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Sullivan AP˙Sullivan˙Chapter01 October 8, 2016 17:4 Sullivan SullivanAP 88 84 Chapter 1 • Limits and Continuity 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 x 3 (a) Find the slope m if x < −1 sec of the secant line containing the both one-sided limits exist at c and both one-sided limits are equal. That is, 34. f (x) = x 2 at c =−1 points P = (2, f (2)) and Q = (2 + h, f (2 + h)). − 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 (e) On the same set of axes, graph f and the tangent line to f at properties. Answers will vary. TABLE 5 1 P = (2, f (2)). 37. lim f (x) = 3; lim f (x) = 3; lim f (x) = 1; x approaches 1 from the right x→2 x→3− x→3 + −−−−−−−−−−−−−−−−−−−−−→ f (2) = 3; f (3) = 1 54. Slope of a Tangent Line For f (x) = x 2 − 1: 1 2 3 4 5 6 x x 1.009 1.0009 1.00009 1.000009 1.0000009 → 1 38. lim f (x) = 0; lim f (x) =−2; lim =−2; (a) Find the slope m sec of the secant line containing the 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 + points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)). 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 + (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 (c) Investigate the limit of the slope of the secant line found EXAMPLE 7 Investigatingin a(a) Limit as h → 0. In Problems 41–50, use either a graph or a table to investigate (d) What is the slope of the tangent line to the graph of f at the 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 (e) On the same set of axes, graph f and the tangent line to f 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 by using a table and evaluating the : 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). (b) Find the slope of the secant line containing the points (2, 12) 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) Compare the results from (a) and (b). What do you conclude (c) Create a table to investigate the slope of the tangent Now suppose line to the we let x approach zero as follows: about the limit? Why do you think this happens? What is graph of f at 2 using the result from (b). 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). x approaches 0 from the right −−−−−−−−−−−−−−−−−−−−→ (d) Use technology to graph f . Begin with the x-window 52. Slope of a Tangent Line For f (x) = x 3 ←−−−−−−−−−−−−−−−−−−−−− : [−2π, 2π] and the y-window [−1, 1]. If you were finding x − 2 − 2 − 2 − 2 − 2 2 2 2 2 2 → 0 (a) Find the slope of the secant line containing the points (2, 8) 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) calculator is set to the radian mode.) and (x, f (x)), x = 2. 56. (a) Investigate lim cos π by using a table and evaluating the (c) Create a table to investigate the slope of the tangent line to the x→0 x2 Table 7 suggests that lim sin π √ 2 graph of f at 2 using the result from (b). function f (x) = cos π x→0 x = 2 2 ≈ 0.707. at x =−0.1, −0.01, −0.001, (d) On the same set of axes, graph f , the tangent line to the graph x2 In fact, by carefully selecting x, we can make f appear to approach any number in of f at the point (2, 8), and the secant line from (a). −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. WEB SITE 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 Chapter 1 • Limits and Continuity TE_Sullivan_Chapter01_PART 0.indd 13 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 (d) Use the graph to investigate lim C(w) and lim C(w). Do the choice of lim sin π = 0 seems reasonable. But in Figure w→115(b), − it appears w→1 + that x→0 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 . 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 about the limit? Why do you think this happens? What is your 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 + (d) Use technology to graph f . Begin with the x-window (g) Use the graph to investigate lim C(w). w→13 [−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.) 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 Figure (5 − 2x). 15 x→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? 59. First-Class Mail As of April 2016, the U.S. Postal Service charged $0.47 postage for EXAMPLE 8 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 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 (a) Find a function C that models the first-class postage charged, in dollars, for a letter weighing w ounces. Assume w>0. NEED TO REVIEW? Inequalities (b) What is the domain of C? involving absolute values are discussed in Appendix (c) A.1, Graph p. A-7. the function C. (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 + y (f) Use the graph to investigate lim C(w). 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 and including 1 ounce, plus a flat fee of $0.21 for each additional 6 or partial ounce up to and including 13 ounces. First-class rates do not apply to large envelopes weighing more than 13 ounces. 4 Source: U.S. Postal Service Notice 123 Source: Submitted by the students of Millikin University. 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) y = f (x) at the point (c, f (c)) is m tan = lim . 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? your reasoning. 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 (a) What straight line and what point? |2x − 4| ≤0.1 (b) Use the graph of f to investigate the one-sided limits of f as |2(xx− approaches 2)| ≤0.13. (c) Does the graph suggest that lim f (x) exists? If so, what is it? |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. CAS (d) Find lim(1 + x) 1/x . (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 in dollars, for a large envelope weighing w ounces. Assume w>0. 1 2 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 Notice that the closer we require f to be to the limit 9, the narrower the interval for (b) What is the domain of C? DF Figure 16 f (x) = 2x + 5 x becomes. See Figure 16. Kathryn Sidenstricker /Dreamstime.com 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) 24p ≤ x ≤ 4p (b) 2p ≤ x ≤ p (c) 21 ≤ x ≤ 1 (a) Write Professor Smith’s claim symbolically as a limit. ■ (b) Write Professor Smith’s claim using the ε-δ definition of limit. NOW WORK Problem 55. NOW WORK Problem 57. 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. 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