable, define (or redefine) f at c to 

make f continuous at c. 







30. Discontinuous at c =− 1. 



33. f (2) = 4 

34. f (3) = 7 

35. f (1) = 2 

36. f ( − 1) =−4 

(3, 1) 


y 

4 

2 

4 2 2 4 

2 (3, 1) 

4 

(2, 3) 

In Problems 19–32, determine whether the function f is continuous 

at c. 

PAGE 

104 19. f (x) = x 2 + 1 at c =−1 20. f (x) = x 3 − 5 at c = 5 

PAGE 

104 21. f (x) = x 

at c =−2 22. f (x) = x 

x − 2 

x 2 + 4 

 

2x + 5 if x ≤ 2 

23. f (x) = 

at c = 2 

4x + 1 if x > 2 

 

2x + 1 if x ≤ 0 

24. f (x) = 

at c = 0 

2x if x > 0 

⎧ 

3x − 1 if x < 1 

⎪⎨ 

PAGE 

105 25. f (x) = 4 if x = 1 at c = 1 

⎪⎩ 

2x if x > 1 

⎧ 

3x − 1 if x < 1 

⎪⎨ 

26. f (x) = 2 if x = 1 at c = 1 

⎪⎩ 

2x if x > 1 

 

3x − 1 if x < 1 

27. f (x) = 

at c = 1 

2x if x > 1 

⎧ 

3x − 1 ⎪⎨ 

if x < 1 

28. f (x) = 2 if x = 1 at c = 1 

⎪⎩ 

3x if x > 1 

 

x 2 if x ≤ 0 

29. f (x) = 

at c = 0 

2x if x > 0 

⎧ 

⎪⎨ 

x 2 if x < −1 

30. f (x) = 2 if x =−1 

⎪⎩ 

−3x + 2 if x > −1 

at c =−1 

37. Continuous on the given interval. 


39. Not continuous on the given interval. 

Continuous on { xx | 3}. 


41. Continuous on { xx | ≠ 0} . 

42. Continuous on the set of all real numbers. 


44. Continuous on { xx | ≥ 0} . 

45. Continuous on { xx | ≥0, x ≠ 9} . 

46. Continuous on { xx | ≥0, x ≠ 4} . 

47. Continuous on { xx | < 2} . 

48. Continuous on { x|| x| > 1} . 

x 

at c = 2 

112 



11/01/17 9:58 am



⎧ 

4 − 3x 

⎪⎨ 

2 if x < 0 

4 if x = 0 at c = 0 

31. f (x) = 

⎪⎩ 16 − x 2 

if 0 < x < 4 

4 − x 

⎧√ ⎨ 4 + x if −4 ≤ x ≤ 4 

 

32. f (x) = 

⎩ x 2 − 3x − 4 

at c = 4 

if x > 4 

x − 4 

In Problems 33–36, each function f has a removable 

discontinuity at c. Define f (c) so that f is continuous at c. 

33. f (x) = x2 − 4 

x − 2 , c = 2 

34. f (x) = x2 + x − 12 

, c = 3 

x − 3 

⎧ 

⎨ 1 + x if x < 1 

PAGE 

105 35. f (x) = 4 if x = 1 c = 1 

⎩ 

2x if x > 1 

⎧ 

⎨ x 2 + 5x if x < −1 

36. f (x) = 0 if x =−1 c =−1 

⎩ 

x − 3 if x > −1 

In Problems 37–40, determine if each function f is 

continuous on the given interval. If the answer is no, 

state the interval, if any, on which f is continuous. 

PAGE 

107 37. f (x) = x2 − 9 

on the interval [−3, 3) 

x − 3 

38. f (x) = 1 + 1 on the interval [−1, 0) 

x 

39. 

1 

f (x) = on the interval [−3, 3] 

x 2 − 9 

 

40. f (x) = 9 − x 2 on the interval [−3, 3] 

In Problems 41–50, determine where each function f is continuous. 

First determine the domain of the function. Then support your decision 

using properties of continuity. 

41. f (x) = 2x 2 + 5x − 1 42. f (x) = x + 1 + 2x 

x 

x 2 + 5 

43. f (x) = (x − 1)(x 2 + x + 1) 44. f (x) = √ x(x 3 − 5) 

PAGE 

109 45. f (x) = √ x − 9 

46. f (x) = √ x − 4 

x − 3 x − 2 

 

PAGE 

109 47. f (x) = 

x 2 + 1 

2 − x 

48. f (x) = 

 

4 

x 2 − 1 

49. f (x) = (2x 2 + 5x − 3) 2/3 50. f (x) = (x + 2) 1/2 

In Problems 51–56, use the function 

⎧ √ 

15 − 3x if x < 2 

⎪⎨ √ 

5 if x = 2 

f (x) = 

9 − x ⎪⎩ 

2 if 2 < x < 3 

x − 2 if 3 ≤ x 

51. Is f continuous at 0? Why or why not? 


PAGE 

105 53. Is f continuous at 3? Why or why not? 


50. Continuous on { xx | ≥− 2} . 

51. f is continuous at 0 because lim fx ( ) = f (0). 

x→0 

52. f is discontinuous at 4 because lim fx ( ) 

x→4 


53. f is discontinuous at 3 because lim fx ( ) does 

x → 3 

not exist. 

54. f is discontinuous at 2 because lim fx ( ) does 

x→2 

not exist. 

55. f is continuous at 1 because lim fx ( ) = f (1). 

x→1 

56. f is continuous at 2.5 because 

lim fx ( ) = f (2.5). 

x→2.5 



56. Is f continuous at 2.5? Why or why not? 

In Problems 57 and 58: 

(a) Use technology to graph f using a suitable scale on each axis. 

(b) Based on the graph from (a), determine where f is continuous. 

(c) Use the definition of continuity to determine where f is continuous. 

(d) What advice would you give a fellow student about using 

technology to determine where a function is continuous? 

57. f (x) = x3 − 8 

x − 2 

58. f (x) = x2 − 3x + 2 

3x − 6 

In Problems 59–64, use the Intermediate Value Theorem to determine 

which of the functions must have zeros in the given interval. Indicate 

those for which the theorem gives no information. Do not attempt to 

locate the zeros. 

PAGE 

111 59. f (x) = x 3 − 3x on [−2, 2] 

60. f (x) = x 4 − 1on[−2, 2] 

61. 

x 

f (x) = − 1 on [10, 20] 

(x + 1) 2 

62. f (x) = x 3 − 2x 2 − x + 2 on [3, 4] 

63. f (x) = x3 − 1 

on [0, 2] 

x − 1 

64. f (x) = x2 + 3x + 2 

x 2 on [−3, 0] 

− 1 

In Problems 65–72, verify that each function has a zero in the indicated 

interval. Then use the Intermediate Value Theorem to approximate the 

zero correct to three decimal places by repeatedly subdividing the 

interval containing the zero into 10 subintervals. 

PAGE 

112 65. f (x) = x 3 + 3x − 5; interval: [1, 2] 

66. f (x) = x 3 − 4x + 2; interval: [1, 2] 

67. f (x) = 2x 3 + 3x 2 + 4x − 1; interval: [0, 1] 

68. f (x) = x 3 − x 2 − 2x + 1; interval: [0, 1] 

69. f (x) = x 3 − 6x − 12; interval: [3, 4] 

70. f (x) = 3x 3 + 5x − 40; interval: [2, 3] 

71. f (x) = x 4 − 2x 3 + 21x − 23; interval: [1, 2] 

72. f (x) = x 4 − x 3 + x − 2; interval: [1, 2] 

In Problems 73 and 74, 

(a) Use the Intermediate Value Theorem to show that f has a zero in 

the given interval. 

(b) Use technology to find the zero rounded to three decimal places. 

 

73. f (x) = x 2 + 4x − 2 in [0, 1] 

74. f (x) = x 3 − x + 2in[−2, 0] 


Heaviside Functions In Problems 75 and 76, determine whether the 

given Heaviside function is continuous at c. 

0 if t < 1 

75. u 1(t) = 

1 if t ≥ 1 

c = 1 

0 if t < 3 

76. u 3(t) = 

1 if t ≥ 3 

c = 3 

57. (a) 

y 

18 

14 

10 

6 

2 

3 2 1 1 2 3 x 

(b) Based on the graph, it appears that the 

function is continuous for all real numbers. 

(c) f is actually continuous at all real numbers 

except x = 2. 

(d) Answers will vary. Sample answer: 

Conclusions drawn from graphing technology 

should always be confirmed using basic 

analysis. 

58. (a) 

23 22 21 

y 

1.5 

1 

0.5 

21 

21.5 

1 2 3 

(b) Based on the graph, it appears 

that the function is continuous on all 

real numbers. 

(c) f is continuous on { xx | ≠ 2}. 

(d) Answers will vary. Sample 

answer: Conclusions drawn from 

graphing technology should always be 

confirmed using analysis. 

59. Yes. A zero exists on the given 

interval. 

60. IVT gives no information. 





65. 1.154 

66. 1.675 

67. 0.211 

68. 0.445 

69. 3.134 

70. 2.137 

71. 1.157 

72. 1.308 

73. (a) Since f is continuous on [0,1], 

f (0) < 0, and f (1) > 0, the IVT 

guarantees that f has a zero on the 

interval (0,1). 

(b) 0.828 

74. (a) Since f is continuous on [ −2,0], 

f ( − 2) < 0, and f (2) > 0, the IVT 

guarantees that f has a zero on the 

interval (0,1). 

(b) −1.521 



x 


113 


11/01/17 9:58 am




77. Continuous on the set of all real 

numbers. 

y 

8 

6 

4 

2 

3 2 1 1 2 3 

78. Continuous on the set of all real 

numbers. 

y 

4 

3 

2 

1 

23 22 21 1 2 3 

⎧0.47 if 0< w ≤1 

⎪ 

⎪0.68 if 1< w ≤2 

79. (a) C( w)= 

⎨ 

⎪0.89 if 2< w ≤3 

⎪ 

⎩⎪ 

1.10 if 3< w ≤3.5 

{ } 

(b) w 0< w ≤3.5 

(c) Continuous on the intervals (0,1], 

(1,2], (2,3], and (3,3.5]. 

(d) Jump discontinuities at w = 1, 

w = 2, and w = 3. 

(e) Answers will vary. Sample answer: 

It is in the consumer’s best interest 

to have letters weigh as close as 

possible to a whole number of ounces 

without going over. 

⎧ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

80. (a) Cw ( ) = ⎨ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎪ 

⎩ 

0.94 if 0< w ≤1 

1.15 if 1< w ≤2 

1.36 if 2 < w ≤3 

1.57 if 3 < w ≤4 

1.78 if 4 < w ≤5 

1.99 if 5 < w ≤6 

2.20 if 6 < w ≤7 

2.41 if 7 < w ≤8 

2.62 if 8 < w ≤9 

2.83 if 9 < w ≤10 

3.04 if 10< w ≤11 

3.25 if 11< w ≤12 

3.46 if 12< w ≤13 

(b) { w|0< w ≤13} 

(c) C is continuous on the intervals 

(0, 1], (1, 2], (2, 3], (3, 4], (4, 5], 

(5, 6], (6, 7], (7, 8], (8, 9], (9, 10], 

(10, 11], (11, 12], and (12, 13]. 

(d) C has jump discontinuities at 

w = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 

and 12. 

(e) Answers will vary. Sample answer: 

It is in the customer’s best interest 

x 

x 

In Problems 77 and 78, determine where each function is continuous. 

Graph each function. 

1 − x 2 if |x| ≤1 

77. f (x) = 

x 2 − 1 if |x| > 1 

 

78. f (x) = 

4 − x 2 if |x| ≤2 

| x | − 2 if |x| > 2 


charged $0.47 postage for first-class letters weighing up to and 

including 1 ounce, plus a flat fee of $0.21 for each additional or 

partial ounce up to 3.5 ounces. First-class letter rates do not apply 

to letters weighing more than 3.5 ounces. 

Source: U.S. Postal Service Notice 123. 

(a) Find a function C that models the first-class postage charged 

for a letter weighing w ounces. Assume w>0. 


(c) Determine the intervals on which C is continuous. 

(d) At numbers where C is not continuous (if any), what type of 

discontinuity does C have? 

(e) What are the practical implications of the answer to (d)? 


charged $0.94 postage for first-class large envelopes weighing up 

to and including 1 ounce, plus a flat fee of $0.21 for each 

additional or partial ounce up to 13 ounces. First-class rates do not 

apply to large envelopes weighing more than 13 ounces. 

Source: U.S. Postal Service Notice 123. 

(a) Find a function C that models the first-class postage 

charged for a large envelope weighing w ounces. Assume 

w>0. 






81. Cost of Electricity In June 2016, Florida Power and Light 

had the following monthly rate schedule for electric usage in 

single-family residences: 

Monthly customer charge $7.87 

Fuel charge 

≤ 1000 kWH 

0.02173 per kWH 

> 1000 kWH $21.73 + 0.03173 for each 

kWH in excess of 1000 

Source: Florida Power and Light, Miami, FL. 

(a) Find a function C that models the monthly cost of 

using x kWH of electricity. 






82. Cost of Water The Jericho Water District determines quarterly 

water costs, in dollars, using the following rate schedule: 

to have packages that weigh as close as 

possible to a whole number of ounces 

without going over. 

81. (a) 

⎧⎪ 

C x 

7.87 0.02173x 

if 0 x 1000 

( )= ⎨ 

+ ≤ ≤ 

⎩⎪ − 2.13+ 0.03173x 

if x> 

1000 

(b) { x| x≥ 

0} 

(c) C is continuous on its domain. 

(d) See (c). 

(e) Answers will vary. Sample answer: To 

minimize the monthly cost of electricity, it is in 

the consumer’s best interest to minimize the 

amount of electricity used. 

Water used 

(in thousands of gallons) Cost 

0 ≤ x ≤ 10 $9.00 





x > 100 $143.50 + 2.20 for each thousand 


Source: Jericho Water District, Syosset, NY. 

(a) Find a function C that models the quarterly cost of 

using x thousand gallons of water. 






83. Gravity on Europa Europa, one of 

the larger satellites of Jupiter, has an icy 

surface and appears to have oceans 

beneath the ice. This makes it a 

candidate for possible extraterrestrial 

life. Because Europa is much smaller 

than most planets, its gravity is weaker. 

If we think of Europa as a sphere with 

uniform internal density, then inside the sphere, the gravitational 

field g is given by g(r) = Gm r, 0≤ r < R, where R is the 

R3 radius of the sphere, r is the distance from the center of the 

sphere, and G is the universal gravitation constant. Outside a 

uniform sphere of mass m, the gravitational field g is given by 

g(r) = Gm 

r 2 , R < r 

(a) For the gravitational field of Europa to be continuous 

at its surface, what must g(r) equal? 

[Hint: Investigate lim r→R 

g(r).] 

(b) Determine the gravitational field at Europa’s surface. This 

will indicate the type of gravity environment organisms will 

experience. Use the following measured values: Europa’s 

mass is 4.8 × 10 22 kilograms, its radius is 1.569 × 10 6 

meters, and G = 6.67 × 10 −11 . 

(c) Compare the result found in (b) to the gravitational field on 

Earth’s surface, which is 9.8 meter/second 2 . Is the gravity 

on Europa less than or greater than that on Earth? 

84. Find constants A and B so that the function below is continuous 

for all x. Graph the resulting function. 

⎧ 

⎨ (x − 1) 2 if −∞ < x < 0 

f (x) = (A − x) 2 if 0 ≤ x < 1 

⎩ 

x + B if 1 ≤ x < ∞ 

85. Find constants A and B so that the function below is continuous 

for all x. Graph the resulting function. 

⎧ 

⎨ x + A if −∞ < x < 4 

f (x) = (x − 1) 2 if 4 ≤ x ≤ 9 

⎩ 

Bx + 1 if 9 < x < ∞ 

82. (a) ⎧ 

⎪ 

⎪ 

Cx ( ) = ⎨ 

⎪ 

⎪ 

⎩ 

9.00 if 0 ≤x 

≤10 

9.00 + 0.95( x− 10) if 10< x < 30 

28.00 + 1.65( x− 30) if 30 < x ≤100 

143.50 + 2.20( x− 100) if x > 100 

(b) { x| x≥ 

0} 

NASA/JPL/DLR 

(c) C is continuous on its domain. 

(d) See (c). 

(e) Answers will vary. Sample answer: There 

is no penalty for going just a little over 10,000 

or 30,000 or 100,000 gallons. 

83. (a) Gm 

R 2 

(b) 1.3 m/s 2 

(c) The gravity on Europa is less than the 

gravity on Earth. 


114 



11/01/17 9:58 am



86. For the function f below, find k so that f is continuous at 2. 

⎧ √ √ 

⎨ 2x + 5 − x + 7 

if x ≥− 5 f (x) = x − 2 

2 , x = 2 

⎩ 

k if x = 2 

x 2 − 6x − 16 

87. Suppose f (x) = 

. 

(x 2 − 7x − 8) x 2 − 4 

(a) For what numbers x is f defined? 

(b) For what numbers x is f discontinuous? 

(c) Which discontinuities found in (b) are removable? 

88. Intermediate Value Theorem 

(a) Use the Intermediate Value Theorem to show that the 

function f (x) = sin x + x − 3 has a zero in the 

interval [0,π]. 

(b) Approximate the zero rounded to three decimal places. 



function f (x) = e x + x − 2 has a zero in the interval [0, 2]. 

(b) Approximate the zero rounded to three decimal places. 

In Problems 90–93, verify that each function intersects the given line in 

the indicated interval. Then use the Intermediate Value Theorem to 

approximate the point of intersection correct to three decimal places by 

repeatedly subdividing the interval into 10 subintervals. 

90. f (x) = x 3 − 2x 2 − 1; line: y =−1; interval: (1, 4) 

91. g(x) =−x 4 + 3x 2 + 3; line: y = 3; interval: (1, 2) 

92. h(x) = x3 − 5 

x 2 ; line: y = 1; interval: (1, 3) 

+ 1 

93. r(x) = x − 6 

x 2 ; line: y =−1; interval: (0, 3) 

+ 2 

94. Graph a function that is continuous on the closed interval [5, 12], 

that is negative at both endpoints and has exactly three distinct 

zeros in this interval. Does this contradict the Intermediate Value 

Theorem? Explain. 

95. Graph a function that is continuous on the closed interval [−1, 2], 

that is positive at both endpoints and has exactly two zeros in this 

interval. Does this contradict the Intermediate Value Theorem? 

Explain. 


is positive at −2 and negative at 3 and has exactly two zeros in 

this interval. Is this possible? Does this contradict the 

Intermediate Value Theorem? Explain. 


is negative at −5 and positive at 0 and has exactly three zeros in 

the interval. Is this possible? Does this contradict the Intermediate 

Value Theorem? Explain. 

98. (a) Explain why the Intermediate Value Theorem gives no 

information about the zeros of the function f (x) = x 4 − 1 on 


(b) Use technology to determine whether or not f has a zero on 


99. (a) Explain why the Intermediate Value Theorem gives no 

information about the zeros of the function 

f (x) = ln(x 2 + 2) on the interval [−2, 2]. 

84. A = 1 and B =−1 

or A =− 1 and B = 3 

y 

6 

4 

2 

22 21 1 2 3 

y 

6 

4 

2 

22 21 1 2 3 

x 

x 

(b) Use technology to determine whether or not f has a zero on 




functions y = x 3 and y = 1 − x 2 intersect somewhere 

between x = 0 and x = 1. 

(b) Use technology to find the coordinates of the point of 

intersection rounded to three decimal places. 

(c) Use technology to graph both functions on the same set of 

axes. Be sure the graph shows the point of intersection. 

101. Intermediate Value Theorem An airplane is travelling at a 

speed of 620 miles per hour and then encounters a slight 

headwind that slows it to 608 miles per hour. After a few 

minutes, the headwind eases and the plane’s speed increases to 

614 miles per hour. Explain why the plane’s speed is 610 miles 

per hour on at least two different occasions during the flight. 


102. Suppose a function f is defined and continuous on the closed 

interval [a, b]. Is the function h(x) = 1 also continuous on 

f (x) 

the closed interval [a, b]? Discuss the continuity of h 

on [a, b]. 

103. Given the two functions f and h: 

f (x) 

f (x) = x 3 − 3x 2 if x = 3 

− 4x + 12 h(x) = x − 3 

p if x = 3 

(a) Find all the zeros of the function f . 

(b) Find the number p so that the function h is continuous at 

x = 3. Justify your answer. 

(c) Determine whether h, with the number found in (b), is even, 

odd, or neither. Justify your answer. 

104. The function f (x) = |x| is not defined at 0. Explain why it is 

x 

impossible to define f (0) so that f is continuous at 0. 

f 

105. Find two functions f and g that are each continuous at c, yet 

g 

is not continuous at c. 

106. Discuss the difference between a discontinuity that is removable 

and one that is nonremovable. Give an example of each. 

Bisection Method for Approximating Zeros of a Function Suppose 

the Intermediate Value Theorem indicates that a function f has a zero 

in the interval (a, b). The bisection method approximates the zero by 

evaluating f at the midpoint m 1 of the interval (a, b). If f (m 1) = 0, 

then m 1 is the zero we seek and the process ends. If f (m 1) = 0, then 

the sign of f (m 1) is opposite that of either f (a) or f (b) (but not both), 

and the zero lies in that subinterval. Evaluate f at the midpoint m 2 of 

this subinterval. Continue bisecting the subinterval containing the zero 

until the desired degree of accuracy is obtained. 

In Problems 107–114, use the bisection method three times to 

approximate the zero of each function in the given interval. 

107. f (x) = x 3 + 3x − 5; interval: [1, 2] 

108. f (x) = x 3 − 4x + 2; interval: [1, 2] 

109. f (x) = 2x 3 + 3x 2 + 4x − 1; interval: [0, 1] 

110. f (x) = x 3 − x 2 − 2x + 1; interval: [0, 1] 

111. f (x) = x 3 − 6x − 12; interval: [3, 4] 

85. A = 5 , B = 7 

80 

60 

40 

20 

y 

5 5 10 x 

86. 

1 

k = 

6 

87. (a) { x| x< 2} ∪ { x| x> 2, x ≠8} 

(b) f is discontinuous at x = 8 and on the 

interval [ − 2,2] . 

(c) The discontinuity at x = 8 is removable. 

88. (a) Since f is continuous on [0, π ], f (0) < 0, 

and f ( π ) > 0, the IVT guarantees that f has a 

zero on the interval (0, π ). 

(b) x ≈ 2.180 

89. (a) Since f is continuous 

on [0,2], f (0) < 0, and f (2) > 0, the 

IVT guarantees that f must have a zero 

on the interval (0,2). 

(b) x ≈ 0.443 

90. 2 

91. 1.732 

92. 2.219 

93. 1.561 

94. Graphs will vary. Sample graph: 

y 

4 

2 

22 

24 

2 4 6 8 10 12 

Does not contradict IVT because IVT 

provides no information when endpoint 

values have the same sign. 


y 

3 

2 

1 

22 21 

21 

1 2 

Does not contradict IVT. f ( −1) and f (2) are 

both positive, so IVT gives no information 

about zeros in the interval ( −1,2). 


y 

10 

5 

23 22 21 

25 

210 

215 

220 

1 2 3 

Does not contradict IVT, which guarantees 

that f must have at least 1 zero in the 

interval ( −2,3). 


25 24 23 22 

y 

20 

15 

10 

5 

21 1 

25 

210 

Does not contradict IVT, which guarantees 

that f must have at least 1 zero on the 

interval ( −5,0). 

98. (a) Since f ( −2) and f (2) are both 

positive, IVT gives no information 

about whether the function has a zero 

in the interval ( −2,2). 


x 

x 

x 

x 


115 


11/01/17 9:58 am




y 

3 

2 

1 

22 21 

21 

1 2 

The graph indicates that f has zeros 

at x =− 1 and x = 1 in the interval 

( −2,2). 

99. (a) Since f ( −2) and f (2) are both 

positive, there is no guarantee that 

the function has a zero in the interval 

( −2,2). 

(b) 

y 

2 

1 

x 

112. f (x) = 3x 3 + 5x − 40; interval: [2, 3] 

113. f (x) = x 4 − 2x 3 + 21x − 23; interval [1, 2] 

114. f (x) = x 4 − x 3 + x − 2; interval: [1, 2] 

115. Intermediate Value Theorem Use the Intermediate Value 

 

Theorem to show that the function f (x) = x 2 + 4x − 2 has a 

zero in the interval [0, 1]. Then approximate the zero correct to 

one decimal place. 

116. Intermediate Value Theorem Use the Intermediate Value 

Theorem to show that the function f (x) = x 3 − x + 2 has a zero 

in the interval [−2, 0]. Then approximate the zero correct to two 

decimal places. 

117. Continuity of a Sum If f and g are each continuous at c, 

prove that f + g is continuous at c.(Hint: Use the Limit of a 

Sum Property.) 

118. Intermediate Value Theorem Suppose that the functions f 

and g are continuous on the interval [a, b]. If f (a) g(b), prove that the graphs of y = f (x) and 

y = g(x) intersect somewhere between x = a and x = b. 

[Hint: Define h(x) = f (x) − g(x) and show h(x) = 0 for some 

x between a and b.] 


119. Intermediate Value Theorem Let f (x) = 1 

x − 1 + 1 

x − 2 . 

Use the Intermediate Value Theorem to prove that there is a real 

number c between 1 and 2 for which f (c) = 0. 

120. Intermediate Value Theorem Prove that there is a real 

number c between 2.64 and 2.65 for which c 2 = 7. 

f (a + h) − f (a) 

121. Show that the existence of lim 

implies f is 

h→0 h 

continuous at x = a. 

122. Find constants A, B, C, and D so that the function below is 

continuous for all x. Sketch the graph of the resulting function. 

⎧ 

x 2 + x − 2 

if −∞ < x < 1 

x − 1 

⎪⎨ A if x = 1 

f (x) = 

B (x − C) 2 if 1 < x < 4 

D 

⎪⎩ 

if x = 4 

2x − 8 if 4 < x < ∞ 

123. Let f be a function for which 0 ≤ f (x) ≤ 1 for all x in [0, 1]. 

If f is continuous on [0, 1], show that there exists at least one 

number c in [0, 1] such that f (c) = c. 

[Hint: Let g(x) = x − f (x).] 

2 1 1 2 

x 

The graph indicates that f does not 

have a zero in the interval ( −2,2). 

100. If the functions intersect, then any 

point of intersection must be a solution 

3 2 

of fx ( ) = x + x − 1= 

0. This f is 

continuous on [0,1] , with f (0) < 0 and 

f (1) > 0, so IVT guarantees a solution, 

and hence an intersection of the two 

original functions, on (0,1). 

(b) (0.755,0.430) 

(c) 

y 

1.0 

0.8 

0.6 

0.4 

0.2 

101. See TSM. 

(0.755, 0.430) 

0.2 0.4 0.6 0.8 1.0 

102. h is continuous anywhere on 

[ ab , ] except where fx ( ) = 0 . If 

fx ( ) ≠ 0 anywhere on [ ab , ] , then 

h is continuous everywhere on that 

interval. 

103. (a) x =− 2, x = 2, x = 3 

(b) p = 5 

(c) h is an even function. 

104. This is because lim fx ( ) does not 

x → 0 

exist (because lim fx ( ) ≠ lim fx ( )). 

− 

x→0 x→+ 

0 


fx ( ) = x 

2 −1, 

gx ( ) = x − 3, c = 3. 


2 

x −9 

fx ( ) = has a removable 

x −3 

discontinuity at x = 3, and 

⎧ 

⎪ x+ 3, x≤3 

gx ( ) = ⎨ 

has a 

2 

⎩⎪ 9 − x , x> 

3 

nonremovable discontinuity at x = 3. 

x 

AP® Practice Problems 

PAGE 

103 1. The graph of a function f is shown below. Where on the open 

interval (−3, 5) is f discontinuous? 

23 

21 

(A) 3 only (B) −1 and 3 only 

(C) 1 only (D) −1, 2, and 3 

y 

3 

1 

y 5 f (x) 

2 3 5 x 

PAGE 

103 2. The graph of a function f is shown below. 

22 21 

y 

3 

1 

y 5 f (x) 

1 2 3 4 

If lim x→c 

f (x) exists and if f is not continuous at c, then c = 

(A) −1 (B) 1 (C) 2 (D) 3 

107. (1.125,1.25) 108. (1.625,1.75) 

109. (0.125,0.25) 110. (0.375,0.5) 

111. (3.125,3.25) 112. (2.125,2.25) 

113. (1.125,1.25) 114. (1.25,1.375) 

115. Since f is continuous on [0, 1], 

f (0) < 0, and f (1) > 0, IVT guarantees that f 

has a zero on (0, 1). Correct to one decimal 

place, the zero is x = 0.8. 

116. Since f is continuous on [−2, 0], f ( − 2) < 0, 

and f (0) > 0, IVT guarantees that f has a zero 

on ( −2,0). Correct to two decimal places, the 

zero is x =− 1.52. 

117. See TSM. 

118. See TSM. 

119. See TSM. 

120. See TSM. 

x 

PAGE 

104 3. If the function f (x) = x2 − 25 

then f (−5) = 

x + 5 

is continuous at −5, 

(A) −10 (B) −5 (C) 0 (D) 10 

⎧ √ √ 

⎨ 2x + 5 − x + 15 

if x = 10 

PAGE 

105 4. If f (x) = x − 10 

⎩ 

k if x = 10 

and if f is continuous at x = 10, then k = 

1 

(A) 0 (B) (C) 1 (D) 10 

10 

PAGE 

105 5. If lim f (x) = L, where L is a real number, which of the 

x→c 

following must be true? 

(A) f is defined at x = c. (B) f is continuous at x = c. 

(C) f (c) = L. (D) None of the above. 

PAGE 

111 6. If f (x) = x 3 − 2x + 5 and if f (c) = 0 for only one real 

number c, then c is between 

(A) −4 and −2 (B) −2 and −1 (C) −1 and 1 (D) 1 and 3 

PAGE 

111 7. The function f is continuous at all real numbers, and f (−8) = 3 

and f (−1) =−4. If f has only one real zero (root), then which 

number x could satisfy f (x) = 0? 

(A) −10 (B) −5 (C) 0 (D) 2 

121. See TSM. 

122. A = 3, B = 1 , C = 4, D = 0. 

3 

123. See TSM. 

y 

6 

4 

2 

22 2 4 6 

Answers to AP® Practice 

Problems 

1. D 2. C 3. A 4. B 5. D 

6. A 7. B 

x 

116 



11/01/17 9:58 am


Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions 117 

PAGE 

105 8. Let f be the function defined by 

⎧ 

x 

⎪⎨ 

2 if x < 0 

√ x if 0 ≤ x < 1 

f (x) = 

2 − x if 1 ≤ x < 2 

⎪⎩ 

x − 3 if x ≥ 2 

For what numbers x is f NOT continuous? 

(A) 1 only 

(C) 0 and 2 only 

(B) 2 only 

(D) 1 and 2 only 

PAGE 

111 9. The function f is continuous on the closed interval [−2, 6]. 

If f (−2) = 7 and f (6) =−1, then the Intermediate Value 

Theorem guarantees that 

(A) f (0) = 0. 

(B) f (c) = 2 for at least one number c between −2 and 6. 

(C) f (c) = 0 for at least one number c between −1 and 7. 

(D) −1 ≤ f (x) ≤ 7 for all numbers in the closed 

interval [−2, 6]. 

PAGE 

111 10. The function f is continuous on the closed interval [−2, 2]. 

Several values of the function f are given in the table below. 

x −2 0 2 

f (x) 3 c 2 

The equation f (x) = 1 must have at least two solutions in the 

interval [−2, 2] if c = 

(A) 

1 

2 

(B) 1 (C) 3 (D) 4 

PAGE 

105 11. The function f is defined by f (x) = 

 

x 2 − 2x + 3 if x ≤ 1 

−2x + 5 if x > 1 

(a) Is f continuous at x = 1? 

(b) Use the definition of continuity to explain your answer. 

8. B 

9. B 

10. A 

11. (a) No (b) Answers will vary. 

1.4 Limits and Continuity of Trigonometric, 

Exponential, and Logarithmic Functions 


1 Use the Squeeze Theorem to find a limit (p. 117) 

2 Find limits involving trigonometric functions (p. 119) 

3 Determine where the trigonometric functions are continuous (p. 122) 

4 Determine where an exponential or a logarithmic function is continuous (p. 124) 

Until now we have found limits using the basic limits 

lim 

x→c A = A 

lim x = c 

x→c 

and properties of limits. But there are many limit problems that cannot be found by 

directly applying these techniques. To find such limits requires different results, such as 

the Squeeze Theorem ∗ , or basic limits involving trigonometric and exponential functions. 





Teaching Tip 

This section can be taught in conjunction 

with Section 1.2 when the students learn 

how to find limits analytically (that is, 

algebraically). 

y 

L 

c 

y h(x) 

y g(x) 

y f (x) 

lim f(x) L, lim h(x) L, lim g(x) L 


x 

1 Use the Squeeze Theorem to Find a Limit 

To use the Squeeze Theorem to find lim 

x→c 

g(x), we need to know, or be able to find, two 

functions f and h that “sandwich” the function g between them for all x close to c. That 

is, in some interval containing c, the functions f, g, and h satisfy the inequality 

f (x) ≤ g(x) ≤ h(x) 

Then if f and h have the same limit L as x approaches c, the function g is “squeezed” 

to the same limit L as x approaches c. See Figure 38. 

We state the Squeeze Theorem here. The proof is given in Appendix B. 

Figure 38 

∗ The Squeeze Theorem is also known as the Sandwich Theorem and the Pinching Theorem. 

Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions 

117 


11/01/17 9:58 am





for example 1 

Using the Squeeze Theorem to Find a 

Limit 

⎛ ⎞ 

Find lim 

⎝ 

⎜ x cos 1 

x→0 

x ⎠ 

⎟ . 

Solution 

Recall that −1≤cos x ≤ 1. We will use that 

fact as a springboard to write inequalities 

involving the original function. The original 

problem is not simply about cos x, but 

⎛ ⎞ 

about cos 1 ⎝ 

⎜ x ⎠ 

⎟ . However, the cosine of any 

value is still a number between −1 and 1. 

⎛ ⎞ 

−1≤cos 1 ⎝ 

⎜ x ⎠ 

⎟ ≤ 1 

We continue to build inequalities with 

⎛ ⎞ 

x cos 1 ⎝ 

⎜ x ⎠ 

⎟ in the center. Next, we multiply 

all three parts by x. 

− x ⎛ 

≤ x ⎞ 

cos 1 

⎝ 

⎜ x ⎠ 

⎟ ≤ x 

We have now built inequalities with the 

original expression in the center. Now we 

take the limit of all three parts. 

⎛ ⎞ 

lim ( −x) ≤ lim xcos 1 ⎝ 

⎜ 

⎠ 

⎟ ≤ lim ( x) 

x→0 x→0 x x→0 

Finally, use the Squeeze Theorem. 

lim ( − x) = 0 

x→0 

lim ( x) = 0 

x→0 

⎛ ⎞ 

Therefore lim x cos 1 

x→0 

⎝ 

⎜ x ⎠ 

⎟ = 0. 

Teaching Tip 

Often, students do not know how to begin 

a Squeeze Theorem problem. Most of the 

time, we begin with the fact that sin x or 

cos x falls between −1 and 1. Then 

consider guiding the students through 

building the original problem as shown in 

AP ® Calc Skill Builder for Example 1. 

y 

2 

1 

y x 2 

2 1 

1 2 3 

1 

2 

y x 2 

y f (x) 

x 

THEOREM Squeeze Theorem 

Suppose the functions f , g, and h have the property that for all x in an open interval 

containing c, except possibly at c, 

If 

then 

f (x) ≤ g(x) ≤ h(x) 

lim f (x) = lim h(x) = L 

x→c x→c 

lim 

x→c g(x) = L 

For example, suppose we wish to find lim f (x), and we know that −x 2 ≤ f (x) ≤ x 2 

x→0 

for all x = 0. Since lim(−x 2 ) = 0 and lim x 2 = 0, the Squeeze Theorem tells us that 

x→0 x→0 

lim f (x) = 0. Figure 39 illustrates how f is “squeezed” between y = x 2 and y =−x 2 

x→0 

near 0. 

Figure 39 EXAMPLE 1 Using the Squeeze Theorem to Find a Limit 

Use the Squeeze Theorem to find lim 

(x sin 1 ) 

. 

x→0 x 

g 

g 

Solution If x = 0, then g(x) = x sin 1 is defined. We seek two functions that “squeeze” 

x 

g(x) = x sin 1 near 0. Since −1 ≤ sin x ≤ 1 for all x, we begin with the inequality 

x ∣ sin 1 x ∣ ≤ 1 x = 0 

Since x = 0 and we seek to squeeze g(x) = x sin 1 , we multiply both sides of the 

x 

inequality by |x|, x = 0. Since |x| > 0, the direction of the inequality is preserved. [Note 

that if we multiply 

∣ sin 1 x ∣ ≤ 1 by x, we would not know whether the inequality symbol 

would remain the same or be reversed since we do not know whether x > 0 or x < 0.] 

|x| 

∣ sin 1 x ∣ ≤|x| Multiply both sides by |x| > 0. 

∣ x sin 1 x ∣ ≤|x| 

|a|·|b| =|ab|. 

−|x| ≤x sin 1 ≤|x| |a|≤b is equivalent to −b ≤ a ≤ b. 

x 

Now use the Squeeze Theorem with f (x) = −|x|, g(x) = x sin 1 , and h(x) =|x|. 

x 

Since f (x) ≤ g(x) ≤ h(x) and 

lim 

x→0 

it follows that 

f (x) = lim 

x→0 

(−|x|) = 0 and lim 

x→0 

h(x) = lim |x| =0 

x→0 

( 

lim g(x) = lim x · sin 1 ) 

= 0 ■ 

x→0 x→0 x 

2 

p, 

p 

4 4 3 f 20.6, 0.6g 

Figure 40 g(x) = x sin 1 is squeezed 

x 

Figure 40 illustrates how g(x) = x sin 1 is squeezed between y = −|x| and y =|x|. 

x 

between f (x) = −|x| and h(x) =|x|. NOW WORK Problem 5 and AP® Practice Problem 11. 


Section 1.4 




118 


TE_Sullivan_Chapter01_PART II.indd 1 

11/01/17 9:55 am



2 Find Limits Involving Trigonometric Functions 

Knowing lim A = A and lim x = c helped us to find the limits of many algebraic 

x→c x→c 

functions. Knowing several basic trigonometric limits can help to find many limits 

involving trigonometric functions. 

THEOREM Two Basic Trigonometric Limits 

lim 

x→0 

lim sin x = 0 

x→0 

lim cos x = 1 

x→0 

The graphs of y = sin x and y = cos x in Figure 41 suggest that lim sin x = 0 and 

x→0 

cos x = 1. The proofs of these limits both use the Squeeze Theorem. Problem 63 

provides an outline of the proof that lim 

x→0 

sin x = 0. 

sin θ 

A third basic trigonometric limit, lim = 1, is important in calculus. In 

θ→0 θ 

sin θ 

Section 1.1, a table suggested that lim = 1. The function f (θ) = sin θ , whose 

θ→0 θ 

θ 

graph is shown in Figure 42, is defined for all real numbers θ = 0. The graph suggests 

lim 

θ→0 

sin θ 

θ 

= 1. 

AP® Exam Tip 

Students should memorize this limit to 

save time on the exam. 

ax 

lim sin( ) = a 

x→0 

x 

Consider working with variations of this 

problem as well, such as 

ax a 

lim sin( ) = . 

x→0 

bx b 

These limits can be found using the 

Squeeze Theorem, but memorization 

will be a big time saver. 

y 

y 

y 

1.0 

1 

y sin x 

1 

y cos x 

0.5 

2π 3π 

π 

2 

π 

 

2 

π 

2 

π 

3π 

2 

2π x 

2π 

π 

π 

2π 

x 

2π π π 2π θ 

1 

1 

0.5 

Figure 41 

Figure 42 f (θ) = sin θ 

θ 

THEOREM 

If θ is measured in radians, then 

sin θ 

lim = 1 

θ→0 θ 

Proof Although sin θ 

sin θ 

is a quotient, we have no way to divide out θ. To find lim , 

θ 

θ→0 + θ 

we let θ be a positive acute central angle of a unit circle, as shown in Figure 43(a) on 

page 120. Notice that COP is a sector of the circle. We add the point B = (cos θ,0) to 

the graph and form triangle BOP. Next we extend the terminal side of angle θ until it 

intersects the line x = 1 at the point D, forming a second triangle COD. The x-coordinate 

of D is 1. Since the length of the line segment OC is OC = 1, then the y-coordinate of 

D is CD = CD = tan θ. So, D = (1, tan θ). See Figure 43(b). 

OC 


119 


11/01/17 9:55 am




x 2 y 2 1 

y 

1 

P (cos θ, sin θ) 

x 2 y 2 1 

y 

1 

D (1, tan θ) 

P 

1 

O 

θ 

C (1, 0) 

1 x 

1 

θ 

O 

B (cos θ, 0) 

1 

C 

x 

1 

1 

Figure 43 

π 

(a) 0 θ 

2 

(b) 

We see from Figure 43(b) that 

area of triangle BOP ≤ area of sector COP ≤ area of triangle COD (1) 

Each of these areas can be expressed in terms of θ as 

• area of triangle BOP = 1 2 cos θ · sin θ area of a triangle = 1 base × height 

2 

• area of sector COP = θ 2 

• area of triangle COD = 1 tan θ 

· 1 · tan θ = 

2 2 

area of a sector = 1 2 r 2 θ; r = 1 

area of a triangle = 1 base × height 

2 

So, for 0



Figure 44 

So, 

sin θ sin (−θ) sin θ 

lim = lim = lim = 1 

θ→0 − θ θ→0 + −θ θ→0 + θ 

sin θ 

It follows that lim = 1. ■ 

θ→0 θ 

sin θ 

Since lim = 1, the ratio sin θ is close to 1 for values of θ close to 0. That is, 

θ→0 θ 

θ 

sin θ ≈ θ for values of θ close to 0. Figure 44 illustrates this property. 

sin θ 

The basic limit lim = 1 can be used to find the limits of similar expressions. 

θ→0 θ 

EXAMPLE 2 

Find: 

(a) lim 

θ→0 

sin(3θ) 

θ 

Finding the Limit of a Trigonometric Function 

(b) lim 

θ→0 

sin(5θ) 

sin(2θ) 

sin(3θ) 

sin θ 

Solution (a) Since lim is not in the same form as lim , we multiply the 

θ→0 θ 

θ→0 θ 

numerator and the denominator by 3, and make the substitution t = 3θ. 

( 

sin(3θ) 3 sin(3θ) 

lim = lim 

lim 3 sin t ) 

sin t 

= 3 lim = (3)(1) = 3 

θ→0 θ θ→0 3θ 

t 

t→0 t ↑ 

= 

↑ t→0 

t = 3θ 

t → 0 as θ → 0 

sin t 

lim 

t→0 t 

(b) We begin by dividing the numerator and the denominator by θ. Then 

sin(5θ) 

sin(5θ) 

sin(2θ) = θ 

sin(2θ) 

θ 

Now we follow the approach in (a) on the numerator and on the denominator. 

sin(5θ) 

lim = lim 

θ→0 θ 

sin(2θ) 

lim = lim 

θ→0 θ 

5 sin(5θ) 

θ→0 5 θ 

2 sin(2θ) 

θ→0 2θ 

lim 

= 

↑ t→0 

t = 5θ 

t → 0 as θ → 0 

= 

↑ t→0 

t = 2θ 

t → 0 as θ → 0 

5 sin t 

t 

( 2 sin t 

lim 

t 

= 1 

( ) sin t 

= 5 lim = 5 

t→0 t 

) 

sin t 

= 2 lim = 2 

t→0 t 

sin(5θ) 

sin(5θ) 

lim 

lim 

θ→0 sin(2θ) = θ→0 θ 

= 5 sin(2θ) 2 ■ 

lim 

θ→0 θ 

NOW WORK Problems 23 and 25 and AP® Practice Problems 1, 2, 4, 6, 7, and 10. 


for example 3 

Finding the Limit of a Trigonometric 

Function 

⎛ (cot x− 

csc x)sinx 

⎞ 

Find lim 

x→0⎝ 

⎜ x ⎠ 

⎟ 

Solution 

To find this limit, we first rewrite the 

trigonometric expressions in terms of 

sines and cosines. For x ≠ 0, 

(cot x− 

csc x)sinx 

x 

⎛ cos x 1 ⎞ 

− 

⎝ 

⎜ x x⎠ 

⎟ sin x 

sin sin 

= 

x 

cosx 

−1 

= 

x 

Since we know 

x − 

lim cos 1 = 0, then 

x→0 

x 

x− 

x x 

lim (cot csc )sin = 0. 

x→0 

x 

Example 3 establishes an important limit used in Chapter 2. 

EXAMPLE 3 


for example 2 

Establish the formula 


Function 

x 

Find lim sin(3 ) 

xcos(4 x) . 

x→0 

where θ is measured in radians. 

Solution 

x 

x 

lim sin(3 ) lim sin(3 ) 1 

= ⋅lim 

x→0 xcos(4 x) 

x→0 x x→0 

cos(4 x) 

x 

= lim 3sin(3 ) ⋅limsec(4 x) 

x→0 3x 

x→0 

x 

= 3 lim sin(3 ) ⋅sec(0) 

x→0 

3x 

= 311 ⋅ ⋅ = 3 

Finding a Basic Trigonometric Limit 

cos θ − 1 

lim = 0 

θ→0 θ 


for example 3 


Function 

x x+ 

x 

Find lim 2 csc 2 

csc . 

x→0 

2 

xcsc 

x 

Solution 

x x+ 

x 

lim 2 csc 2 

csc 

x→0 

2 

xcsc 

x 

⎛ 1 ⎞ 

= lim + 

→ ⎝ 

⎜ 2 

x 0 xcsc 

x⎠ 

⎟ 

⎛ sin x ⎞ 

= lim + 

⎝ 

⎜ 2 

⎠ 

⎟ = 2 + 1 = 3 

x→0 

x 


121 


11/01/17 9:55 am

Teaching Tip 



Function 

x x+ 

x 

Find lim 2 csc 2 

csc . 

x→0 

2 

xcsc 

x 

Solution 

This time, we rewrite in terms of sines: 

x x+ 

x 

lim 2 csc 2 

csc 

x→0 

2 

xcsc 

x 

⎛ 2x 

1 ⎞ 

⎜ 2 ⎟ 

= lim sin x 

⎜ + sin x 

⎟ 

x→0 

⎜ 

x x 

⎟ 

⎝ 

2 2 

sin x sin x ⎠ 

⎛ x x x ⎞ 

= lim ⎜ 

2 sin 2 2 

sin 

+ 

x→0 

2 ⎟ 

⎝ xsin 

x xsin 

x⎠ 

⎛ sin x ⎞ 

= lim + 

⎝ 

⎜ 2 

⎠ 

⎟ = 2 + 1 = 3 

x→0 

x 

Chapter objective 3: Determine where the 

trigonometric functions are continuous is a 

good review. This knowledge is a calculus 

prerequisite, so if the students are well 

prepared, you may be able to address this 

objective briefly to save time. 


for example 4 


Function 



2π π π π 

 

π 

2 2 

y 

1 

1 

y 

1 

1 

y cos x 

y sin x 

3π 

2 

2π π π π π 3π 

 

2π x 

2 2 

2 

Figure 45 

x 


Solution First we rewrite the expression cos θ − 1 as the product of two terms whose 

θ 

limits are known. For θ = 0, 

( )( ) 

cos θ − 1 cos θ − 1 cos θ + 1 

= 

θ 

θ cos θ + 1 

= cos2 θ − 1 

θ(cos θ + 1) = − sin 2 ( ) 

θ sin θ (− sin θ) 

↑ θ(cos θ + 1) = θ cos θ + 1 

sin 2 θ+ cos 2 θ = 1 

Now we find the limit. 

[( )( )] [ ][ 

] 

cos θ − 1 sin θ (− sin θ) 

sin θ − sin θ 

lim = lim 

= lim lim 

θ→0 θ θ→0 θ cos θ + 1 θ→0 θ θ→0 cos θ + 1 

lim(− sin θ) 

θ→0 

= 1 · 

lim (cos θ + 1) = 0 2 = 0 

■ 

θ→0 


3 Determine Where the Trigonometric Functions 

Are Continuous 

The graphs of f (x) = sin x and g(x) = cos x in Figure 45 suggest that f and g are 

continuous on their domains, the set of all real numbers. 

EXAMPLE 4 Showing f(x) = sin x Is Continuous at 0 

• f (0) = sin 0 = 0, so f is defined at 0. 

• lim f (x) = lim sin x = 0, so the limit at 0 exists. 

x→0 x→0 

• lim sin x = sin 0 = 0. 

x→0 

Since all three conditions of continuity are satisfied, f (x) = sin x is continuous at 0. ■ 

In a similar way, we can show that g(x) = cos x is continuous at 0. That is, 

lim cos x = cos 0 = 1. 

x→0 

You are asked to prove the following theorem in Problem 67. 

THEOREM 

• The sine function y = sin x is continuous on its domain, all real numbers. 

• The cosine function y = cos x is continuous on its domain, all real numbers. 

Based on this theorem the following two limits can be added to the list of basic 

limits. 

lim sin x = sin c for all real numbers c 

x→c 

lim cos x = cos c for all real numbers c 

x→c 

Following are some function values for two 

continuous functions, f and g. 

x fx ( ) gx ( ) 

1 2 −1 

2 1 0 

sin( gx ( )) + fx ( ) 

Find lim . 

x→2 

gf (( x)) 

Solution 

Here we will use the continuity of f, g, and 

the sine function to find the limit. 

sin( gx ( )) + fx ( ) 

lim 

x→2 

gf (( x)) 



Theorems 

Use the continuity of a function to find a limit. 

Ask students to use the definition of continuity 

to explain why lim cosx 

=−1. 

Have students 

x→π 

explain by using a graph how the continuity 

of a function f at a point x = c shows that 

lim f( x) = f( c). 

x→c 

sin( g(2)) + f(2) 

= lim 

x→2 

gf ((2)) 

+ + 

= lim sin(0) 1 0 1 = 

− =− 1 

x→2 

g(1) 

1 

122 



11/01/17 9:55 am



Using the facts that the sine and cosine functions are continuous for all real numbers, 

we can use basic trigonometric identities to determine where the remaining four 

trigonometric functions are continuous: 

• y = tan x: Since tan x = sin x , from the Continuity of a Quotient property, 

cos x 

y = tan x is continuous at all real numbers except those for which cos x = 0. 

That is, y = tan x is continuous on its domain, all real numbers except odd 

multiples of π 2 . 

• y = sec x: Since sec x = 1 , from the Continuity of a Quotient property, 

cos x 

y = sec x is continuous at all real numbers except those for which cos x = 0. 

That is, y = sec x is continuous on its domain, all real numbers except odd 

multiples of π 2 . 

• y = cot x: Since cot x = cos x , from the Continuity of a Quotient property, 

sin x 

y = cot x is continuous at all real numbers except those for which sin x = 0. 

That is, y = cot x is continuous on its domain, all real numbers except integer 

multiples of π. 

• y = csc x: Since csc x = 1 , from the Continuity of a Quotient property, 

sin x 

y = csc x is continuous at all real numbers except those for which sin x = 0. 

That is, y = csc x is continuous on its domain, all real numbers except integer 

multiples of π. 

Teaching Tip 

The abbreviation of the function as “cos” 

is derived from the word “cosine,” which is 

also an abbreviation—of the longer name 

“complement of sine.” In a right triangle, 

the cosine of an angle is the same as the 

sine of the complementary angle. The 

same explanation holds for cotangent and 

cosecant. The cotangent of an angle is the 

same as the tangent of the complementary 

angle, and the cosecant of an angle is the 

same as the secant of the complementary 

angle. 

NEED TO REVIEW? Inverse 

trigonometric functions are discussed in 

Section P.7, pp. 61--66. 

Recall that a one-to-one function that is continuous on its domain has an inverse 

function that is continuous on its domain. 

Since each of the six trigonometric functions is continuous on its domain, then each 

is continuous on the restricted domain used to define its inverse trigonometric function. 

This means the inverse trigonometric functions are continuous on their domains. These 

results are summarized in Table 10. 

TABLE 10 Continuity of the Trigonometric Functions and Their Inverses 

Function Domain Properties 

Sine all real numbers continuous on the interval (−∞, ∞) 

Cosine all real numbers continuous on the interval (−∞, ∞) 

{ 

Tangent 

x|x = odd multiples of π } 

continuous at all real numbers except odd multiples of π 2 

2 

Cosecant {x|x = multiples of π} continuous at all real numbers except multiples of π 

{ 

Secant 

x|x = odd multiples of π } 

continuous at all real numbers except odd multiples of π 2 

2 

Cotangent {x|x = multiples of π} continuous at all real numbers except multiples of π 

Inverse sine −1 ≤ x ≤ 1 continuous on the closed interval [−1, 1] 

Inverse cosine −1 ≤ x ≤ 1 continuous on the closed interval [−1, 1] 

Inverse tangent all real numbers continuous on the interval (−∞, ∞) 

Inverse cosecant |x| ≥1 continuous on the set (−∞, −1] ∪ [1, ∞) 

Inverse secant |x| ≥1 continuous on the set (−∞, −1] ∪ [1, ∞) 

Inverse cotangent all real numbers continuous on the interval (−∞, ∞) 


123 


11/01/17 9:55 am




Teaching Tip 

Section objective 4: Determine where 

an exponential or a logarithmic function 

is continuous is a good review. This 

knowledge is a calculus prerequisite, so if 

the students are well prepared, you may 

be able to address this objective briefly to 

save time. 

4 Determine Where an Exponential or a Logarithmic 

Function Is Continuous 

The graphs of an exponential function y = a x and its inverse function y = log a x are 

shown in Figure 46. The graphs suggest that an exponential function and a logarithmic 

function are continuous on their domains. We state the following theorem without 

proof. 

y a x 

1 

(1, ) 

a 

y 

3 

(0, 1) 

(a, 1) 

(1, a) 

y x 

y y a x 

3 

(1, a) 

(0, 1) 1 

(1, ) 

a 

(a, 1) 

(1, 0) 

y x 

y log a x 

Students should be familiar with the 

graphs of y = e x and y = ln x, so that 

they can easily evaluate the following 

limits: 

x 

lim e = 0 

x→−∞ 

x 

lim e = 1 

x→0 

x 

lim e 

x→∞ 

=∞ 

lim lnx 

=−∞ 

+ 

x→0 

lim lnx 

= 0 

x→1 

lim lnx 

=∞ 

x→∞ 

AP® Exam Tip 


for example 5 

Show a Composite Function Is 

Continuous 

Which of the following functions is (are) 

continuous at x = 0? 

2 

x /2 

I. fx ( ) = e 

II. gx ( ) = 

3 cos x 

III. hx ( ) = e −2x 

tanx 

Solution 

2 

x /2 

I. fx ( ) = e is continuous for all real 

numbers, so the function is 

continuous at x = 0. 

II. gx ( ) = 3 cos x is continuous on its 

domain, the set of all real numbers. 

Since the composition of two 

continuous functions is a continuous 

function, g is continuous at x = 0. 

III. The function hx ( ) = e −2x 

tanx 

is 

defined and continuous on 

π π 

− < x < , so h is continuous at 

2 2 

x = 0. 

CALC 

CLIP 

3 (1, 0) 

3 x 

Figure 46 

3 

(a) 0 a 1 

1 

( , 1) 

a 

y log a x 

3 3 

3 

(b) a 1 

1 

( , 1) 

a 

THEOREM Continuity of Exponential and Logarithmic Functions 

• An exponential function is continuous on its domain, all real numbers. 

• A logarithmic function is continuous on its domain, all positive real numbers. 

Based on this theorem, the following two limits can be added to the list of basic 

limits. 

lim 

x→c ax = a c for all real numbers c, a > 0, a = 1 

and 

lim log 

x→c 

a x = log a c for any real number c > 0, a > 0, a = 1 

EXAMPLE 5 

Show that: 

Showing a Composite Function Is Continuous 

(a) f (x) = e 2x is continuous for all real numbers. 

(b) F(x) = 3√ ln x is continuous for x > 0. 

Solution (a) The domain of the exponential function is the set of all real numbers, so 

f is defined for any number c. That is, f (c) = e 2c . Also for any number c, 

[ 


x→c x→c e2x = lim(e x ) 2 = lim ex] 2 

= (e c ) 2 = e 2c = f (c) 

x→c x→c 

Since lim f (x) = f (c) for any number c, then f is continuous at all numbers c. 

x→c 

(b) The logarithmic function f (x) = ln x is continuous on its domain, the set of 

all positive real numbers. The function g(x) = 3√ x is continuous on its domain, the 

set of all real numbers. Then for any real number c > 0, the composite function 

F(x) = (g ◦ f )(x) = 3√ ln x is continuous at c. That is, F is continuous at all real 

numbers x > 0. ■ 

x 

124 



11/01/17 9:55 am



Figures 47(a) and 47(b) illustrate the graphs of f and F. 

y 

8 

4 

y 

1 

1 

2 

4 

x 



AB: 9–45 odd, AP ® Practice Problems 

BC: 11, 23, 25, 35, 43, 45, 55, all AP ® 


Summary Basic Limits 

• lim 

x→c 

A = A, where A is a constant 

• lim sin x = sin c • lim cos x = cos c 

x→c x→c 

• lim a x = a c , a > 0, a = 1 

x→c 



1. lim x→0 

sin x = 

Figure 47 

• lim 

x→c 

x = c 

2 2 

(a) f (x) e 2x 

sin θ 

cos θ − 1 

• lim = 1 • lim = 0 

θ→0 θ 

θ→0 θ 

• lim 

x→c 

log a x = log a c, c > 0, a > 0, a = 1 

cos x − 1 

2. True or False lim = 1 

x→0 x 

3. The Squeeze Theorem states that if the functions f, g, and h 

have the property f (x) ≤ g(x) ≤ h(x) for all x in 

an open interval containing c, except possibly at c, and 

if lim f (x) = lim h(x) = L, then lim g(x) = . 


4. True or False f (x) = csc x is continuous for all real numbers 

except x = 0. 


In Problems 5–8, use the Squeeze Theorem to find each limit. 

PAGE 

118 5. Suppose −x 2 + 1 ≤ g(x) ≤ x 2 + 1 for all x in an open 

interval containing 0. Find lim g(x). 

x→0 

6. Suppose −(x − 2) 2 − 3 ≤ g(x) ≤ (x − 2) 2 − 3 for all x 

in an open interval containing 2. Find lim x→2 

g(x). 

7. Suppose cos x ≤ g(x) ≤ 1 for all x in an open interval 

containing 0. Find lim x→0 

g(x). 

8. Suppose −x 2 + 1 ≤ g(x) ≤ sec x for all x in an open interval 

containing 0. Find lim x→0 

g(x). 


9. lim x→0 

(x 3 + sin x) 10. lim x→0 

(x 2 − cos x) 

x 

2 

3 

(b) F(x) ln x 

NOW WORK Problem 45 and AP® Practice Problems 3, 5, and 9. 

11. lim (cos x + sin x) 12. lim (sin x − cos x) 

x→π/3 x→π/3 

13. 

cos x 

sin x 

lim 

14. lim 

x→0 1 + sin x 

x→0 1 + cos x 

15. 

3 

e x − 1 

lim x→0 1 + e x 16. lim x→0 1 + e x 

17. lim(e x sin x) 18. lim(e −x tan x) 

x→0 x→0 

( ) e x 

( x 

) 

19. lim ln 

20. lim ln 

x→1 x 

x→1 e x 

e 2x 

1 − e x 

21. lim x→0 1 + e x 22. lim x→0 1 − e 2x 


PAGE 

121 23. 

sin(7x) 

lim x→0 x 

PAGE 

121 25. 

θ + 3 sin θ 

lim θ→0 2θ 

27. 

sin θ 

lim θ→0 θ + tan θ 

29. 

5 

lim θ→0 θ · csc θ 

PAGE 

122 31. 

1 − cos 2 θ 

lim θ→0 θ 

24. lim x→0 

sin x 3 

x 

26. 

2x − 5 sin(3x) 

lim x→0 x 

28. 

tan θ 

lim θ→0 θ 

30. 

sin(3θ) 

lim θ→0 sin(2θ) 

[ 

33. lim(θ · cot θ) 34. lim sin θ 

θ→0 θ→0 

cos(4θ)− 1 

32. lim θ→0 2θ 

( )] 

cot θ − csc θ 

θ 



Problems 


Problems 

1. 0 

2. False. 

3. L 

4. False. 

5. 1 

6. −3 

7. 1 

8. 1 

9. 0 

10. −1 

3 1 

11. + 

2 2 

3 1 

12. − 

2 2 

13. 1 

14. 0 

15. 3 2 

16. 0 

17. 0 

18. 0 

19. 1 

20. −1 

21. 1 2 


This worksheet includes 9 limit questions to be 

solved analytically. 

22. 1 2 

23. 7 

24. 1 3 

25. 2 

26. −13 

27. 1 2 

28. 1 

29. 5 

30. 3 2 

31. 0 

32. 0 

33. 1 

34. 0 


125 


11/01/17 9:55 am



35. f is continuous at c = 0. 

36. f is discontinuous at c = 0. 

37. f is continuous at c = π 

4 . 


39. f is continuous on { x| x ≠ 4}. 

40. f is continuous on { x| x ≠ 0} 

41. f is continuous on 

⎧⎪ 

3π 

⎫⎪ 

⎨xx≠ + 2 kπ 

⎬ , k any integer. 

⎩⎪ 2 

⎭⎪ 

42. f is continuous on the set of all real 

numbers. 

43. f is continuous on { x| x > 0, x ≠3}. 


numbers. 


numbers. 


numbers. 

47. 0 

48. 0 

49. 0 

50. 0 

51. See TSM. 

52. See TSM. 

53. See TSM. 

54. See TSM. 

55. See TSM. 

56. See TSM. 

57. See TSM. 

58. f (0) =π 

59. f (0) = π, f (1) = π 

60. Yes. 

61. Yes. 

62. See TSM. 

63. See TSM. 

64. See TSM. 

65. Explanations will vary. For sample 

explanation, see TSM. 

66. Explanations will vary. For sample 

explanation, see TSM. 

67. See TSM. 

68. 0 

69. See TSM. 

70. See TSM. 


In Problems 35–38, determine whether f is continuous at the number c. 

⎧ 

⎨ 3 cos x if x < 0 

35. f (x) = 3 if x = 0 at c = 0 

⎩ 

x + 3 if x > 0 

⎧ 

⎨ cos x if x < 0 

36. f (x) = 0 if x = 0 at c = 0 

⎩ 

e x if x > 0 

⎧ 

⎨ sin θ if θ ≤ π 4 

37. f (θ) = 

⎩ cos θ if θ> π at c = π 4 

4 

tan −1 x if x < 1 

38. f (x) = 

at c = 1 

ln x if x ≥ 1 

In Problems 39–46, determine where f is continuous. 

 

 

x 2 − 4x 

x 2 − 5x + 1 

39. f (x) = sin 

x − 4 

40. f (x) = cos 

2x 

41. 

1 

1 

f (θ) = 

42. f (θ) = 

1 + sin θ 

1 + cos 2 θ 

43. f (x) = ln x 

x − 3 

44. f (x) = ln(x 2 + 1) 

PAGE 

125 45. f (x) = e −x sin x 46. 

e x 

f (x) = 

1 + sin 2 x 


In Problems 47–50, use the Squeeze Theorem to find each limit. 

47. lim 

x 2 sin 1 

 

48. lim x 1 − cos 1 

x→0 x 

x→0 x 

49. 

 

lim x 2 1 − cos 1 

 

1 

50. lim x x→0 x 

+ 3x 2 sin x→0 x 

In Problems 51–54, show that each statement is true. 

sin(ax) 

51. lim x→0 sin(bx) = a cos(ax) 

; b = 0 52. lim 

b x→0 cos(bx) = 1 

sin(ax) 

53. lim = a x→0 bx b ; b = 0 

1 − cos(ax) 

54. lim 

= 0; a = 0, b = 0 

x→0 bx 

1 − cos x 

55. Show that lim x→0 x 2 = 1 2 . 

56. Squeeze Theorem If 0 ≤ f (x) ≤ 1 for every number x, show 

that lim[x 2 f (x)] = 0. 

x→0 

57. Squeeze Theorem If 0 ≤ f (x) ≤ M for every x, show that 

lim 

x→0 [x2 f (x)] = 0. 

sin(π x) 

58. The function f (x) = is not defined at 0. Decide how to 

x 

define f (0) so that f is continuous at 0. 

sin(π x) 

59. Define f (0) and f (1) so that the function f (x) = 

x(1 − x) is 

continuous on the interval [0, 1]. 

⎧ 

⎨ sin x 

if x = 0 

60. Is f (x) = x 

continuous at 0? 

⎩ 

1 if x = 0 

⎧ 

⎨ 1 − cos x 

if x = 0 

61. Is f (x) = x 

continuous at 0? 

⎩ 

0 if x = 0 

1 

62. Squeeze Theorem Show that lim x n sin = 0, where n 

x→0 x 

is a positive integer. (Hint: Look first at Problem 56.) 

63. Prove lim sin θ = 0. 

θ→0 

(Hint: Use a unit circle as 

shown in the figure, first 

assuming 0


Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes 127 


⎧ 

⎨ sin(2x) 

if x = 0 

PAGE 

121 1. The function g(x) = 2x 

⎩ 

k if x = 0 

is continuous at x = 0. What is the value of k? 

1 

(A) 0 (B) (C) 1 (D) 2 

2 

PAGE 

sin(4x) 

121 2. lim x→0 2x 

(A) 0 

= 

(B) 

1 

2 

PAGE 

125 3. The function f (x) = 

(C) 1 (D) 2 

x 3 + 2x 2 if x ≤−2 

e 2x+4 if x > −2 . 

Find lim f (x) if it exists. 

x→−2 

(A) 0 (B) 1 (C) 16 (D) The limit does not exist. 

PAGE 

1 − cos 2 (3x) 

121 4. lim x→0 x 2 = 

(A) 0 (B) 1 (C) 3 (D) 9 

PAGE 

125 5. Which of the following functions are continuous for all real 

numbers x? 

I. f (x) = x 1/3 

II. g(x) = sec x 

III. h(x) = e −x 

(A) I only 

(B) I and II only 

(C) I and III only (D) I, II, and III 

RECALL The symbols ∞ (infinity) 

and −∞ (negative infinity) are not 

numbers. The symbol ∞ expresses 

unboundedness in the positive direction, 

and −∞ expresses unboundedness in 

the negative direction. 

PAGE 

1 

121 6. Find lim if it exists. 

x→0 x csc x 

(A) −1 (B) 0 (C) 1 (D) The limit does not exist. 

 

sin x − π 

PAGE 

121 7. lim 

3 

x→π/3 x − π = 

3 

(A) − π π 

(B) 0 (C) 1 (D) 

3 

3 

PAGE 

1 − cos x 

122 8. lim x→0 3 sin 2 = 

x 

(A) 

1 

6 

PAGE 

125 9. If f (x) = 

(B) 

 

1 

3 

(C) 

1 

2 

(D) 1 

ln x if 0 < x < 3 

, 

(2x − 3) ln 3 if x ≥ 3 

then lim f (x) = 

x→3 

(A) ln 3 (B) 3 (C) ln 9 (D) The limit does not exist. 

PAGE 

tan(2x) 

121 10. lim = 

x→0 3x 

(A) 

1 

3 

PAGE 

118 11. lim 

x 3 sin 1 x→0 x 

1 

(B) 

2 

 

= 

(C) 

2 

3 

(D) 2 

(A) −1 (B) 0 (C) 1 (D) The limit does not exist. 

1.5 Infinite Limits; Limits at Infinity; Asymptotes 


1 Investigate infinite limits (p. 128) 

2 Find the vertical asymptotes of a graph (p. 131) 

3 Investigate limits at infinity (p. 131) 

4 Find the horizontal asymptotes of a graph (p. 137) 

5 Find the asymptotes of the graph of a rational function (p. 138) 

We have described lim f (x) = L by saying if a function f is defined everywhere in 

x→c 

an open interval containing c, except possibly at c, then the value f (x) can be made as 

close as we please to L by choosing numbers x sufficiently close to c. Here c and L are 

real numbers. In this section, we extend the language of limits to allow c to be ∞ or 

−∞ (limits at infinity) and to allow L to be ∞ or −∞ (infinite limits). These limits are 

useful for locating asymptotes that aid in graphing some functions. 

We begin with infinite limits. 

Answers to AP® Practice 

Problems 

1. C 

2. D 

3. D 

4. D 

5. C 

6. C 

7. C 

8. A 

9. D 

10. C 

11. B 

AP® Exam Tip 

Limits at infinity, as well as vertical and 

horizontal asymptotes, are concepts 

that commonly appear on the multiplechoice 

portion of the exam. 

Teaching Tip 

Students may be familiar with finding 

horizontal and vertical asymptotes from 

prior math classes. In this section, they will 

learn the formal definition of a horizontal 

and vertical asymptote and how to use 

limits to find all horizontal and vertical 

asymptotes of a function. 






Section 1.5 




Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes 

127 


11/01/17 9:55 am




Teaching Tip 

1 

Note that when finding the limit of 

x 

2 

as x approaches zero, the authors take 

a tabular and graphical approach. There 

is no algebraic manipulation that can be 

1 

performed to simplify the function , so a 

x 

2 

table or graph is the preferred method for 

this problem. Some students may prefer to 

create a table mentally when a written table 

is not required for the answer. For example, 

since zero cannot be substituted in this 

situation, the students may find it helpful 

to think, What does 1 divided by a really 

small positive number squared equal? Then 

they should think, What does 1 divided by 

a really small negative number squared 

equal? 

In both cases, the answer comes out to 

be an exceptionally large number, which 

grows increasingly large as x gets closer 

and closer to zero. 

Therefore. lim 1 =∞. 

x→0 

2 

x 

Teaching Tip 

It may be helpful for the students to keep 

these two limits in mind. They appear often 

as the students find one-sided limits and 

define vertical asymptotes. 

1 

lim 

=∞ 

x→ 0 

+ very small positive value 

1 

lim 

=−∞ 

x→0 

− very small negative value 

Numerical examples help here as well: 

1/0.1 = 10, 1/0.01 = 100, and so forth. 

For example, 

1 

= 

1 1 

1 

= 

0.1 10 

1 

= 

0.01 100 

1 

= 

0.001 1000 

1 

= 

0.0001 10,000 

y 

4 

1 Investigate Infinite Limits 

Consider the function f (x) = 1 . Table 11 lists values of f for selected numbers x that 

x 

2 

are close to 0 and Figure 48 shows the graph of f . 

Figure 48 f (x) = 1 x 2 As x approaches 0, the value of 1 x increases without bound. Since the value of 1 2 x is 2 

not approaching a single real number, the limit of f (x) as x approaches 0 does not exist. 

TABLE 11 


2 

from the left 

−−−−−−−−−−−−−−−→ 

from the right 

←−−−−−−−−−−−−−−−− 

x −0.01 −0.001 −0.0001 → 0 ← 0.0001 0.001 0.01 

f (x) = 1 10,000 10 6 10 8 f (x) becomes unbounded 10 8 10 6 10,000 

4 2 

2 4 x 

x 2 

x 0 

y 

4 

2 

2 

4 2 4 

4 

Figure 49 f (x) = 1 x 

2 

x 

However, since the numbers are increasing without bound, we describe the behavior of 

the function near zero by writing 

1 

lim 

x→0 x =∞ 2 

and say that f (x) = 1 has an infinite limit as x approaches 0. 

x 

2 

In other words, a function f has an infinite limit at c if f is defined everywhere in 

an open interval containing c, except possibly at c, and f (x) becomes unbounded when 

x is sufficiently close to c. ∗ 

EXAMPLE 1 

1 1 

Investigating lim and lim 

x→0 − x x→0 + x 

Consider the function f (x) = 1 . Table 12 shows values of f for selected numbers x 

x 

that are close to 0 and Figure 49 shows the graph of f . 

TABLE 12 


from the left 

−−−−−−−−−−−−−−−−→ 

from the right 

←−−−−−−−−−−−−−−−−− 

x −0.01 −0.001 −0.0001 → 0 ← 0.0001 0.001 0.01 0.1 

f (x) = 1 x 

−100 −1000 −10,000 f (x) becomes unbounded 10,000 1000 100 10 

Here as x gets closer to 0 from the right, the value of f (x) = 1 can be made as 

x 

large as we please. That is, 1 becomes unbounded in the positive direction. So, 

x 

Similarly, the notation 

1 

lim 

x→0 + x =∞ 

1 

lim 

x→0 − x = −∞ 

is used to indicate that 1 becomes unbounded in the negative direction as x approaches 0 

x 

from the left. ■ 

∗ A precise definition of an infinite limit is given in Section 1.6. 

128 



11/01/17 9:55 am



y 

2 

2 

4 

Figure 51 f (x) = ln x 

y f (x) 

lim 

x→c 

c x 

c x 

x→c x→c 

(a) lim f (x) ∞ (b) lim f(x) ∞ 

Figure 50 

2 4 

x 

So, there are four possible one-sided infinite limits of a function f at c: 

x→c 

x→c 

x→c + 

f (x) =∞, lim f (x) = −∞, lim f (x) =∞, lim 

− − + 

See Figure 50 for illustrations of these possibilities. 

y f (x) 

c 

y f (x) 

(c) lim f (x) ∞ 

x→c 

x 

c 

f (x) = −∞ 

(d) lim f (x) ∞ 

x→c 

y f (x) 

When we know the graph of a function, we can use it to investigate infinite limits. 

EXAMPLE 2 

Investigate lim ln x. 

x→0 + 

Investigating an Infinite Limit 

Solution The domain of f (x) = ln x is {x|x > 0}. Notice that the graph of f (x) = ln x 

in Figure 51 decreases without bound as x approaches 0 from the right. The graph 

suggests that 

lim ln x = −∞ 

x→0 + 

■ 

x 


Based on the graphs of the trigonometric functions in Figure 52, we have the 

following infinite limits: 

lim tan x =∞ lim 

x→π/2− lim csc x = −∞ lim 

x→0− tan x = −∞ lim 

x→π/2 + 

csc x =∞ lim 

x→0 + 

sec x =∞ lim 

x→π/2− cot x = −∞ lim 

x→0− sec x = −∞ 

x→π/2 + 

cot x =∞ 

x→0 + 

Teaching Tip 

Students should have sound knowledge 

of the graphical representations of the 

basic functions. In particular, they should 

be able to quickly sketch and identify the 

asymptotes of the following functions: 

y = tanx 

y = csc x 

y = sec x 

y = cot x 

y = 

y 

e x 

= e −x 

y = lnx 

1 

y = 

x 

Teaching Tip 

Consider presenting the students with the 

following graph: 

y 

y 

5 

5 

π 

x 2 

Figure 52 

y 

y 

y 

y tan x y sec x 

y csc x y cot x 

π 

x 

5 

5 

π 

x 

π 

 

2 

5 

5 

π 

x x 0 

x 0 

2 

π 

2 

Limits of quotients, in which the limit of the numerator is a nonzero number and 

the limit of the denominator is 0, often result in infinite limits. (If the limits of both 

the numerator and the denominator are 0, the quotient is called an indeterminate form. 

Limits of indeterminate forms are discussed in Section 4.5.) 

x 

π 

 

2 

5 

5 

π 

2 

x 

Ask them to draw a function such that 

1. lim fx ( ) =∞ 

x→c 

− 

2. lim fx ( ) =∞ 

x→ c 

+ 

3. lim fx ( ) =−∞ 

x→c 

− 

4. lim fx ( ) =−∞ 

x→ c 

+ 

A few possible answers are at the top of 

page 129 of the student edition. Students 

can show their graphs to students around 

them and troubleshoot misconceptions. 

c 

x 


129 


11/01/17 9:55 am




Teaching Tip 

Ask your students to think through the 

limits of rational functions near vertical 

asymptotes mentally. For Example 3(a), 

the thought process would be: We would 

like to substitute a number in place of x 

that is slightly larger than 3. When we do 

so, the numerator will be essentially 7. The 

denominator will be a very small positive 

number because, for example, 3.00000001 

− 3 = 0.00000001. Therefore, 

x 

lim 2 + 1 

7 

. 

x − 3 

≈ A very small positive number 

=∞ 

→ + 

x 3 

Teaching Tip 

Remind students of the graph of y = ln x 

when working on Example 3b. 

y 

The graph of y = ln x crosses the x-axis at 

x = 1. Therefore, 

and 

2 

1 

21 

22 

lim lnx 

= a small negative number 

x→1 

− 

lim lnx 

= a small positive number. 

x→ 1 

+ 

2 4 6 8 x 

EXAMPLE 3 

Investigate: (a) 


2x + 1 

lim 

x→3 + x − 3 

(b) 

x 2 

lim 

x→1 − ln x 

2x + 1 

Solution (a) lim is the right-hand limit of the quotient of two functions, 

x→3 + x − 3 

f (x) = 2x + 1 and g(x) = x − 3. Since lim 

x→3 +(x 

− 3) = 0, we cannot use the Limit of a 

Quotient. So we need a different strategy. As x approaches 3 + , we have 

lim 

x→3 

+(2x 

+ 1) = 7 and lim 

+(x 

− 3) = 0 

From arithmetic, we know that a fraction p is arbitrarily large if p is a fixed nonzero 

q 

number and q is arbitrarily close to 0. Here, the limit of the numerator is 7. The 

denominator is positive and approaching 0. So the quotient is positive and becoming 

unbounded. We conclude that 

2x + 1 

lim 

x→3 + x − 3 =∞ 

(b) As in (a), we cannot use the Limit of a Quotient because lim ln x = 0. Here we 

x→1− have 

lim x 2 = 1 lim ln x = 0 

x→1 − x→1 − 

We are seeking a left-hand limit, so for numbers close to 1 but less than 1, the denominator 

ln x < 0. So the quotient is negative and becoming unbounded. We conclude that 

x 2 

lim 

x→1 − ln x = −∞ 

■ 

x→3 

The discussion below may prove useful when finding lim 

x→c − 

is a nonzero number and lim g(x) = 0. 

x→c− • lim f (x) = L, L > 0, and lim g(x) = 0 

x→c − 

x→c − 

If g(x) 0, and lim g(x) = 0 

x→c − 

x→c − 

If g(x) >0 for numbers x close to c, but less than c, then 

f (x) 

lim 

x→c − g(x) =∞ 

• lim f (x) = L, L < 0, and lim g(x) = 0 

x→c − 

x→c − 

If g(x) 0 for numbers x close to c, but less than c, then 

f (x) 

lim 

x→c − g(x) = −∞ 

f (x) 

, where lim 

g(x) f (x) 

x→c − 

Similar arguments are valid for right-hand limits. 



for example 3 


⎛ −1 

⎞ 

Find lim 

⎝ 

⎜ − ⎠ 

⎟ . 

x→5 

− x 5 

Solution 

Think x ≈ 4.999999999 

⎛ −1 

⎞ 

⎝ 

⎜ − ⎠ 

⎟ ≈ −1 

lim 

→ − x 5 4.999999999 − 5 

x 5 

−1 

= 

a very small negative number 

As the values of x get closer to 5 from the 

left, the denominator decreases without 

⎛ −1 

⎞ 

bound and lim 

⎝ 

⎜ − ⎠ 

⎟ =∞ . 

x→5 

− x 5 

Graphical Approach 

Solution 

Also, consider how this limit may be 

explained using the graph. Explain that since 

⎛ −1 

⎞ 

fx ( ) = 

⎝ 

⎜ x − 5⎠ 

⎟ is a rational function, its graph 

has asymptotes. Show the graph and observe that 

there is a vertical asymptote at x = 5 because as 

x approaches 5 from the right or from the left, the 

function is infinite. 

y 

4 

2 

22 

24 

2 4 6 8 

Also note that as x approaches infinity, the limit 

approaches zero. 

x 

130 



11/01/17 9:55 am



y 

x c 

y f (x) 

c x 

(a) lim f(x) ∞ 

x→c 

Figure 53 

y 

y f (x) 

x c 

c 

x 

(b) lim f(x) ∞ 

x→c 

2 Find the Vertical Asymptotes of a Graph 

Figure 53 illustrates the possibilities that can occur when a function has an infinite limit 

at c. In each case, notice that the graph of f has a vertical asymptote x = c. 

y 

x c 

y f (x) 

c 

x 

(c) lim f(x) ∞ 

x→c 

lim 

x→c 

y 

x c y x c y x c 

c x 

c x 

c 

y f (x) 

y f (x) 

(d) lim f(x) ∞ 

x→c 

x→c 

x→c 

(e) lim f(x) ∞ 

x→c 

y f (x) 

(f) lim f(x) ∞ 

x→c 

DEFINITION Vertical Asymptote 

The line x = c is a vertical asymptote of the graph of the function f if any of the 

following is true: 

f (x) =∞ lim f (x) =∞ lim f (x) = −∞ lim f (x) = −∞ 

− + − 

x→c + 

For rational functions, a vertical asymptote may occur where the denominator 

equals 0. 

EXAMPLE 4 Finding a Vertical Asymptote 

x 

Find any vertical asymptote(s) of the graph of f (x) = 

(x − 3) . 2 

Solution The domain of f is {x|x = 3}. Since 3 is the only number for which the 

denominator of f equals zero, we construct Table 13 and investigate the one-sided limits 

of f as x approaches 3. Table 13 suggests that 

x 

lim 

x→3 (x − 3) =∞ 2 

So, x = 3 is a vertical asymptote of the graph of f . 

x 


for example 4 

Finding a Vertical Asymptote 

Find any vertical asymptote(s) of the graph 

x 

of fx ( ) = . 

2 

x −1 

Solution 

The domain of f is {x | x ≠ ±1}. Since the 

denominator is zero for x = −1 and x = 1, 

we investigate the one-sided limits at x = 

−1 and x = 1. 

For lim , think x ≈ −1.0000001: 

x→−1 

x− 

− 2 

x 1 

x 

− ≈ −1.0000001 

lim 

2 2 

x 1 ( −1.0000001) −1 

− 

x→−1 

−1 

= 

very small positive 

x 

so lim 

− =−∞ . 

− 2 

x→−1 

x 1 

TABLE 13 

x approaches 3 from the left 


−−−−−−−−−−−−−−−−−−→ 

←−−−−−−−−−−−−−−−−−−− 

x 2.9 2.99 2.999 → 3 ← 3.001 3.01 3.1 

x 

f (x) = 

290 29,900 2,999,000 f (x) becomes unbounded 3,001,000 30,100 310 

(x − 3) 2 

y 

12 

8 

4 

Figure 54 shows the graph of f (x) = 

x 

and its vertical asymptote. 

(x − 3) 

2 

NOW WORK Problems 15 and 63 (find any vertical asymptotes) 

and AP® Practice Problem 5. 

x 

3 Investigate Limits at Infinity 

2 

4 

x 3 

Now we investigate what happens as x becomes unbounded in either the positive direction 

or the negative direction. Suppose as x becomes unbounded, the value of a function f 

x 


approaches some real number L. Then the number L is called the limit of f at infinity. 

(x − 3) 2 

common error 

Sometimes, students think that a vertical 

asymptote exists only if both the left-hand and 

right-hand limits are infinite. Note the precise 

definition of a vertical asymptote. If any of the 

one-sided limits are infinite at x = c, then x = c is a 

vertical asymptote of the graph of the function. It 

is not necessary for both sides to be infinite, and it 

is not necessary for the two sides to be equal. 


This worksheet contains 4 rational functions and 

their corresponding graphs to help the students 

identify the vertical asymptotes of each of the 

functions. 

■ 

For 

lim , think x ≈ −0.9999999: 

2 

x 1 

x 

− ≈ −0.9999999 

lim 

+ 2 2 

x→−1 

x 1 ( −0.9999999) −1 

x→− 1 

x− 

+ 

−1 

= 

very small negative 

x 

so lim 

− =∞ . 

+ 2 

x→−1 

x 1 

The line x = −1 is a vertical asymptote of 

the graph of f because the one-sided limits 

are infinite. 

Note: Once one of these limits is found 

to be infinite, the value x = −1 has been 

shown to be a vertical asymptote. It is not 

necessary to show that both one-sided 

limits are infinite. 

Repeating the same procedure for x = 1, 

we get 

x 

lim 

x − 1 

=−∞ and lim 

2 

x→1 

− 

x→ 1 

+ 

x 

x − 1 

=∞ . 

2 

The line x = 1 is a vertical asymptote of the 

graph of f because the one-sided limits are 

infinite. 


131 


11/01/17 9:55 am




Teaching Tip 

When studying the limit of functions as x 

approaches ∞ or −∞, it helps students to 

keep the following generalizations in mind: 

Infinity in the denominator: 

2 

y 

4 

2 

4 2 4 

2 

x 

For example, the graph of f (x) = 1 in Figure 55 suggests that, as x becomes 

x 

unbounded in either the positive direction or the negative direction, the values f (x) get 

closer to 0. Table 14 illustrates this for a few numbers x. 

TABLE 14 

x ±100 ±1000 ±10,000 ±100,000 x becomes unbounded 

f (x) = 1 x 

±0.01 ±0.001 ±0.0001 ±0.00001 f (x) approaches 0 

1. lim 1 = 0 

x→∞ x 

4 

2. lim 1 = 0 

x→−∞ x 

Infinity in the numerator: 

x 

3. =∞ 

xlim 

→∞ 2 

x 

4. =−∞ 

xlim 

→−∞ 2 

Square root and natural log of infinity: 

5. ∞≈∞ lim x =∞ 

x→∞ 

Figure 55 f (x) = 1 x 

Since f can be made as close as we please to 0 by choosing x sufficiently large, we 

write 

1 

lim 

x→∞ x = 0 

and say that the limit as x approaches infinity of f is equal to 0 . Similarly, as x becomes 

unbounded in the negative direction, the function f (x) = 1 can be made as close as we 

x 

please to 0, and we write 

lim 

x→−∞ 

1 

x = 0 

1 

Although we do not prove lim = 0 here, it can be proved using the ε-δ 

x→−∞ x 

Definition of a Limit at Infinity, and is left as an exercise in Section 1.6. 

The limit properties discussed in Section 1.2 hold for infinite limits. Although these 

properties are stated below for limits as x → ∞, they are also valid for limits as 

x → −∞. 

6. ln( ∞) ≈∞ lim lnx =∞ 

x→∞ 

THEOREM Properties of Limits at Infinity 

If k is a real number, n ≥ 2 is an integer, and the functions f and g approach real 

numbers as x →∞, then the following properties are true: 


Finding Limits at Infinity 

Following are alternate ways to solve 

Example 5: 

a. lim 4 = 4 lim 1 

x→−∞ 

2 

x x→−∞ 

2 

x 

= 4 lim 1 ⋅ lim 1 = 400 ⋅ ⋅ = 0 

x→−∞ 

x x→−∞ 

x 

• lim A = A where A is a constant 

x→∞ 

• lim [kf(x)] = k lim f (x) 

x→∞ x→∞ 

• lim [ f (x) ± g(x)] = lim f (x) ± lim g(x) 

x→∞ x→∞ x→∞ 

[ ][ ] 

• lim [ f (x)g(x)] = lim f (x) lim g(x) 

x→∞ x→∞ x→∞ 

f (x) lim 

• f (x) 

lim 

x→∞ g(x) = x→∞ 

, provided lim g(x) = 0 

lim g(x) x→∞ 

x→∞ 

[ ] n 

• lim [ f 

x→∞ (x)]n = lim f (x) 

x→∞ 

√ √ 

n 

• lim f (x) = n lim f (x), where f (x) >0 if n is even 

x→∞ 

x→∞ 

b. ⎛ 10 ⎞ 

⎜ − ⎟ 

⎝ ⎠ 

=− ⎛ 

lim 

10 lim 

x→∞ 

x→∞⎜ 

x ⎝ 

=−10⋅ 0= 

0 

1 ⎞ 

⎟ 

x ⎠ 

EXAMPLE 5 

Find: 

(a) 

lim 

x→−∞ 


( 

4 

(b) lim − 10 ) 

√ 

x 2 x→∞ x 

(c) 

( 

lim 2 + 3 ) 

x→∞ x 

⎛ 3 ⎞ 

c. + 

⎝ 

⎜ 

⎠ 

⎟ = + ⎛ ⎞ 

lim 2 lim 2 3 lim 1 

x→∞ x x→∞ x→∞⎝ 

⎜ x⎠ 

⎟ 

⎛ ⎞ 

→∞ 

= 2+ 

3 lim lim 1 

x 

⎝ 

⎜ 

⎠ 

⎟ = 2 + 3 ⋅ 0 = 2 

x→∞ 

lim x 

x→∞ 

132 



11/01/17 9:55 am



Solution We use the properties of limits at infinity. 

 

4 

1 2 

(a) lim 

x→−∞ x = 4 lim 

1 2 

= 4 lim = 4 · 0 = 0 

2 x→−∞ x 

x→−∞ x ↑ 

1 

lim 

x→−∞ x = 0 

 

(b) lim − 10 

1 

1 

√ =−10 lim √ =−10 · lim 

x→∞ x x→∞ x x→∞ x 

 

1 

=−10 · lim 

x→∞ x =−10 ↑ 

· 0 = 0 

1 

lim 

x→∞ x = 0 

 

(c) lim 2 + 3 

= lim 

x→∞ x 

2 + lim 3 

x→∞ x→∞ x = 2 + 3 · lim 1 

x→∞ x = 2 + 3 · 0 = 2 

EXAMPLE 6 

Find: 

(a) 

lim 

3x 2 − 2x + 8 

x→∞ x 2 + 1 


(b) 

lim 

4x 2 − 5x 

x→−∞ x 3 + 1 

■ 


Solution (a) We find this limit by dividing each term of the numerator and the 

denominator by the term with the highest power of x that appears in the denominator, in 

this case, x 2 . Then 

3x 2 − 2x + 8 

lim 

x→∞ x 2 + 1 

(b) 

= 

lim 

x→∞ 

⏐ 

Divide the numerator and 

denominator by x 2 

3x 2 − 2x + 8 

x 2 

x 2 + 1 

x 2 

3 − 2 

= lim 

x + 8 x 2 

x→∞ 

1 + 1 = ⏐⏐⏐ 

x 2 

Limit of a 

Quotient 

lim 

3 − 2 

x→∞ x + 8 

x 

1 2 + 1 

x 2 

lim 

x→∞ 

 

lim 3 − lim 2 

x→∞ x→∞ 

= 

x + lim 8 

1 

3 − 2 lim 

x→∞ x 2 

x→∞ x + 8 1 

lim 

x→∞ x 

= 

lim 1 + lim 1 

 

1 2 

x→∞ x→∞ x 2 1 + lim 

x→∞ x 

= 3 − 0 + 0 = 3 

↑ 1 + 0 

1 

lim 

x = 0 

x→∞ 

lim 

4x 2 − 5x 

= lim 

x→−∞ x 3 + 1 ↑ 

x→−∞ 

Divide the numerator and 

denominator by x 3 


This worksheet contains 6 questions in which the 

student finds the limit at infinity analytically. 

4x 2 − 5x 4 

x 3 

x 3 = lim 

x − 5 x 2 

+ 1 x→−∞ 

1 + 1 = 

x 3 x 3 

lim 

x→−∞ 

lim 

x→−∞ 

4 

x − 5 x 2 

2 

1 + 3 x 3 = 0 1 = 0 

■ 


for example 6 


x − x+ 

Find lim 2 3 

10 . 

x→∞ 

4 

5− 

x 

Solution 

To find the limit at infinity, divide each term 

by the highest term in the denominator, that 

is, x 4 : 

3 

2x 

x 10 

− + 

x − x+ 

lim 2 3 

10 

4 4 4 

= lim x x x 

x→∞ 

4 

5 − x 

x→∞ 

4 

5 x 

− 

4 4 

x x 

= lim 

x→∞ 

2 1 10 

− + 

3 4 

x x x 

5 

−1 

4 

x 

lim 2 − lim 1 + lim 10 

x→∞ →∞ →∞ 

= 

x x 

3 

x x 

4 

x 

lim 5 − lim 1 

x→∞ 

4 

x x→∞ 

0 

= − 0 + 0 

= 0 

0−1 



Find 

x − x+ 

a. lim 3 2 

2 8 

x→∞ 

2 

x + 1 

x − x 

b. 

xlim 4 2 

5 

→−∞ 

3 

x + 1 

Solution 

Another way to evaluate the limits in 

Example 6 is to consider only the leading 

terms in the numerator and denominator. 

For instance, in (a), consider only 3x 2 in the 

numerator and x 2 in the denominator: 

x − x+ 

x 

lim 3 2 

2 8 = lim 3 

x→∞ 

2 

x + 1 x→∞ 

2 

x 

= lim 3= 

3 

x→∞ 

x − x x 

lim 4 2 

5 = lim 4 2 

= lim 4 = 0 

→−∞ 

3 

x + 1 →−∞ 

3 

x →−∞ x 

x x x 

Your students may wonder why this method 

works. Go back to Example 6 and observe 

that every expression drops out except for 

the leading terms. This solution method is 

presented in Example 8 on page 135. 

2 


133 


11/01/17 9:55 am




The graphs of the functions g(x) = 3x 2 − 2x + 8 

and f (x) = 4x 2 − 5x 

are 

x 2 + 1 

x 3 + 1 

shown in Figures 56(a) and 56(b), respectively. In each graph, notice how the graph of 

the function behaves as x becomes unbounded. 

y 

y 

8 

6 

4 

g(x) 

3x 2 2x 8 

x 2 1 

y 3 

y 0 

4 

4 

f(x) 

4x 2 5x 

x 3 1 

4 8 

x 

2 

4 


Finding the Limit at Infinity 

Find 

x + 

lim 4 2 

10 

x→∞ 

x − 5 

Solution 

Another way to find this limit at infinity is to 

consider the variable terms only, because 

the constant 10 in the numerator and −5 in 

the denominator contribute very little as 

x → ∞: 

x + 

x 

lim 4 2 10 = lim 4 2 

x→∞ 

x − 5 x→∞ 

x 

x 

= lim 2| | = lim 2= 

2 

x→∞ 

x x→∞ 


Finding the Limit at Infinity 

x 

Find lim sin . 

x→∞ 

x 

Solution 

The numerator, sin x, fluctuates between 

−1 and 1 over its entire domain. 

Since the numerator is bounded, we can 

use the following argument to show the 

limit exists. 

sinx 

Some number between −1 and 1 

lim ≈ 

x→∞ 

x a very large number 

x 

and thus lim sin = 0 

x→∞ 

x 

This problem can also be solved formally 

using the Squeeze Theorem. 

We can also take a graphical approach. 

y 

1 

20 

10 

y 

20 

10 

10 


lim f (x) 2 

x →∞ 

y 2 

10 20 x 

√ 

4x 2 + 10 

x − 5 

NEED TO REVIEW? The end behavior 

of a polynomial function is discussed in 

Section P.2, p. 21. 

CALC 

CLIP 

Figure 56 

4 

8 

x 

(a) lim g(x) 3 (b) lim 

f(x) 0 

x→∞ x→ ∞ 

Limits at infinity have the following property. 

If p > 0 is a rational number and k is any real number, then 

provided x p is defined when x < 0. 

k 

lim = 0 and lim 

x→∞ x 

p 

5 

−6 

For example, lim = 0, lim = 0, and 

x→∞ x 

3 x→∞ x lim 

2/3 

EXAMPLE 7 Finding the Limit at Infinity 

√ 

4x 

2 

+ 10 

Find lim 

. 

x→∞ x − 5 

NOW WORK Problems 47 and 49. 

x→−∞ 

x→−∞ 

k 

x = 0 p 

4 

= 0. 

x 

8/3 

Solution Divide each term of the numerator and the denominator by x, the term with the 

highest power of x that appears in the denominator. But remember, since √ x 2 =|x| =x 

when x ≥ 0, in the numerator the divisor in the square root will be x 2 . 

lim 

x→∞ 

√ 

4x 

2 

+ 10 

x − 5 

= lim 

x→∞ 

√ 

4x 2 + 10 

x 2 

x − 5 

= lim 

x→∞ 

x 

√ 

√ 

lim 4 + 10 

x→∞ 

= [ x 2 

lim 1 − 5 ] = 

x→∞ x 

lim 

x→∞ 

√ 

4x 2 

x + 10 

2 x 2 

1 − 5 x 

[ 

4 + 10 ] 

x 2 

1 

= √ 4 = 2 

√ 

4x 

2 

+ 10 

The graph of f (x) = 

x − 5 

and its behavior as x →∞is shown in Figure 57. 


An alternative method for finding limits at infinity for rational functions uses the 

end behavior of a polynomial function. 

■ 

220 

20 

x 

21 

x→∞ 

x 

x 

lim sin 

= 0. 

134 



11/01/17 9:56 am



Recall that for large values of x, either positive or negative, the graph of the 

polynomial function 

f (x) = a n x n + a n−1 x n−1 +···+a 1 x + a 0 

resembles the graph of the power function y = a n x n . 

From this fact, we conclude that 

As examples, 

lim f (x) = lim a nx n (1) 

x→±∞ x→±∞ 

Teaching Tip 

When finding limits at infinity of rational 

functions, the terms with the largest 

degree in the numerator and denominator 

determine the limit. 

• lim (3x 3 − x 2 + 5x + 1) = lim 3x 3 = 3 lim x 3 =∞ 

x→∞ ↑ x→∞ x→∞ 

(1) 

• lim (−x 5 + 2x) = lim (−x 5 ) = −∞ 

x→∞ ↑ x→∞ 

(1) 

• lim (2x 4 − x 2 + 2x + 8) = lim (2x 4 ) =∞ 

x→−∞ x→−∞ 

• lim (x 3 + 10x + 7) = lim (x 3 ) = −∞ 

x→−∞ x→−∞ 

We conclude that for a polynomial function, as x becomes unbounded in either the 

positive direction or the negative direction, the polynomial function is also unbounded. 

In other words, polynomial functions have an infinite limit at infinity. 

Since a rational function R is the ratio of two polynomial functions p and q where 

q is not the zero polynomial, we can find a limit at infinity for a rational function using 

only the leading terms of p and q. That is, if the leading term of p(x) is a n x n and the 

leading term of q(x) is b m x m , then 

lim R(x) = lim p(x) 

x→±∞ x→±∞ q(x) = lim a n x n 

(2) 

x→±∞ b m x m 

EXAMPLE 8 

Find: 

(a) lim 

x→∞ 

3x 2 − 2x + 8 

x 2 + 1 

Finding Limits at Infinity of a Rational Function 

(b) 

lim 

4x 2 − 5x 

x→−∞ x 3 + 1 

(c) 

5x 4 − 3x 2 

lim 

x→∞ 2x 2 + 1 

Solution We use the alternative method of finding a limit at infinity of a rational function. 

y 

3x 2 − 2x + 8 3x 2 

(a) lim 

= lim 

x→∞ x 2 + 1 ↑ x→∞ x 2 

(2) 

= lim 

x→∞ 3 = 3 

1 

R(x) 5x4 3x 2 

2x 2 1 

(b) 

lim 

x→−∞ 

4x 2 − 5x 

x 3 + 1 = ↑ 

lim 

(2) 

x→−∞ 

4x 2 

x 3 

x 2 

x→−∞ 

= 4 lim 

x 3 = 4 lim 

x→−∞ 

1 

x = 4 · 0 = 0 

2 

2 

x 

5x 4 − 3x 2 

(c) lim 

x→∞ 2x 2 + 1 

= lim 5x 4 

x→∞ 2x = 5 2 2 lim x 4 

x→∞ x = 5 2 2 lim x 2 =∞ 

↑ 

x→∞ 

(2) 

■ 

1 

Figure 58 lim R(x) =∞ 

x→∞ 

Observe 

The function R(x) = 5x 4 − 3x 2 

has an infinite limit at infinity. The graph of R is 

2x 2 + 1 

shown in Figure 58. 

that the limits found in Example 8(a) and 8(b) are the same as the limits 

found in Example 6(a) and 6(b), as we should expect. Compare the two methods and 

decide which you prefer to use. 



135 


11/01/17 9:56 am




Teaching Tip 

Students may not find it intuitive that the 

x 

lim e = 0. 

x→−∞ 

Consider showing them the following steps: 

lim e 

x 

x→−∞ 

= lim 1 Rewritethe expression. 

x→−∞ 

e 

− x 

= 0 As x approaches negative 

infinity,the limit approaches 

zero. 

y 

4 

2 

(0, 1) 

2 

Figure 59 f (x) = e x 

2 

x 

Other Infinite Limits at Infinity 

We have seen that all polynomial functions have infinite limits at infinity, and some 

rational functions have infinite limits at infinity. Other functions also have an infinite 

limit at infinity. 

For example, consider the function f (x) = e x . 

The graph of the exponential function, shown in Figure 59, suggests that 

lim 

x→−∞ ex = 0 

lim 

x→∞ ex =∞ 

These limits are supported by the information in Table 15. 

TABLE 15 

x −1 −5 −10 −20 x approaches −∞ 

f (x) = e x 0.36788 0.00674 0.00005 −2 × 10 −9 f (x) approaches 0 

x 1 5 10 20 x approaches ∞ 

f (x) = e x e ≈ 2.71828 148.41 22,026 4.85 × 10 8 f (x) becomes unbounded 

y 

2 

EXAMPLE 9 

Find lim ln x. 

x→∞ 

Finding the Limit at Infinity of g(x) =ln x 

Solution Table 16 and the graph of g(x) = ln x in Figure 60 suggest that g(x) = ln x 

has an infinite limit at infinity. That is, 

2 4 

x 

lim ln x =∞ 

x→∞ 

2 

4 

Figure 60 g(x) = ln x 

y 

4 

TABLE 16 

x e 10 e 100 e 1000 e 10,000 e 100,000 → x becomes unbounded 

g(x) = ln x 10 100 1000 10,000 100,000 → g(x) becomes unbounded 

Now let’s compare the graph of f (x) = e x in Figure 59, the graph of g(x) = ln x 

in Figure 60, and the graph of h(x) = x 2 in Figure 61. As x becomes unbounded in 

the positive direction, all three of the functions increase without bound. But in Table 17, 

f (x) = e x approaches infinity more rapidly than h(x) = x 2 , which approaches infinity 

more rapidly than g(x) = ln x. 

■ 

24 

22 

2 

21 

2 

4 

x 

TABLE 17 

x 10 50 100 1000 10,000 

f (x) = e x 22,026 5.185 × 10 21 2.688 × 10 43 e 1000 e 10,000 

h(x) = x 2 100 2500 = 2.5 × 10 3 1.0 × 10 4 1.0 × 10 6 1.0 × 10 8 

g(x) = ln x 2.303 3.912 4.605 6.908 9.210 

Figure 61 h(x) = x 2 

EXAMPLE 10 

Application: Decomposition of Salt in Water 

Salt (NaCl) dissolves in water into sodium (Na + ) ions and chloride (Cl − ) ions according 

to the law of uninhibited decay 

A(t) = A 0 e kt 

where A = A(t) is the amount (in kilograms) of undissolved salt present at time t (in 

hours), A 0 is the original amount of undissolved salt, and k is a negative number that 

represents the rate of dissolution. 

136 



11/01/17 9:56 am



NEED TO REVIEW? Solving exponential 

equations is discussed in Section P.5, pp. 

48--49. 

(a) If initially there are 25 kilograms (kg) of undissolved salt and after 10 hours (h) 

there are 15 kg of undissolved salt remaining, how much undissolved salt is left 

after one day? 

(b) How long will it take until 1 kg of undissolved salt remains? 

2 

(c) Find lim A(t). 

t→∞ 

(d) Interpret the answer found in (c). 

Solution (a) Initially, there are 25 kg of undissolved salt, so A(0) = A 0 = 25. To 

find the number k in A(t) = A 0 e kt , we use the fact that at t = 10, then A(10) = 15. 

That is, 

A(10) = 15 = 25e 10k A(t) = A 0e kt , A 0 = 25; A(10) = 15 

e 10k = 3 5 

10k = ln 3 5 

k = 1 ln 0.6 

10 

So, A(t) = 25e ( 1 10 ln 0.6)t . The amount of undissolved salt that remains after one day 

(24 h) is 

A(24) = 25e ( 1 

10 ln 0.6)24 ≈ 7.337 kilograms 

(b) We want to find t so that A(t) = 25e ( 1 10 ln 0.6)t = 1 kg. Then 

2 

1 

2 = 25e( 1 

e ( 1 

10 ln 0.6)t = 1 50 

( ) 1 

10 ln 0.6 t = ln 1 

50 

t ≈ 76.582 

10 ln 0.6)t 

common error 

Sometimes students think that a function 

cannot cross a horizontal asymptote. A 

function cannot cross a vertical asymptote; 

however, horizontal asymptotes describe 

the function’s behavior as x approaches 

positive or negative infinity. Horizontal 

asymptotes give no information about 

the function’s behavior anywhere else. A 

function may cross a horizontal asymptote 

for values of x that are not near infinity; 

however, as x becomes infinitely large (or 

small), the function will approach, but will 

not cross, a limiting value. That limiting 

value defines the horizontal asymptote. 

Here is a graph of the function 

x 

fx=− + 2 

( ) 

x + 1 . This function has a 

2 

horizontal asymptote at y = 0. 

The function crosses that asymptote at 

x = −2, then slowly approaches the 

horizontal asymptote from above as x 

approaches negative infinity. 

y 

After approximately 76.6 h, 1 kg of undissolved salt will remain. 

2 

(c) Since 1 

25 

ln 0.6 ≈−0.051, we have lim A(t) = lim 

10 t→∞ t→∞ (25e−0.051t ) = lim = 0 

t→∞ e 0.051t 

(d) As t becomes unbounded, the amount of undissolved salt in the water approaches 

0 kg. Eventually, there will be no undissolved salt. ■ 

1 

28 

26 

24 

22 

21 

2 

x 


22 

4 Find the Horizontal Asymptotes of a Graph 

Limits at infinity have an important geometric interpretation. When lim f (x) = M, 

x→∞ 

it means that as x becomes unbounded in the positive direction, the value of f can 

be made as close as we please to a number M. That is, the graph of y = f (x) is 

as close as we please to the horizontal line y = M by choosing x sufficiently large. 

Similarly, lim f (x) = L means that the graph of y = f (x) is as close as we please 

x→−∞ 

to the horizontal line y = L for x unbounded in the negative direction. These lines are 

horizontal asymptotes of the graph of f. 


This worksheet contains 4 rational 

functions and their corresponding graphs 

to help the students identify the horizontal 

asymptotes of each of the functions. 

common error 

Functions may have infinitely many vertical 

asymptotes. Students may incorrectly believe 

that functions can have infinitely many horizontal 

asymptotes as well. Remind them that functions may 

have 0, 1, or 2 horizontal asymptotes. A function will 

not have more than 2 horizontal asymptotes, but a 

function can have infinitely many vertical asymptotes. 

Why can a function have no more than 2 horizontal 

asymptotes? Recall that a horizontal asymptote 

is found by finding the limit of the function as x 

approaches positive infinity and as x approaches 

negative infinity. If both of those limits yield unique 

constants, the function has 2 horizontal asymptotes. 

If one of those two limits is constant, or if they 

have the same value, the function has 1 horizontal 

asymptote. If neither of those limits is constant, the 

function has 0 horizontal asymptotes. 


137 


11/01/17 9:56 am




Teaching Tip 

Examples of functions that have 0, 1, or 2 

horizontal asymptotes: 

Zero Horizontal Asymptotes 

y = x 2 

y = sinx 

One Horizontal Asymptote 

y = e x 

This function has the horizontal asymptote 

y = 0. 

2x 

+ 1 

y = 

x 

This function has the horizontal asymptote 

y = 2. 

Two Horizontal Asymptotes 

− 

y = tan 1 x 

This function has horizontal asymptotes at 

π 

y = ± . 2 

| x| 

y = 

x 

This function has horizontal asymptotes at 

y = ± 1. 

Teaching Tip 

Functions may have 0, 1, or 2 horizontal 

asymptotes. Use this fact to remind 

students that they have to find the limit 

of the function as x approaches positive 

infinity and as x approaches negative 

infinity. 

lim f(x) L 

x→ ∞ 

Horizontal 

asymptote 

y M 

Figure 62 

y f (x) 

y 

y 

4 

2 

2 

Horizontal 

asymptote 

y L 

x 

lim f(x) M 

x→ ∞ 

3 

y 

4 

4 2 

2 4 

Figure 63 f (x) = 3x − 2 

4x − 1 

x 

DEFINITION Horizontal Asymptote 

The line y = L is a horizontal asymptote of the graph of a function f for x unbounded 

in the negative direction if lim f (x) = L. 

x→−∞ 

The line y = M is a horizontal asymptote of the graph of a function f for x 

unbounded in the positive direction if lim f (x) = M. 

x→∞ 

In Figure 62, y = L is a horizontal asymptote as x → −∞ because lim f (x) = L. 

x→−∞ 

The line y = M is a horizontal asymptote as x →∞because lim f (x) = M. To 

x→∞ 

identify horizontal asymptotes, we find the limits of f at infinity. 

EXAMPLE 11 Finding the Horizontal Asymptotes of a Graph 

Find the horizontal asymptotes, if any, of the graph of f (x) = 3x − 2 

4x − 1 . 

3x − 2 3x − 2 

Solution We examine the two limits at infinity: lim and lim 

x→−∞ 4x − 1 x→∞ 4x − 1 . 

3x − 2 

Since lim 

x→−∞ 4x − 1 = 3 4 , the line y = 3 is a horizontal asymptote of the graph 

4 

of f for x unbounded in the negative direction. 

3x − 2 

Since lim 

x→∞ 4x − 1 = 3 4 , the line y = 3 is a horizontal asymptote of the graph 

4 

of f for x unbounded in the positive direction. ■ 

Figure 63 shows the graph of f (x) = 3x − 2 

4x − 1 and the horizontal asymptote y = 3 4 . 

NOW WORK Problem 63 (find any horizontal asymptotes) 

and AP®Practice Problems 1, 4, and 7. 

5 Find the Asymptotes of the Graph of a Rational Function 

In the next example we find the horizontal asymptotes and vertical asymptotes, if any, 

of the graph of a rational function. 

EXAMPLE 12 

Finding the Asymptotes of the Graph of a 


Find any asymptotes of the graph of the rational function R(x) = 3x 2 − 12 

2x 2 − 9x + 10 . 

Solution We begin by factoring R. 

R(x) = 3x 2 − 12 3(x − 2)(x + 2) 

= 

2x 2 − 9x + 10 (2x − 5)(x − 2) 

The domain of R is 

{x| x = 5 } 

2 and x = 2 . Since R is a rational function, it is 

continuous on its domain, that is, all real numbers except x = 5 and x = 2. 

2 

To check for vertical asymptotes, we find the limits as x approaches 5 and 2. First 

2 

we consider lim 

x→ 5 2 

lim 

x→ 5 − 

2 

R(x) = lim 

x→ 5 − 

2 

− R(x). 

[ 3(x − 2)(x + 2) 

(2x − 5)(x − 2) 

] 

= lim 

x→ 5 − 

2 

[ ] 3(x + 2) 

(2x − 5) 

= 3 lim 

x→ 5 − 

2 

x + 2 

2x − 5 = −∞ 

That is, as x approaches 5 from the left, R becomes unbounded in the negative direction. 

2 

The graph of R has a vertical asymptote on the left at x = 5 2 . 


for example 12 

Finding the Asymptotes of the Graph of a 


2x 

+ 1 

Find any asymptotes of the function fx ( ) = 

x − 2 . 

Solution 

Vertical Asymptote(s) 

The function is undefined at x = 2, so we will 

check the one-sided limits at x = 2. 

x + 

lim 2 1 

− ≈ a number close to 5 

− 

→ x 2 very small negative and 

x 2 

x + 

thus lim 2 1 

− 

≈−∞ . 

− 

x→2 

x 2 

We do not have to do further verification. This 

function has a vertical asymptote at x = 2. 

Horizontal Asymptote(s) 

We must find the limit of the function as x 

approaches positive and negative infinity. 

x + 

− = x 

lim 2 1 lim 2 = lim 2 = 2 

x →∞ x 2 x →∞ x x →∞ 

The function has a horizontal asymptote at y = 2. 

x + 

− = x 

lim 2 1 lim 2 = lim 2 = 2 

x →−∞ x 2 x →−∞ x x →−∞ 

Since these two limits are the same, this function 

has only one horizontal asymptote: y = 2. 

138 



11/01/17 9:56 am



3 

y 

2 

y 

8 

4 

8 4 

4 8 12 

4 

8 

12 

5 

x 

2 

Figure 64 R(x) = 3x2 − 12 

2x 2 − 9x + 10 



1. True or False ∞ is a number. 

1 

2. (a) lim = 

x→0 − x 

1 

; (b) lim 

x→0 + x = ; 

(c) lim x→0 + 

x 

To determine the behavior to the right of x = 5 , we find the right-hand limit. 

2 

lim 

x→ 5 + 

2 

R(x) = lim 

x→ 5 + 

2 

[ ] 3(x + 2) 

(2x − 5) 

= 3 lim 

x→ 5 + 

2 

x + 2 

2x − 5 =∞ 

As x approaches 5 from the right, R becomes unbounded in the positive direction. The 

2 

graph of R has a vertical asymptote on the right at x = 5 2 . 

Next we consider lim 

x→2 

R(x). 

lim R(x) = lim 

x→2 x→2 

3. True or False The graph of a rational function has a vertical 

asymptote at every number x at which the function is not defined. 

4. If lim f (x) =∞, then the line x = 4 is a(n) 

x→4 

asymptote 

of the graph of f. 

1 

1 

5. (a) lim = ; (b) lim 

x→∞ x x→∞ x 2 = ; 

(c) lim x→∞ 

6. True or False lim x→−∞ 

7. (a) lim 

x→−∞ ex = ; (b) lim 

x→∞ ex = ; (c) lim 

x→∞ e−x = 

8. True or False The graph of a function can have at most two 

horizontal asymptotes. 

3(x − 2)(x + 2) 

(2x − 5)(x − 2) = lim 

x→2 

3(x + 2) 

2x − 5 

= 

3(2 + 2) 

2 · 2 − 5 = 12 

−1 =−12 

Since the limit is not infinite, the function R does not have a vertical asymptote at 2. 

Since 2 is not in the domain of R, the graph of R has a hole at the point (2, −12). 

To check for horizontal asymptotes, we find the limits at infinity. 

3x 2 − 12 

lim R(x) = lim 

x→∞ x→∞ 2x 2 − 9x + 10 = lim 3x 2 

x→∞ 2x = lim 3 

2 x→∞ 2 = 3 ↑ 

2 

(2) 

lim R(x) = lim 3x 2 − 12 

x→−∞ x→−∞ 2x 2 − 9x + 10 = lim 3x 2 

x→−∞ 2x = lim 3 

2 x→−∞ 2 = 3 ↑ 

2 

(2) 

The line y = 3 is a horizontal asymptote of the graph of R for x unbounded in the 

2 

negative direction and for x unbounded in the positive direction. ■ 

The graph of R and its asymptotes are shown in Figure 64. Notice the hole in the 

graph at the point (2, −12). 



In Problems 9–16, use the accompanying graph of y = f (x). 

9. Find lim f (x). 

x→∞ 


x→−∞ 

PAGE 

129 11. Find lim f (x). 

x→−1 − 


x→−1 + 


x→3 − 


x→3 + 

PAGE 

131 15. Identify all 

vertical 

asymptotes. 

16. Identify all 

horizontal 

asymptotes. 

y 

12 

8 

4 

4 4 8 12 

x 1 x 3 

y 2 

x 

Must-Do Exercises for 


AB: 2–26, 27–59 odd, 67–71 odd, AP ® 


BC: 17–26, 30–35, 45, 53, 55, 63, 73, 78, 

all AP ® Practice Problems 



Problems 


Problems 

1. False. 

2. (a) −∞ (b) ∞ (c) − ∞ 

3. False. 

4. Vertical. 

5. (a) 0 (b) 0 (c) ∞ 

6. False. 

7. (a) 0 (b) ∞ (c) 0 

8. True. 

9. 2 

10. 0 

11. ∞ 

12. ∞ 

13. ∞ 

14. ∞ 

15. x = −1, x = 3 

16. y = 0, y = 2 


139 


11/01/17 9:56 am




17. −3 18. 0 

19. ∞ 20. − ∞ 

21. 0 22. ∞ 

23. ∞ 24. ∞ 

25. x = −3, x = 0, x = 4 

26. y = 0, y =−3 

27. − ∞ 28. − ∞ 29. ∞ 

30. − ∞ 31. ∞ 32. ∞ 

33. ∞ 34. ∞ 35. − ∞ 

36. ∞ 37. − ∞ 38. ∞ 

39. − ∞ 40. − ∞ 41. − ∞ 

42. − ∞ 43. 0 44. 0 

45. 2 5 

46. 1 47. 1 

48. 2 5 

49. 0 50. 0 

51. 5 4 

52. − 1 3 

53. 0 

54. 0 55. 0 56. 0 

57. 3 58. 4 59. −∞ 

60. −∞ 

61. x = 0 is a vertical asymptote. y = 3 is 

a horizontal asymptote. 



63. x = −1 and x = 1 are vertical 

asymptotes. y = 1 is a horizontal 

asymptote. 



asymptote. 

3 

65. x = is a vertical asymptote. 

2 

2 

2 

y =− and y = are 

2 

2 

horizontal asymptotes. 



67. (a) { x| x ≠−2, x ≠ 0} 

(b) y = 0 is a horizontal asymptote. 

(c) x = −2 and x = 0 are vertical 

asymptotes. 

(d) See TSM. 

68. (a) { x| x ≠± 1} 


(c) x = −1 and x = 1 are vertical 

asymptotes. 

(d) See TSM. 

⎧ 3 ⎫ 

69. (a) ⎨xx 

| ≠ x ≠ ⎬ 

⎩ 2 , 2 

⎭ 

1 

(b) y = is a horizontal asymptote. 

2 

3 

(c) x = is a vertical asymptote. 

2 

(d) See TSM. 

In Problems 17–26, use the graph below of y = f (x). 

17. Find lim f (x). 18. Find lim f (x). 

x→∞ x→−∞ 

19. Find lim f (x). 20. Find lim f (x). 

x→−3− x→−3 + 

21. Find lim 

x→0 − f (x). 

23. Find lim 

x→4 − f (x). 

25. Identify all vertical asymptotes. 

26. Identify all horizontal asymptotes. 

y 3 

y 


x→0 + 


x→4 + 

y 0 4 2 

2 4 6 

4 

2 

2 

4 


PAGE 

130 27. lim 

x→2 − 

3x 

x − 2 

x→2 + x 2 − 4 

29. lim 

5 

31. 

5x + 3 

lim 

x→−1 + x(x + 1) 

x→−3 − x 2 − 9 

33. lim 

1 

35. lim 

1 − x 

x→3 

x 3 x 0 x 4 

(3 − x) 2 36. lim 

2x + 1 

28. lim 

x→−4 + x + 4 

2x 

30. lim 

x→1 − x 3 − 1 

5x + 3 

32. lim 

x→0 − 5x(x − 1) 

x 

34. lim 

x→2 + x 2 − 4 

x + 2 

x→−1 (x + 1) 2 

37. lim cot x 38. lim tan x 

x→π− x→−π/2 − 

39. lim csc(2x) 

x→π/2 + 

40. lim x→−π/2 − 

41. lim ln(x + 1) 

x→−1 + 

42. lim ln(x − 1) 

x→1 + 


5 

43. lim 

x→∞ x 2 + 4 

PAGE 

133 45. 

2x + 4 

lim 

x→∞ 5x 

PAGE 

134 47. 

x 3 + x 2 + 2x − 1 

lim 

x→∞ x 3 + x + 1 

PAGE 

134 49. 

x 2 + 1 

lim 

x→−∞ x 3 − 1 

51. 

[ ] 

3x 

lim 

x→∞ 2x + 5 − x2 + 1 

4x 2 + 8 

1 

44. lim 

x→−∞ x 2 − 9 

x + 1 

46. lim 

x→∞ x 

2x 2 − 5x + 2 

48. lim 

x→∞ 5x 2 + 7x − 1 

x 2 − 2x + 1 

50. lim 

x→∞ x 3 + 5x + 4 

[ 

52. lim 

x→∞ 

70. (a) { x| x ≠−3} 


(c) x =−3 is a vertical asymptote. 

(d) See TSM. 

71. (a) { x| x ≠0, x ≠1} 

(b) There are no horizontal asymptotes. 

(c) x = 0 is a vertical asymptote. 

(d) See TSM. 

72. (a) { x| x ≠ 1} 

(b) There are no horizontal asymptotes. 

(c) x = 1 is a vertical asymptote. 

(d) See TSM. 

x 

PAGE 

131 

1 

x 2 + x + 4 − x + 1 

3x − 1 

] 

[ ( )] 5x + 1 

53. lim 2e x 

x→−∞ 3x 

√ x + 2 

55. lim 

PAGE 

134 57. lim 

x→∞ 

x→∞ 3x − 4 

√ 

3x 2 − 1 

x 2 + 4 

PAGE 

135 59. 

5x 3 

lim 

x→−∞ x 2 + 1 

54. lim 

x→−∞ 

[ 

( )] 

x 

e x 2 + x − 3 

2x 3 − x 2 

√ 

3x 

56. lim 

3 + 2 

x→∞ x 2 + 6 

( ) 16x 3 2/3 

+ 2x + 1 

58. lim 

x→∞ 2x 3 + 3x 

x 4 

60. lim 

x→−∞ x − 2 

In Problems 61–66, find any horizontal or vertical asymptotes of the 

graph of f . 

61. f (x) = 3 + 1 x 

PAGE 

138 63. f (x) = x2 

x 2 − 1 

65. f (x) = 

√ 

2x 2 − x + 10 

2x − 3 

62. f (x) = 2 − 1 x 2 

64. f (x) = 2x2 − 1 

x 2 − 1 

√ 

3 

x 

66. f (x) = 

2 + 5x 

x − 6 

In Problems 67–72, for each rational function R: 

(a) Find the domain of R. 

(b) Find any horizontal asymptotes of R. 

(c) Find any vertical asymptotes of R 

(d) Discuss the behavior of the graph at numbers where R is not 

defined. 

67. R(x) = −2x2 + 1 

2x 3 + 4x 2 68. R(x) = x3 

x 4 − 1 

PAGE 

139 69. R(x) = x2 + 3x − 10 

2x 2 − 7x + 6 

70. R(x) = 

x(x − 1)2 

(x + 3) 3 

71. R(x) = x3 − 1 

x − x 2 72. R(x) = 4x5 

x 3 − 1 


In Problems 73 and 74: 

(a) Sketch a graph of a function f that has the given properties. 

(b) Define a function that describes the graph. 

73. f (3) = 0, lim f (x) = 1, lim 

x→∞ 

lim f (x) =∞, lim 

x→1− 74. f (2) = 0, lim 

x→∞ 

f (x) =∞, lim 

lim 

x→0 

f (x) = 1, 

x→−∞ 

f (x) = −∞ 

x→1 + 

f (x) = 0, lim 

f (x) = −∞, lim 

x→5− f (x) = 0, 

x→−∞ 

f (x) =∞ 

x→5 + 

75. Newton’s Law of Cooling Suppose an object is heated to a 

temperature u 0. Then at time t = 0, the object is put into a 

medium with a constant lower temperature T causing the object to 

cool. Newton’s Law of Cooling states that the temperature u of 

the object at time t is given by u = u(t) = (u 0 − T )e kt + T , 

where k < 0 is a constant. 

(a) Find lim u(t). Is this the value you expected? Explain why 

t→∞ 

or why not. 

(b) Find lim u(t). Is this the value you expected? Explain why 

t→0 + 

or why not. 



(a) 

26 24 22 

y 

6 

4 

2 

24 

26 

x 

(b) fx= − 3 

( ) 

x −1 

2 4 6 

x 


140 



11/01/17 9:56 am



76. Environment A utility company burns coal to generate 

electricity. The cost C, in dollars, of removing p% of the 

pollutants emitted into the air is 

Find the cost of removing: 

C = 70,000p 

100 − p , 0 ≤ p < 100 

(a) 45% of the pollutants. 

(b) 90% of the pollutants. 

(c) Find lim C. 

p→100 − 


77. Pollution Control The cost C, in thousands of dollars, to 

remove a pollutant from a lake is 

C(x) = 

5x 

100 − x , 0 ≤ x < 100 

where x is the percent of pollutant removed. Find lim C(x). 

x→100 − 

Interpret your answer. 

78. Population Model A rare species of insect was discovered in 

the Amazon Rain Forest. To protect the species, entomologists 

declared the insect endangered and transferred 25 insects to a 

protected area. The population P of the new colony t days after 

the transfer is 

50(1 + 0.5t) 

P(t) = 

2 + 0.01t 

(a) What is the projected size of the colony after 1 year 

(365 days)? 

(b) What is the largest population that the protected area can 

sustain? That is, find lim 

t→∞ P(t). 

(c) Graph the population P as a function of time t. 

(d) Use the graph from (c) to describe the regeneration of the 

insect population. Does the graph support the answer to (b)? 

PAGE 

137 79. Population of an Endangered Species Often environmentalists 

capture several members of an endangered species and transport 

them to a controlled environment where they can produce 

offspring and regenerate their population. Suppose six American 

bald eagles are captured, tagged, transported to Montana, and set 

free. Based on past experience, the environmentalists expect the 

population to grow according to the model 

where t is measured in years. 

500 

P(t) = 

1 + 82.3e −0.162t 

(a) If the model is correct, how many bald eagles can the 

environment sustain? That is, find lim 

t→∞ P(t). 

(b) Graph the population P as a function of time t. 

(c) Use the graph from (b) to describe the growth of the 

bald eagle population. Does the graph support the answer 

to (a)? 

80. Hailstones Hailstones typically originate at an altitude of about 

3000 meters (m). If a hailstone falls from 3000 m with no air 

resistance, its speed when it hits the ground would be about 

240 meters/second (m/s), which is 540 miles/hour (mi/h)! That 

would be deadly! But air resistance slows the hailstone 

considerably. Using a simple model of air resistance, the speed 

v = v(t) of a hailstone of mass m as a function of time t is given 

by v(t) = mg 

k (1 − e−kt/m ) m/s, where g = 9.8m/s 2 and k is a 

constant that depends on the size of the hailstone, its mass, and the 

conditions of the air. For a hailstone with a diameter 

d = 1 centimeter (cm) and mass m = 4.8 × 10 −4 kg, k has been 

measured to be 3.4 × 10 −4 kg/s. 

(a) Determine the limiting speed of the hailstone by finding 

lim v(t). Express your answer in meters per second and 

t→∞ 

miles per hour, using the fact that 1 mi/h ≈ 0.447 m/s. This 

speed is called the terminal speed of the hailstone. 

(b) Graph v = v(t). Does the graph support the answer to (a)? 

81. Damped Harmonic Motion The motion of a spring is given by 

the function 

x(t) = 1.2e −t/2 cos t + 2.4e −t/2 sin t 

where x is the distance in meters from the the equilibrium position 

and t is the time in seconds. 

(a) Graph y = x(t). What is lim x(t), as suggested by the 

t→∞ 

graph? 

(b) Find lim x(t). 

t→∞ 

(c) Compare the results of (a) and (b). Is the answer to (b) 

supported by the graph in (a)? 

82. Decomposition of Chlorine in a Pool Under certain water 

conditions, the free chlorine (hypochlorous acid, HOCl) in a 

swimming pool decomposes according to the law of uninhibited 

decay, C = C(t) = C(0)e kt , where C = C(t) is the amount 

(in parts per million, ppm) of free chlorine present at time t 

(in hours) and k is a negative number that represents the rate 

of decomposition. After shocking his pool, Ben immediately 

tested the water and found the concentration of free chlorine to 

be C 0 = C(0) = 2.5 ppm. Twenty-four hours later, Ben 

tested the water again and found the amount of free chlorine 

to be 2.2 ppm. 

(a) What amount of free chlorine will be left after 72 hours? 

(b) When the free chlorine reaches 1.0 ppm, the pool should be 

shocked again. How long can Ben go before he must shock 

the pool again? 

(c) Find lim 

t→∞ C(t). 


83. Decomposition of Sucrose Reacting with water in an 

acidic solution at 35 ◦ C, the amount A of sucrose (C 12H 22O 11) 

decomposes into glucose (C 6H 12O 6) and fructose (C 6H 12O 6) 

according to the law of uninhibited decay A = A(t) = A(0)e kt , 

where A = A(t) is the amount (in moles) of sucrose present at 

time t (in minutes) and k is a negative number that represents the 

rate of decomposition. An initial amount A 0 = A(0) = 0.40 mole 

of sucrose decomposes to 0.36 mole in 30 minutes. 

(a) How much sucrose will remain after 2 hours? 

(b) How long will it take until 0.10 mole of sucrose remains? 

(c) Find lim 

t→∞ A(t). 


84. Macrophotography A camera lens can be approximated by a 

thin lens. A thin lens of focal length f obeys the thin-lens 

equation 1 f 

78. (a) 1624 insects 

(b) 2500 insects 

(c) 

y 

2500 

2000 

1500 

1000 

t 5 1 year 

500 

1000 2000 3000 4000 5000 x 

Time (days) 

(d) The graph supports the answer 

to (b). 

79. (a) 500 bald eagles 

(b) 

y 

500 

400 

300 

200 

100 

20 40 60 80 100 x 

Time (years) 

(c) Answers will vary. See TSM for 

sample answer. The graph supports the 

answer to (a). 

80. (a) 13.84 m/s or 30.96 mi/h 

(b) 

y 

13.84 

12 

10 

8 

6 

4 

2 

2 4 6 8 10 x 

Time (seconds) 

The graph supports the answer to (a). 

81. (a) 

x 

2.0 

1.5 

= 1 p + 1 q , where p > f is the distance from the 1.0 

Number of insects 

Number of bald eagles 

Speed (meters/second) 

0.5 


(a) 

76. (a) $57,272.73 

(b) $630,000 

y 

4 

(c) ∞ 

2 

24 22 2 4 6 8 x 

22 

24 

x − 2 

(b) fx ( ) = 

2 

x ( x− 

5) 

(b) u 0 

75. (a) T 

(d) Interpretations will vary. See TSM for 

sample interpretation. 

77. ∞. Interpretations will vary. Sample 

interpretation: As the percentage of the 

pollutants removed from the air increases 

toward 100%, the cost of removing those 

pollutants increases without bound. 

0.5 

1 2 3 4 5 6 7 8 9 

The graph suggests that lim xt () = 0. 

t→∞ 

(b) 0 

(c) The graph in (a) supports the 

answer in (b). 

82. (a) Approx. 1.7 ppm 

(b) Roughly 172 h 

(c) 0 ppm 

(d) In the long run, all of the free 

chlorine in the pool will decompose. 

t 



141 


11/01/17 9:56 am




83. (a) 0.26 moles 

(b) 395 min 

(c) 0 moles 

(d) In the long run, all of the sucrose 

will decompose. See TSM. 

84. (a) Not continuous. 

(b) A camera cannot focus on an object 

placed close to its focal length because 

the distance of the image from the lens 

becomes unbounded. 

85. a 

≠ 0; b can be any real number; 

c = 0; and d ≠ 0. See TSM. 

86. a ≠ 0; b can be any real number; 

c = 0; and d ≠ 0. 

87. See TSM. 

88. See TSM. 

89. See TSM. 

90. See TSM. 

91. (a) For table, see TSM. 

(b) e ≈ 2.718281828 

(c) Answers will vary. Sample answer: 

The results from (a) and (b) agree to 

five decimal places. 

92. The property requires the exponent to 

be a constant, independent of x, but in 

x 

⎛ 1 ⎞ 

lim + 

→∞⎝ 

⎜1 

x ⎠ 

⎟ the exponent is x. 

x 

93. (a) ∞ 

(b) The result suggests that it is not 

possible to reach the speed of light. 

lens to the object being photographed and q is the distance from 

the lens to the image formed by the lens. See the figure below. 

To photograph an object, the object’s image must be formed on 

the photo sensors of the camera, which can only occur if q is 

positive. 

Object 

p 

Lens 

q 

Photo 

sensors 

(a) Is the distance q of the image from the lens continuous as the 

distance of the object being photographed approaches the 

focal length f of the lens? (Hint: First solve the thin-lens 

equation for q and then find lim q.) p→ f + 

(b) Use the result from (a) to explain why a camera (or any lens) 

cannot focus on an object placed close to its focal length. 

In Problems 85 and 86, find conditions on a, b, c, and d so that the 

graph of f has no horizontal or vertical asymptotes. 

85. f (x) = ax3 + b 

cx 4 + d 

86. f (x) = ax + b 

cx + d 

87. Explain why the following properties are true. Give an example of 

each. 

1 

(a) If n is an even positive integer, then lim x→c (x − c) n =∞. 

1 

(b) If n is an odd positive integer, then lim 

x→c − (x − c) n = −∞. 

1 

(c) If n is an odd positive integer, then lim 

x→c + (x − c) n =∞. 

88. Explain why a rational function, whose numerator and 

denominator have no common zeros, will have vertical 

asymptotes at each point of discontinuity. 

89. Explain why a polynomial function of degree 1 or higher cannot 

have any asymptotes. 


PAGE 

138 1. For x > 0, the line y = 1 is an asymptote of the graph of a 

function f . Which of the following statements must be true? 

(A) f (x) = 1 for x > 0. (B) lim f (x) =∞ 

x→1 

(C) 

lim 

x→∞ 

PAGE 

3x 3 + 4x 2 − x + 10 

135 2. lim 

x→∞ 2x 4 − x 3 + 2x 2 − 2 = 

3 

(A) –5 (B) 0 (C) 

2 

PAGE 

5x 3 − x 

135 3. lim 

x→∞ 8 − x 3 = 

5 

(A) –5 (B) (C) 5 

8 

f (x) = 1 (D) lim 

x→−∞ f (x) = 1 

(D) ∞ 

(D) ∞ 

90. If P and Q are polynomials of degree m and n, respectively, 

P(x) 

discuss lim 

x→∞ Q(x) when: 

(a) m > n (b) m = n (c) m < n 

 

91. (a) Use a table to investigate lim 1 + 1 x 

. 

x→∞ x 

 

CAS (b) Find lim 1 + 1 x 

. 

x→∞ x 

(c) Compare the results from (a) and (b). Explain the possible 

causes of any discrepancy. 


 

92. lim 1 + 1 

x→∞ x 

= 1, but lim 

x→∞ 

 

1 + 1 x 

> 1. Discuss why the 

x 

 

lim 

x→∞ f (x) n 

cannot be used to find the 

property lim [ f x→∞ (x)]n = 

second limit. 

93. Kinetic Energy At low speeds the kinetic energy K , that is, the 

energy due to the motion of an object of mass m and speed v, is 

given by the formula K = K (v) = 1 2 mv2 . But this formula is 

only an approximation to the general formula, and works only for 

speeds much less than the speed of light, c. The general formula, 

which holds for all speeds, is 

⎡ 

⎤ 

Kgen(v) = mc 2 ⎢ 

1 

⎣ 

− 1⎥ 

⎦ 

1 − v2 

c 2 

(a) As an object is accelerated closer and closer to the speed of 

light, what does its kinetic energy Kgen approach? 

(b) What does the result suggest about the possibility of reaching 

the speed of light? 

PAGE 

138 4. Find all the horizontal asymptotes of the graph of y = 2 + 3x 

4 − 3 x . 

(A) y = –1 only 

(B) y = 1 2 only 

(C) y = –1 and y = 0 (D) y = –1 and y = 1 2 

PAGE 

131 5. Find all the vertical asymptotes of the graph of 

r(x) = x2 + 5x + 6 

. 

x 3 − 4x 

(A) x = 0 and x = –2 (B) x = 0 and x = 2 

(C) x = –2 and x = 2 (D) x = 0, x = –2 and x = 2 

PAGE 

134 6. lim 

x→−∞ 

√ 

8x 2 − 4x 

x + 2 

= 

(A) –∞ (B) −2 √ 2 (C) 4 (D) 2 √ 2 

Answers to AP® Practice Problems 

1. C 

2. B 

3. A 

4. D 

5. B 

6. B 

142 



11/01/17 9:56 am


Section 1.6 • The ε-δ Definition of a Limit 143 

PAGE 

138 7. The graph of which of the following functions has an 

asymptote of y = 1? 

(A) y = cos x (B) y = x − 1 

x 

(C) y = e −x (D) y = ln x 

PAGE 

139 8. If the graph of f (x) = ax − b has a vertical asymptote 

x + c 

x = –5 and horizontal asymptote y = –3, then a + c = 

3 

(A) –8 (B) –2 (C) (D) 2 

5 

PAGE 

x 

130 9. lim 

x→1 − ln x is 

(A) –∞ (B) –1 (C) 1 (D) ∞ 

PAGE 

139 10. The function f (x) = 2x 

|x|−1 has 

(A) no vertical asymptote and one horizontal asymptote. 

(B) one vertical asymptote and one horizontal asymptote. 

(C) two vertical asymptotes and one horizontal asymptote. 

(D) two vertical asymptotes and two horizontal asymptotes. 

PAGE 

5x + 1 

130 11. lim 

x→2 − 2x − 4 = 

(A) –∞ (B) − 5 2 

(C) 

5 

2 

(D) ∞ 

7. B 

8. D 

9. A 

10. D 

11. A 

AP® EXAM INSIGHT The ε-δ definition 

of a limit is not assessed on the AP® 

Calculus AB or BC Exam. However, your 

teacher may include this topic in the 

course if time permits. The ε-δ definition 

is an essential piece of calculus. The limit 

rules and theorems we have used in this 

chapter are proved using the ε-δ 

definition, which makes the ε-δ 

definition part of the foundation of 

calculus. 

y 

10 

8 

7.3 

6.7 

6 

4 

(2, 10) 

1.6 The ε-δ Definition of a Limit 


1 Use the ε-δ definition of a limit (p. 145) 

Throughout the chapter, we stated that we could be sure a limit was correct only if it 

was based on the ε-δ definition of a limit. In this section, we examine this definition and 

how to use it to prove a limit exists, to verify the value of a limit, and to show that a limit 


Consider the function f defined by 

{ 

3x + 1 if x = 2 

f (x) = 

10 if x = 2 

whose graph is given in Figure 65. 

As x gets closer to 2, the value f (x) gets closer to 7. If in fact, by taking x close 

enough to 2, we can make f (x) as close to 7 as we please, then lim f (x) = 7. 

x→2 

Suppose we want f (x) to differ from 7 by less than 0.3; that is, 

−0.3 < f (x) − 7 < 0.3 

6.7 < f (x)




That is, whenever x = 2 and x differs from 2 by less than 0.1, then f (x) differs from 7 

by less than 0.3. 

Now, generalizing the question, we ask, for x = 2, how close must x be to 2 to 

guarantee that f (x) differs from 7 by less than any given positive number ε? (ε might 

be extremely small.) The statement “ f (x) differs from 7 by less than ε” means 

−ε < f (x) − 7



DEFINITION Limit of a Function 

Let f be a function defined everywhere in an open interval containing c, except possibly 

at c. Then the limit as x approaches c of f(x) is L, written 

lim f (x) = L 

x→c 

if, given any number ε>0, there is a number δ>0 so that 

whenever 0 < |x − c| 0 so that 

whenever 0 < |x − (−1)| =|x + 1|




EXAMPLE 2 

Using the ε-δ Definition of a Limit 

Use the ε-δ definition of a limit to prove that: 

(a) lim 

x→c 

A = A, where A and c are real numbers 

(b) lim 

x→c 

x = c, where c is a real number 

Solution (a) f (x) = A is the constant function whose graph is a horizontal line. Given 

any ε>0, we must find δ>0 so that whenever 0 < |x − c| < δ, then | f (x) − A|



To ensure that both inequalities are satisfied, we select δ to be the smaller of the 

numbers 1 and ε { 

5 , abbreviated as δ = min 1, ε } 

. Now, 

5 

{ 

whenever |x − 2| 0, we need to find a positive number δ so that whenever 0 < |x −c| 0, then x > c 2 > 0, and 1 x < 2 c . Now 

1 

∣ x − 1 |x − c| 

c ∣ = 

c|x| 

< 2 1 

·|x − c| Substitute 

c2 |x| < 2 c . 

We can make 

1 

∣ x − 1 c2 

c ∣




( ) c 

So, given any ε>0, we choose δ = min 

2 , c2 

2 · ε . Then whenever 0 < |x−c| 0. ■ NOW WORK Problem 25. 

The ε-δ definition of a limit can be used to show that a limit does not exist, or that a 

limit is not equal to a specific number. Example 5 illustrates how the ε-δ definition of a 

limit is used to show that a limit is not equal to a specific number. 

NOTE In a proof by contradiction, we 

assume that the conclusion is not true 

and then show this leads to a 

contradiction. 

EXAMPLE 5 

Showing a Limit Is Not Equal to a Specific Number 

Use the ε-δ definition of a limit to prove the statement lim 

x→3 

(4x − 5) = 10. 

Solution We use a proof by contradiction. Assume lim 

x→3 

(4x −5) = 10 and choose ε = 1. 

(Any smaller positive number ε will also work.) Then there is a number δ>0, so that 

whenever 0 < |x − 3|



Suppose x 1 is a rational number satisfying 0 < |x 1 − c| 1 2 , and from the right inequality, we have L < 1 2 . 

Since it is impossible for both inequalities to be satisfied, we conclude that lim 

x→c 

f (x) 

does not exist. ■ 

EXAMPLE 7 

Using the ε-δ Definition of a Limit 

Prove that if lim 

x→c 

f (x) >0, then there is an open interval around c, for which f (x) >0 

everywhere in the interval except possibly at c. 

Solution Suppose lim 

x→c 

f (x) = L > 0. Then given any ε>0, there is a δ>0 so that 

whenever 0 < |x − c|




Figures 68 and 69 illustrate limits at infinity. 

y 

y 5 f (x) 

y 

y 5 f (x) 

L 1 

L 

L 2 

L 1 

L 

L 2 

M 

x . M 

x 

x , N 

N 

x 

lim f(x) 5 L 

x→∞ 

Figure 68 For any ε>0, there is a positive 

number M so that whenever x > M, then 

| f (x) − L| 0, there is a 

negative number N so that whenever x < N, 

then | f (x)−L| 0, suppose there is a δ>0 

so that whenever 0 < |x − c| < δ, then | f (x) − L|



6. True or False A function f has a limit L at infinity, if for 

any given ε>0, there is a positive number M so that 

whenever x > M, then | f (x) − L| > ε. 


In Problems 7–12, for each limit, find the largest δ that “works” for 

the given ε. 

7. lim x→1 

(2x) = 2, ε = 0.01 8. lim x→2 

(−3x) =−6, ε = 0.01 

9. lim(6x − 1) = 11 10. lim (2 − 3x) = 11 

x→2 x→−3 

ε = 1 ε = 1 2 

3 

( 

11. lim − 1 ) ( 

x→2 2 x + 5 = 4 12. lim 3x + 1 ) 

= 3 

x→ 5 2 

6 

ε = 0.01 ε = 0.3 

13. For the function f (x) = 4x − 1, we have lim x→3 

f (x) = 11. 

For each ε>0, find a δ>0 so that 

whenever 0 < |x − 3| 0, find a δ>0 so that 

whenever 0 < |x + 2| 0, find a δ>0 so that 

∣ whenever 0 < |x + 3| 0, find a δ>0 so that 

∣ whenever 0 < |x − 2| 0, let δ ≤ . See TSM for 

(c) d ≤ 0.00025 

complete proof. 2 

ε 

(d) δ ≤ ε 

4 

20. Given any e > 0, let δ ≤ . See TSM for 

complete proof. 3 

} 

. 

ε 

21. Given any e > 0, let δ ≤ . See 

TSM for complete proof. 5 

ε 


TSM for complete proof. 2 

⎧ ε ⎫ 

23. Given any e > 0, let δ ≤ min⎨1, ⎬ . 

See TSM for complete proof. ⎩ 3 ⎭ 

⎧ ε ⎫ 



⎧ ε ⎫ 

25. Given any e > 0, let δ ≤ min⎨1, 2 ⎬ . 


⎧ ε ⎫ 



27. Given any e > 0, let δ ≤ ε 

3 . See TSM 

for complete proof. 

28. Given any e > 0, let d ≤ e. See TSM 

for complete proof. 

⎧ ε ⎫ 



⎧ ε ⎫ 



31. Given any e > 0, let d ≤ min{1, 6e}. 

See TSM for complete proof. 

⎧ ε ⎫ 



33. See TSM. 

34. See TSM. 

35. Given any e > 0, let 

⎧ ε ⎫ 

δ ≤ min⎨1, 234 ⎬. 

⎩ 7 ⎭ 


⎧ ε ⎫ 


⎩ 5 

See TSM for complete proof. ⎭ 

37. Given any e > 0, let d ≤ min{1, 26e}. 


38. See TSM. 

ε 


TSM for complete proof. 1 + | m| 

40. Given any e > 0, let 

⎧ ε ⎫ 

δ ≤ min⎨ 

1 ⎬ 

⎩ 3 , 3 . See TSM. 

13 ⎭ 

41. x must be within 0.05 of 3. 

42. x must be within approx. 0.087 of 3. 

43. See TSM. 

1 

44. Given any e > 0, let N ≤− . See 

TSM for complete proof. ε 

1 

45. Given any e > 0, let M ≥ 

2 

ε 

. See 

TSM for complete proof. 

46. N =− 10 

47. See TSM. 


151 


11/01/17 9:56 am




48. See TSM. 

49. See TSM. 

50. See TSM. 

51. lim fx ( ) = 0; 

x→0 

lim fx ( ) does not exist. 

x→1 

See TSM for discussion and proof. 

52. lim fx ( ) = 0. See TSM for discussion 

x→0 

and proof. 

⎧ ε ⎫ 

53. Given any e > 0, let δ ≤ min⎨1, ⎬. 

⎩ 47 

See TSM for complete proof. ⎭ 

54. See TSM. 

55. M = 101. 

1 

56. Given any e > 0, let δ ≤ 

1 + | a | . See 

TSM for complete proof. 

57. See TSM. 

58. K = 12. 

48. Explain why in the ε-δ definition of a limit, the inequality 

0 < |x − c| 0 if n is even 

x→c x→c 

(p. 95) 

[ ] m/n 

• lim[ f (x)] m/n = lim f (x) , provided [ f (x)] m/n is 

x→c x→c 

defined for positive integers m and n (p. 95) 

[ ] f (x) 

lim f (x) 

x→c 

• lim = , provided lim g(x) = 0 (p. 96) 

x→c g(x) lim g(x) x→c 

x→c 

• If P is a polynomial function, then lim P(x) = P(c). (p. 96) 

x→c 

• If R is a rational function and if c is in the domain of R, 

then lim R(x) = R(c). (p. 96) 

x→c 

1.3 Continuity 

Definitions 

• Continuity at a number (p. 103) 

• Removable discontinuity (p. 105) 

• One-sided continuity at a number (p. 105) 

• Continuity on an interval (p. 106) 

• Continuity on a domain (p. 107) 

152 



11/01/17 9:56 am


Chapter 1 • Chapter Review 153 

Properties of Continuity 

• A polynomial function is continuous on its domain, all real 

numbers. (p. 107) 

• A rational function is continuous on its domain. (p. 107) 

• If the functions f and g are continuous at a number c, and if k 

is a real number, then the functions f + g, f − g, f · g, and 

kf are also continuous at c. If g(c) = 0, the function f g is 

continuous at c. (p. 108) 

• If a function g is continuous at c and a function f is 

continuous at g(c), then the composite function 

( f ◦ g)(x) = f (g(x)) is continuous at c. (p. 109) 

• If f is a one-to-one function that is continuous on its domain, 

then its inverse function f −1 is also continuous on its domain. 

(p. 110) 

The Intermediate Value Theorem Let f be a function that is 

continuous on a closed interval [a, b] with f (a) = f (b). If N is any 

number between f (a) and f (b), then there is at least one number c in 

the open interval (a, b) for which f (c) = N. (p. 110) 

1.4 Limits and Continuity of Trigonometric, Exponential, 

and Logarithmic Functions 

Basic Limits 

sin θ 

• lim = 1 θ→0 θ 

(p. 119) 

cos θ − 1 

• lim = 0 θ→0 θ 

(p. 121) 

• lim sin x = sin c x→c 

(p. 122) 

• lim cos x = cos c x→c 

(p. 122) 

• lim a x = a c ; x→c 

a > 0, a = 1 (p. 124) 

• lim log a x = log a c; x→c 

a > 0, a = 1, and c > 0 (p. 124) 

Squeeze Theorem If the functions f , g, and h have the property that 

for all x in an open interval containing c, except possibly at c, 

f (x) ≤ g(x) ≤ h(x), and if lim f (x) = lim h(x) = L, 

x→c x→c 

then lim g(x) = L. (p. 118) 

x→c 

Properties of Continuity 

• The six trigonometric functions are continuous on their 

domains. (pp. 122–123) 

• The six inverse trigonometric functions are continuous on their 

domains. (p. 123) 

• An exponential function is continuous on its domain, all real 

numbers. (p. 124) 

• A logarithmic function is continuous on its domain, all 

positive real numbers. (p. 124) 

1.5 Infinite Limits; Limits at Infinity; Asymptotes 

Basic Limits 

1 

• lim 

x→0 − x = −∞ lim 1 

=∞ 

x→0 + x 

(p. 128) 

1 

• lim =∞ x→0 x2 (p. 128) 

• lim ln x = −∞ 

x→0 + 

(p. 129) 

1 

• lim 

x→∞ x = 0 lim 1 

= 0 

x→−∞ x 

(p. 132) 

• lim ln x =∞ 

x→∞ 

(p. 136) 

• lim 

x→−∞ ex = 0 

lim 

x→∞ ex =∞ (p. 136) 

Definitions 

• Vertical asymptote (p. 131) 

• Horizontal asymptote (p. 138) 

Properties of Limits at Infinity (p. 132): If k is a real number, 

n ≥ 2 is an integer, and the functions f and g approach real 

numbers as x →∞, then: 

• lim A = A, where A is a constant 

x→∞ 

• lim [kf(x)] = k lim f (x) 

x→∞ x→∞ 

• lim [ f (x) ± g(x)] = lim f (x) ± lim g(x) 

x→∞ x→∞ x→∞ 

[ ][ ] 

• lim [ f (x)g(x)] = lim f (x) lim g(x) x→∞ x→∞ x→∞ 

f (x) 

lim f (x) 

• lim 

x→∞ g(x) = x→∞ 

provided lim g(x) = 0 

lim g(x) x→∞ 

x→∞ 

[ ] n 

• lim [ f x→∞ (x)]n = lim f (x) x→∞ 

√ √ 

n 

• lim f (x) = n lim f (x), where f (x) >0 if n is even 

x→∞ 

x→∞ 

1.6 The ε-δ Definition of a Limit 

Definitions 

• Limit of a Function (p. 145) 

• Limit at Infinity (p. 149) 

• Infinite Limit (p. 150) 

• Infinite Limit at Infinity (p. 150) 

Properties of limits 

• If lim f (x) >0, then there is an open interval around c, for 

x→c 

which f (x) >0 everywhere in the interval, except possibly 

at c. (p. 149) 

• If lim f (x)




TRM Full Solutions to Chapter 1 

Review Exercises and AP® Review 

Problems 

Answers to Chapter 1 Review 

Exercises 

− x 

1. lim 1 cos = 0. For table, 

x→0 

1+ 

cos x 

see TSM. 

2. lim fx ( ) =−3. 

3. 

x→1 

y 

22 21 

22 

24 

26 

28 

1 2 x 

lim fx ( ) does not exist. 

x→2 

y 

7 

6 

5 

4 

3 

2 

4. (a) 3 

(b) 2 

5. − 

x 

3 

2 

1 

6. 6x + 2 

7. 1 

8. 7 2 

9. −π 

10. 2 

11. 0 

12. 0 

13. 27 

14. 3 

15. 4 

16. 1 2 

17. 2 3 

18. −1 

19. − 1 6 

20. − 1 4 

1 2 3 4 

x 

21. 6 

22. 0 

23. 1 

24. 1 

25. −1 

26. 3 

27. 8 

28. 0 

OBJECTIVES 

AP® Review 

Section You should be able to . . . Example Review Exercises Problems 

1.1 1 Discuss the slope of a tangent line to a graph (p. 78) 1 4 

2 Investigate a limit using a table (p. 80) 2–4 1 5 

3 Investigate a limit using a graph (p. 82) 5–8 2, 3 

1.2 1 Find the limit of a sum, a difference, and a product (p. 91) 1–6 8, 10, 12, 14, 22, 

26, 29, 30, 47, 48 

2 Find the limit of a power and the limit of a root (p. 94) 7–9 11, 18, 28, 55 

3 Find the limit of a polynomial (p. 95) 10 10, 22 

4 Find the limit of a quotient (p. 96) 11–14 13–17, 19–21, 3 

23–25, 27, 56 

5 Find the limit of an average rate of change (p. 98) 15 37 

6 Find the limit of a difference quotient (p. 99) 16 5, 6, 49 

1.3 1 Determine whether a function is continuous at a number (p. 103) 1–4 31–36 11 

2 Determine intervals on which a function is continuous (p. 106) 5, 6 39–42 8 

3 Use properties of continuity (p. 108) 7, 8 39–42 

4 Use the Intermediate Value Theorem (p. 110) 9, 10 38, 44–46 6 

1.4 1 Use the Squeeze Theorem to find a limit (p. 117) 1 7, 69 

2 Find limits involving trigonometric functions (p. 119) 2, 3 9, 51–55 4, 10 

3 Determine where the trigonometric functions are continuous (p. 122) 4 63–65 

4 Determine where an exponential or a logarithmic 5 43 

function is continuous (p. 124) 

1.5 1 Investigate infinite limits (p. 128) 1–3 57, 58 

2 Find the vertical asymptotes of a graph (p. 131) 4 61, 62 1 

3 Investigate limits at infinity (p. 131) 5–10 59, 60 2 

4 Find the horizontal asymptotes of a graph (p. 137) 11 61, 62 1 

5 Find the asymptotes of the graph of a rational function (p. 138) 12 67, 68 7 

1.6 1 Use the ε-δ definition of a limit (p. 145) 1–7 50, 66 

REVIEW EXERCISES 

1 − cos x 

1. Use a table of numbers to investigate lim x→0 1 + cos x . 

In Problems 2 and 3, use a graph to investigate lim f (x). 

{ x→c 

2x − 5 if x < 1 

2. f (x) = 

at c = 1 

6 − 9x if x ≥ 1 

{ 

x 

3. f (x) = 

2 + 2 if x < 2 

at c = 2 

2x + 1 if x ≥ 2 

4. For f (x) = x 2 − 3: 

(a) Find the slope of the secant line joining (1, −2) and (2, 1). 

(b) Find the slope of the tangent line to the graph 

of f at (1, −2). 

In Problems 5 and 6, for each function find the limit of the difference 

f (x + h) − f (x) 

quotient lim 

. 

h→0 h 

5. f (x) = 3 6. f (x) = 3x 2 + 2x 

x 

7. Find lim f (x) if 1 + sin x ≤ f (x) ≤|x|+1 

x→0 


8. lim 

(2x − 1 ) 

x→2 x 

9. lim x→π 

(x cos x) 

29. lim fx ( ) = 7; lim fx ( ) = 7; lim fx ( ) = 7. 

x→2 

− x→ 2 

+ x→2 

( 

10. lim x 3 + 3x 2 − x − 1 ) √ 3 

11. lim x(x + 2) 3 

x→−1 

x→0 

12. lim [(2x + 3)(x 5 x 3 − 27 

+ 5x)] 13. lim 

x→0 x→3 x − 3 

( x 2 

14. lim x→3 x − 3 − 3x ) 

x 2 − 4 

15. lim 

x − 3 

x→2 x − 2 

x 2 + 3x + 2 

x 3 + 5x 2 + 6x 

16. lim 

x→−1 x 2 17. lim 

+ 4x + 3 

x→−2 x 2 + x − 2 

18. lim 

(x 2 − 3x + 1 ) 15 

√ 

3 − x 

19. lim 

2 + 5 

x→1 x 

x→2 x 2 − 4 

20. lim x→0 

{ 1 

x 

[ 

1 

(2 + x) 2 − 1 4 

]} 

21. lim x→0 

(x + 3) 2 − 9 

x 

22. lim[(x 3 − 3x 2 + 3x − 1)(x + 1) 2 ] 

x→1 

In Problems 23–28, find each one-sided limit, if it exists. 

x 2 + 5x + 6 

23. lim 

x→−2 + x + 2 

|x − 5| 

24. lim 

x→5 + x − 5 

x 2 − 16 

26. lim 

27. lim 

x→ 3/2 +2x x→4 − x − 4 

In Problems 29 and 30, find lim f (x) and lim 

x→c− x→c 

given c. Determine whether lim f (x) exists. 

{ x→c 

2x + 3 if x < 2 

29. f (x) = 

at c = 2 

9 − x if x ≥ 2 

|x − 1| 

25. lim 

x→1 − x − 1 

28. lim 

x→1 + √ 

x − 1 

+ 

f (x) for the 

154 



11/01/17 9:57 am


⎧ 

⎨ 3x + 1 if x < 3 

30. f (x) = 10 if x = 3 

⎩ 

4x − 2 if x > 3 

at c = 3 

In Problems 31–36, determine whether f is continuous at c. 

⎧ 

⎨ 5x − 2 if x < 1 

31. f (x) = 5 if x = 1 at c = 1 

⎩ 

2x + 1 if x > 1 

⎧ 

⎨ x 2 if x < −1 

32. f (x) = 2 if x =−1 at c =−1 

⎩ 

−3x − 2 if x > −1 

⎧ 

⎪⎨ 

4 − 3x 2 if x < 0 

33. f (x) = 4 if x = 0 at c = 0 

⎪⎩ 

 

16 − x 2 if 0 < x ≤ 4 

⎧ √ 

⎨ 4 + x if − 4 ≤ x ≤ 4 

 

34. f (x) = 

⎩ x 2 − 16 

at c = 4 

if x > 4 

x − 4 

35. f (x) =2x at c = 1 36. f (x) =|x − 5 | at c = 5 

2 

37. (a) Find the average rate of change of f (x) = 2x 2 − 5x from 1 

to x. 

(b) Find the limit as x approaches 1 of the average rate of change 

found in (a). 

38. A function f is defined on the interval [−1, 1] with the following 

properties: f is continuous on [−1, 1] except at 0, negative at −1, 

positive at 1, but with no zeros. Does this contradict the 

Intermediate Value Theorem? 

In Problems 39–43 find all numbers x for which f is continuous. 

39. f (x) = 

x 

x 3 − 27 

41. f (x) = 

2x + 1 

x 3 + 4x 2 + 4x 

43. f (x) = 2 −x 

40. f (x) = 

x 2 − 3 

x 2 + 5x + 6 

42. f (x) = √ x − 1 

44. Use the Intermediate Value Theorem to determine whether 

2x 3 + 3x 2 − 23x − 42 = 0 has a zero in the 

interval [3, 4]. 

In Problems 45 and 46, use the Intermediate Value Theorem to 

approximate the zero correct to three decimal places. 

45. f (x) = 8x 4 − 2x 2 + 5x − 1 on the interval [0, 1] . 

46. f (x) = 3x 3 − 10x + 9; zero between −3 and −2. 

| x | 

| x | 

47. Find lim (1 − x) and lim (1 − x). 

x→0 + x x→0 − x 

|x| 

What can you say about lim (1 − x)? 

x→0 x 

x 2 

48. Find lim x→2 x − 2 − 2x 

. Then comment on the statement 

x − 2 

x 2 

that this limit is given by lim x→2 x − 2 − lim 2x 

x→2 x − 2 . 

Chapter 1 • Review Exercises 155 

f (x + h) − f (x) 

49. Find lim 

for f (x) = √ x. 

h→0 h 

50. For lim(2x + 1) = 7, find the largest possible δ that 

x→3 

“works” for ε = 0.01. 


sin x 

51. lim cos(tan x) 52. lim 

4 

x→0 x→0 x 

53. lim x→0 

tan (3x) 

tan(4x) 

55. lim x→0 

cos x − 1 

x 

54. lim x→0 

cos x 3 − 1 

x 

10 

56. lim x→0 

e 4x − 1 

e x −1 

2 + x 

57. lim tan x 58. lim 

x→π/2 + x→−3 (x + 3) 2 

3x 3 − 2x + 1 

59. lim 

x→∞ x 3 − 8 

3x 4 + x 

60. lim 

x→∞ 2x 2 

In Problems 61 and 62, find any vertical and horizontal 

asymptotes of f . 

61. f (x) = 4x − 2 

62. f (x) = 2x 

x + 3 

x 2 − 4 

⎧ 

tan x ⎪⎨ if x = 0 

2x 

63. Let f (x) = 

. Is f continuous at 0? 

⎪⎩ 1 

if x = 0 

2 

⎧ 

⎨ sin(3x) 

if x = 0 

64. Let f (x) = x 

. Is f continuous at 0? 

⎩ 

1 if x = 0 

 

cos π x + π 

65. The function f (x) = 

2 

is not defined at 0. 

x 

Decide how to define f (0) so that f is continuous at 0. 

66. Use the ε-δ definition of a limit to prove lim 

x→−3 (x2 − 9) = 18. 

67. (a) Sketch a graph of a function f that has the following 

properties: 

f (−1) = 0 

lim f (x) = −∞ lim 

x→4− lim 

x→∞ f (x) = 2 

f (x) =∞ 

x→4 + 

lim f (x) = 2 

x→−∞ 

(b) Define a function that describes your graph. 

68. (a) Find the domain and the intercepts (if any) of 

R(x) = 2x2 − 5x + 2 

5x 2 − x − 2 . 

(b) Discuss the behavior of the graph of R at numbers 

where R is not defined. 

(c) Find any vertical or horizontal asymptotes of the function R. 

69. If 1 − x 2 ≤ f (x) ≤ cos x for all x in the interval − π 2 < x < π 2 , 

show that lim x→0 

f (x) = 1. 

49. 

2 

1 

x 

50. 0.005 

51. 1 

52. 1 4 

53. 3 4 

54. 0 

55. 0 

56. 4 

57. −∞ 

58. −∞ 

59. 3 

60. ∞ 

61. x = −3 is a vertical asymptote. y = 4 

is a horizontal asymptote. 



asymptote. 

63. Yes 

64. No 

65. f (0) = −π 

66. See TSM. 

67. Answers will vary. One possibility is 

(a) 

y 

8 

6 

4 

2 

8 6 4 2 

2 

2 4 6 8 10 12 14 x 

4 

6 

30. lim fx ( ) = 10; lim fx ( ) = 10; lim fx ( ) = 10. 

x→3 

− x→ 3 

+ x→3 


32. f is discontinuous at c =−1. 



1 

35. f is discontinuous at c = 

2 . 


37. (a) 2x − 3 (b) −1 

38. Does not contradict IVT, because f is not 

continuous on the given interval. 

39. f is continuous on { xx | ≠ 3}. 

40. f is continuous on { xx | ≠−3, x ≠−2}. 

41. f is continuous on { xx | ≠−2, x ≠ 0}. 

42. f is continuous on { xx | ≥ 1}. 

43. f is continuous on the set of all real numbers. 

44. f is continuous on [3, 4], f (3) < 0, and 

f (4) > 0, so f has a zero on (3, 4). 

45. 0.215 

46. −2.171 

x 

x 

lim | | (1 − x) = 1, lim | | (1 − x) =−1, 

+ − 

x→0 x 

x→0 

x 

47. 

x 

lim | | (1 − x) does not exist. 

x→0 

x 

⎛ 

2 

x 

48. ⎜ 

⎝ − − 2x 

⎞ 

lim 

− 

⎟ 

⎠ 

= 2. See TSM for 

x→2 

x 2 x 2 

comment. 

2x 

+ 2 

(b) f( x)= 

x − 4 

⎧ 

⎪ 1 

68. (a) ⎨ 

≠ ± 41 

⎫ 

⎪ 

xx ⎬ , x-intercepts 

10 

⎩⎪ 

⎭⎪ 

1 

and 2, y-intercept −1 

2 

(b) See TSM. 

1 

(c) Vertical asymptotes x = − 41 

10 

1 

and x = + 41 

, horizontal 

10 

2 

asymptote y = 

5 

69. See TSM. 

Chapter 1 • Chapter Review 

155 


11/01/17 9:57 am




Answers to Chapter 1 AP® 

Review Problems 

1. B 

2. B 

3. C 

4. C 

5. C 

6. B 

7. C 

8. C 

9. D 

10. B 

11. D 

TRM Full Solutions for the 

Chapter 1 Project 

CHAPTER 1 AP® REVIEW PROBLEMS 

u 1. Which line is an asymptote to the graph of f (x) = e 2x ? 

u 2. 

(A) x = 0 (B) y = 0 (C) y = 2 (D) y = x 

3x 3 + 4x 

lim 

x→∞ 5 − 2x 4 = 

(A) − 3 2 

(B) 0 

(C) 

3 

5 

(D) 

3 

2 

u 3. If f (x) = 5x 3 − 1, then lim x→0 

f (x) − f (0) 

x 3 = 

(A) 0 (B) 1 (C) 5 (D) The limit does not exist. 

θ 2 

u 4. lim θ→0 1 − cos θ = 

(A) 0 (B) 1 (C) 2 (D) 4 

u 5. The table gives values of three functions: 

x –0.15 –0.1 –0.05 0 0.05 0.1 0.15 

f(x) 0.075 0.05 0.025 –4 0.025 0.05 0.075 

g(x) –8.3 –8.2 –8.1 undefined –7.9 –7.8 –7.7 

h(x) 1.997 1.99 1.9975 1 1.005 1.02 1.045 

For which of these functions does the table suggest that the 

limit as x approaches 0 exists? 

(A) f only (B) h only 

(C) f and g only (D) f and h only 

u 6. If a function f is continuous on the closed interval [1, 4] and if 

f (1) = 6 and f (4) = –1, then which of the following must be 

true? 

(A) f (c) = 0 for some number c in the open interval (–1, 6). 

(B) f (c) = 1 for some number c in the open interval (1, 4). 

(C) f (c) = 1 for some number c in the open interval (–1, 6). 

(D) f (c) = –2 for any number c in the open interval (1, 4). 

u 7. Which are the equations of the asymptotes of the graph of the 

x 

function f (x) = 

x(x 2 − 9) ? 

(A) x = –3, x = 0, x = 3, y = 0 

(B) x = –3, x = 0, x = 3, y = 1 

(C) x = –3, x = 3, y = 0 

(D) x = –3, x = 3, y = 1 

u 8. Find the value of k that makes the function 

{ 

x 

f (x) = 

2 + 2 if x ≤−1 

kx + 4 if x > −1 

continuous for all real numbers. 

(A) –3 (B) –1 (C) 1 (D) 3 

u 9. An odd function f is continuous for all real numbers. 

If lim f (x) =−2, then which of the statements 

x→∞ 

must be true? 

I. f has no vertical asymptotes. 

II. lim f (x) = 0 

x→0 

III. The horizontal asymptotes of the graph of f are y = –2 

and y = 2. 

(A) I only (B) III only 

(C) I and III only (D) I, II, and III 

u 10. lim x→0 

sin(2x) 

tan(3x) = 

(A) 0 

(B) 

2 

3 

(C) 1 

(D) 

u 11. If a function f is continuous for all real numbers, and if 

f (x) = x3 + 8 

when x = −2, then f (–2) = 

x + 2 

(A) 0 (B) 4 (C) 8 (D) 12 

3 

2 

CHAPTER 1 PROJECT 

Pollution in Clear Lake 

AP Photo 

The Toxic Waste Disposal 

Company (TWDC) specializes in 

the disposal of a particularly 

dangerous pollutant, Agent Yellow 

(AY). Unfortunately, instead of 

safely disposing of this pollutant, 

the company simply dumped AY 

in (formerly) Clear Lake. 

Fortunately, they have been caught and are now defending 

themselves in court. 

The facts below are not in dispute. As a result of TWDC’s 

activity, the current concentration of AY in Clear Lake is now 

10 ppm (parts per million). Clear Lake is part of a chain of rivers 

and lakes. Fresh water flows into Clear Lake and the 

contaminated water flows downstream from it. The Department 

of Environmental Protection estimates that the level of 

contamination in Clear Lake will fall by 20% each year. These 

facts can be modeled as 

p(0) = 10 p(t + 1) = 0.80p(t) 

where p = p(t), measured in ppm, is the concentration of 

pollutants in the lake at time t, in years. 

1. Explain how the above equations model the facts. 

2. Create a table showing the values of t for t = 0, 1, 2,...,20. 

3. Show that p(t) = 10(0.8) t . 

4. Use technology to graph p = p(t). 

5. What is lim p(t)? 

t→∞ 

Lawyers for TWDC looked at the results in 1–5 above and argued 

that their client has not done any real damage. They concluded 

that Clear Lake would eventually return to its former clear and 

unpolluted state. They even called in a mathematician, who wrote 

the following on a blackboard: 

lim p(t) = 0 

t→∞ 

and explained that this bit of mathematics means, descriptively, 

that after many years the concentration of AY will, indeed, be 

close to zero. 

156 



11/01/17 9:57 am


Chapter 1 Project • Pollution in Clear Lake 157 

Concerned citizens booed the mathematician’s testimony. Fortunately, 

one of them has taken calculus and knows a little bit about limits. She 

noted that, although “after many years the concentration of AY will 

approach zero,” the townspeople like to swim in Clear Lake and state 

regulations prohibit swimming unless the concentration of AY is 

below 2 ppm. She proposed a fine of $100,000 per year for each full 

year that the lake is unsafe for swimming. She also questioned the 

mathematician, saying, “Your testimony was correct as far as it went, 

but I remember from studying calculus that talking about the eventual 

concentration of AY after many, many years is only a small part of the 

story. The more precise meaning of your statement lim p(t) = 0 is 

t→∞ 

that given some tolerance T for the concentration of AY, there is some 

time N (which may be very far in the future) so that for all t > N, 

p(t) N, 

then p(t) N, 

then p(t)




TRM Full Solutions for AP® 

Practice Exam, Big Idea 1: Limits 

Answers to AP® Practice Exam, 

Big Idea 1: Limits 

1. A 

2. B 

3. C 

4. C 

5. C 

6. B 

7. A 

8. D 

9. C 

10. A 

11. A 

12. D 

13. B 

14. C 

15. C 

AP ® Practice Exam, Big Idea 1: Limits 

Section 1: Multiple Choice 

3x 4 − 6x 3 + 3x 2 

1. lim x→1 x 3 − 3x 2 = 

+ 2x 

(A) 0 (B) 1 (C) 2 (D) nonexistent 

2. lim x→4 

x − 4 

4 − x = 

(A) –4 (B) –1 (C) 0 (D) nonexistent 

 

3x + sin x 

3. lim 

= 

x→0 2x 


4. Suppose a, b, and c are constants. Find 

4ax 4 − 2bx 3 + cx − 1 

lim 

x→∞ x 4 + b − 2 

(A) a (B) 2 (C) 4a (D) nonexistent 

x 2 − 16 

5. lim √ = 

x→4 x − 2 


6. lim x→0 

e x sin x tan x 

x 2 = 

(A) 0 (B) 1 (C) e (D) nonexistent 

7. Let h be defined by 

h(x) = 

 

f (x) · g(x) x ≤ 1 

k + x x > 1 

If lim f (x) = 2 and lim g(x) =−2, then for what value 

x→1 x→1 

of k is h continuous? 

(A) –5 (B) –4 (C) –2 (D) 2 

Use the figure below for Questions 8 and 9. 

y 

a b c d x 

Graph of f 

8. The graph of f is shown in the figure above. Which of the 

following statements is false? 

(A) lim f (x) exists. x→a 

(B) lim f (x) exists. 

x→b 

(C) lim f (x) exists. 

x→c− (D) lim f (x) does not exist. 

x→d− 9. The graph of f is shown in the figure above. If f is defined at k, 

but lim f (x) does not exist, then k = 

x→k 

(A) a (B) b (C) c (D) 0 

10. Which of the following functions has the horizontal asymptotes 

y = 1 and y = –1? 

(A) f (x) = 2 π tan−1 (x) (B) f (x) = e −x + 1 

(C) f (x) = 1 − x2 

1 + x 2 (D) f (x) = 2x2 − 1 

2x 2 + x 

11. Let f be a continuous function for which f (–2) = 1 and 

f (5) = –3. The function g is also continuous and g(x) = f −1 (x) 

for all x. The Intermediate Value Theorem guarantees that 

(A) g(c) = 2 for at least one c between –3 and 1. 

(B) g(c) = 0 for at least one c between –2 and 5. 

(C) f (c) = 0 for at least one c between –3 and 1. 

(D) f (c) = 2 for at least one c between –2 and 5. 

12. The line x = c is a vertical asymptote of the graph of the 

function f . Which of the following statements cannot be true? 

(A) lim f (x) =∞ (B) lim f (x) = c 

x→c− x→∞ 

(C) f (c) is undefined (D) f is continuous at x = c. 

x 2 − 9 

13. The graph of the function f (x) = 

2x 2 has a vertical 

− 5x − 3 

asymptote at x = a and a horizontal asymptote at y = b. What 

are the values of the constants a and b? 

(A) a = –3, b = 2 (B) a =− 1 2 , b = 1 2 

(C) a = 1 2 , b = 1 (D) a = 3, b = 2 

2 

14. The graph of y = 2x2 + 2x + 3 

4x 2 has 

− 4x 

(A) a horizontal asymptote at y = 1 , but no vertical asymptote. 

2 

(B) no horizontal asymptote, but vertical asymptotes at x = 0 

and x = 1. 

(C) a horizontal asymptote at y = 1 and vertical asymptotes at 

2 

x = 0 and x = 1. 

(D) a horizontal asymptote at x = 2, but no vertical asymptote. 

⎧ 

⎨ x 2 + x 

if x = 0 

15. Let f (x) = x 

⎩ 

1 if x = 0 

Which of the following statements is true? 

I. f (0) exists 

II. lim f (x) exists 

x→0 

III. f is continuous at x = 0 

(A) I only (B) II only (C) I, II, III (D) None 

158 



11/01/17 9:57 am


Chapter 1 • AP ® Practice Exam, Big Idea 1: Limits 159 

16. Let f be the function defined by 

⎧ 

5x + 17 if −5 ≤ x < −3 

⎪⎨ 

4 if x =−3 

f (x) = 

⎪⎩ 

8 − x 2 if −3 < x < 3 

x − 4 if x ≥ 3 

For what values of x is f NOT continuous? 

(A) none (B) –3 only (C) 3 only (D) –3 and 3 

k 

17. If lim = 0, which of the following combinations of p and k 

x→∞ x p 

is not possible? 

(A) p = –1, k = 1 (B) p = 0, k = 0 

(C) p = 2, k = –1 (D) p = 3, k = –2 

18. Which of the following functions does not satisfy 

lim f (x) =∞? 

x→∞ 

(A) f (x) = e 2x (B) f (x) = ln(x − 5) 

(C) f (x) = 3e −x (D) f (x) = √ x + 4 

19. If f (x) = x2 − 9 

which of the following is false? 

x − 3 

(A) lim f (x) = 6 

x→3 

(B) f is continuous for all real numbers x. 

(C) lim f (x) =∞ 

x→∞ 

(D) The domain of f is {x|x = 3} 

20. Let f be the function given by f (x) = x2 − 3x 

. For what value 

x − a 

of a is f continuous for all real numbers x? 

(A) None (B) a = –3 (C) a = –1 (D) a = 3 

Section 2: Free Response 

Show all of your work. Be sure to indicate clearly the methods you use 

to obtain your results. 

1. The tax on gross income in a small country is computed using 

the table below. 

Gross Income, i 

Tax, T 

$0 up to but not 

$0 

including $30,000 

$30,000 up to but not 10% of gross income 

including $50,000 

$50,000 and higher $5000 + 20% of gross income 

in excess of $50,000 

(a) Use a piecewise-defined function to represent T as a 

function of i. 

(b) Find lim 

i→0 + T (i). 

(c) Determine the numbers, if any, at which T is discontinuous. 

(d) Use the definition of continuity to explain your answer to (c). 

16. B 

17. A 

18. C 

19. B 

20. A 

Free Response 

1. (a) 

⎧0 if 0≤ i < 30,000 

⎪ 

Ti () = ⎨0.1i 

if 30,000 ≤ i < 50,00 

⎪ 

⎩⎪ 

0.2i− 5000 if i > 50,00 

(b) 0 

(c) T is discontinuous at 30,000. 

(d) Answers will vary. 

Chapter 1 • AP® Practice Exam, Big Idea 1: Limits 

159 


11/01/17 9:57 am

Sullivan Microsite TE SAMPLE

Create successful ePaper yourself

Delete template?

Save as template?