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Sullivan AP˙Sullivan˙Chapter01 October 8, 2016 17:4
Sullivan
88 92 Chapter 1 • Limits and Continuity
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
sec of the secant line containing the
34. f (x) =
x 2 at c =−1
points P = (2, f (2)) and Q = (2 + h, f (2 + h)).
− 1 if x > −1
Solution F(x) = x + 4 is the sum of two functions f (x) = x and g(x) = 4.
(b) Use the result from (a) to complete the following table:
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
■
(d) What is the slope of the tangent line to the graph of f at the
Applications and Extensions
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
properties. Answers will vary.
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:
x→c x→c x→c
of two functions equals
38. f (x) = 0; lim
the difference of
their limits.
f (x) =−2; lim f (x) =−2;
(a) Find the slope m sec of the secant line containing the
x→−1 x→2− x→2 +
points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)).
f (−1) is not defined; f (2) =−2 EXAMPLE 2 Finding the Limit of a Difference
39. lim f (x) = 4; lim f (x) =−1; lim
(b) Use the result from (a) to complete the following table:
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.
40. lim f (x) = 2; lim f (x) = 0; lim f (x) = 1;
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
(d) What is the slope of the tangent line to the graph of f at the
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
(e) On the same set of axes, graph f and the tangent line to f
−
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)
x→c x→c x→c
12 ,..., 1
12 , 1 10 , 1 8 , 1 4 , 1 limits.
2 .
49. lim x−x 50. lim x−x
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
(a) Find the slope of the secant line containing the points (2, 12)
function off a(x) Product = cos π x at
and (3, 27).
Find:
(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 3 , 1.
and (x, f (x)), x = 2.
(a) lim x 2 (b) lim (−4x)
x→3 x→−5
(c) Compare the results from (a) and (b). What do you conclude
(c) Create a table to investigate the slope of the tangent line to the
about the limit? Why do you think this happens? What is
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.
your view about using a table to draw a conclusion about
(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?
of f at the point (2, 12), and the secant line from (a).
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
:
[−2π, 2π] and the y-window [−1, 1]. If you were finding
(b) F(x) =−4x is the product of two functions, f (x) =−4 and g(x) = x. Then,
(a) Find the slope of the secant line containing the points (2, 8)
lim f (x) using a graph, what would you conclude? Zoom in
using the Limit of a Product, wex→0 have
and (3, 27).
on the graph. Describe what you see. (Hint: Be sure your
(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.
56. (a) Investigate lim cos π by using a table and evaluating the
(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,
(d) On the same set of axes, graph f , the tangent line to the graph
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
Chapter 1 • Limits and Continuity
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