C H A P T E R 2
Limits and Their Properties
Section 2.1
A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 71
Section 2.2
Finding Limits Graphically and Numerically . . . . . . . . 72
Section 2.3
Evaluating Limits Analytically
Section 2.4
Continuity and One-Sided Limits . . . . . . . . . . . . . . 94
Section 2.5
Infinite Limits
Review Exercises
. . . . . . . . . . . . . . . 83
. . . . . . . . . . . . . . . . . . . . . . . 105
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
C H A P T E R 2
Limits and Their Properties
Section 2.1
A Preview of Calculus
1. Precalculus: 20 ftsec15 seconds 300 feet
2. Calculus: velocity is not constant
Distance 20 ftsec15 seconds 300 feet
3. Calculus required: slope of tangent line at x 2 is rate of
change, and equals about 0.16.
4. Precalculus: rate of change slope 0.08
1
1
15
5. Precalculus: Area 2 bh 2 53 2 sq. units
6. Calculus required: Area bh
22.5
5 sq. units
7. f x 4x x2
(a)
8. f x x
y
(a)
y
P(4, 2)
4
3
2
P
x
1
x
1
2
2
3
4
5
3
(b) slope m 4x x2 3
x1
(b) slope m x 13 x
3 x, x 1
x1
x 2: m 3 2 1
x 1: m x 1.5: m 3 1.5 1.5
x 3: m x 0.5: m 3 0.5 2.5
(c) At P1, 3 the slope is 2.
x 5: m You can improve your approximation of the slope at
x 1 by considering x-values very close to 1.
x 2
x4
x 2
x 2x 2
1
1 2
1
3 2
1
5 2
(c) At P4, 2 the slope is
1
x 2
, x4
1
3
0.2679
0.2361
1
4 2
1
0.25.
4
You can improve your approximation of the slope at
x 4 by considering x-values very close to 4.
5 5 5
10.417
2 3 4
1
5
5
5
5
5
5
5
Area 5 9.145
2
1.5 2 2.5 3 3.5 4 4.5
9. (a) Area 5 (b) You could improve the approximation by using more rectangles.
71
72
Chapter 2
Limits and Their Properties
10. (a) For the figure on the left, each rectangle has width
Area 3
sin sin sin
sin 4
4
2
4
2
2
1
4 2
2
2 1
4
.
4
1.8961
For the figure on the right, each rectangle has width
Area .
6
2 5
sin sin sin sin sin
6
6
3
2
3
6
3
1
1 3
1
6 2
2
2
2
3 2
6
1.9541
(b) You could obtain a more accurate approximation by using more rectangles. You will learn later that the exact area is 2.
11. (a) D1 5 12 1 52 16 16 5.66
5
5
5
5
5
5
(b) D2 1 2 1 2 3 1 3 4 1 4 1
2
2
2
2
2.693 1.302 1.083 1.031 6.11
(c) Increase the number of line segments.
Section 2.2
1.
x
1.9
1.99
1.999
2.001
2.01
2.1
f x
0.3448
0.3344
0.3334
0.3332
0.3322
0.3226
lim
x→2
2.
x2
0.3333
x2 x 2
Actual limit is 13 .
x
1.9
1.99
1.999
2.001
2.01
2.1
f x
0.2564
0.2506
0.2501
0.2499
0.2494
0.2439
lim
x→2
3.
Finding Limits Graphically and Numerically
x2
0.25
x2 4
x
2.9
f x
lim
x→3
0.0641
Actual limit is 14 .
2.99
2.999
3.001
3.01
3.1
0.0627
0.0625
0.0625
0.0623
0.0610
1x 1 14
0.0625
x3
Actual limit is 161 .
Section 2.2
4.
x
3.1
3.01
3.001
2.999
2.99
2.9
f x
0.2485
0.2498
0.2500
0.2500
0.2502
0.2516
lim
1 x 2
x→3
5.
0.9983
x
0.1
lim
x→0
0.01
0.0500
0.001
0.0050
0.1
f x
1.0000
0.001
0.01
0.1
1.0000
0.99998
0.9983
(Actual limit is 1.) (Make sure you use radian mode.)
cos x 1
0.0000
x
x
0.001
0.99998
sin x
1.0000
x
f x
7.
0.01
lim
x→0
Actual limit is 14 .
0.25
x3
0.1
x
f x
6.
Finding Limits Graphically and Numerically
0.01
0.1
0.0005
0.0050
0.0500
(Actual limit is 0.) (Make sure you use radian mode.)
0.01
0.9516
0.0005
0.001
0.001
0.9950
0.9995
0.001
0.01
0.1
1.0005
1.0050
1.0517
ex 1
1
x→0
x
lim
8.
0.1
x
f x
lim
x→0
9.
0.1
f x
lim
10.
3.99982
0.001
4
4
0.001
0.01
0.1
0
0
0.00018
4
does not exist.
1 e1x
x
x→0
0.01
0.01
1.0536
0.001
1.0050
1.0005
0.001
0.01
0.1
0.9995
0.9950
0.9531
lnx 1
1
x
x
1.9
1.99
1.999
2.001
2.01
2.1
f x
0.5129
0.5013
0.5001
0.4999
0.4988
0.4879
lim
x→2
ln x ln 2
1
x2
2
73
74
Chapter 2
Limits and Their Properties
11. lim 4 x 1
12. lim x2 2 3
x→3
13. lim
x→1
x 3 does not exist. For values of x to the left of 3, x 3x 3 equals 1,
x3
whereas for values of x to the right of 3, x 3 x 3 equals 1.
x→3
14. lim f x lim x2 3 4
x→1
x→1
3
x ln x 2 0
15. lim x→1
17. lim tan x does not exist since the function increases and
x→ 2
16. lim
x→0
4
does not exist.
2 e1x
18. lim 2 cos1x does not exist since the function oscillates
x→0
between 2 and 2 as x approaches 0.
decreases without bound as x approaches 2.
1
does not exist since the function increases and
x2
decreases without bound as x approaches 2.
19. lim sec x 1
20. lim
21. (a) f 1 exists. The black dot at 1, 2 indicates that
f 1 2.
(c) f 4 does not exist. The hollow circle at 4, 2 indicates
that f is not defined at 4.
x→2
x→0
(b) lim f x does not exist. As x approaches 1 from the
x→1
(d) lim f x exists. As x approaches 4, f x approaches 2:
x→4
left, f x approaches 2.5, whereas as x approaches 1
from the right, f x approaches 1.
22. (a) f 2 does not exist. The vertical dotted line indicates
that f is not defined at 2.
(b) lim f x does not exist. As x approaches 2, the
x→2
values of f x do not approach a specific number.
(c) f 0 exists. The black dot at 0, 4 indicates that
f 0 4.
(d) lim f x does not exist. As x approaches 0 from the
x→0
lim f x 2.
x→4
(e) f 2 does not exist. The hollow circle at 2, 12 indicates
that f 2 is not defined.
(f) lim f x exists. As x approaches 2, f x approaches 12 :
x→2
lim f x 12.
x→2
(g) f 4 exists. The black dot at 4, 2 indicates that f 4 2.
(h) lim f x does not exist. As x approaches 4, the values
x→4
of f x do not approach a specific number.
left, f x approaches whereas as x approaches 0
from the right, f x approaches 4.
1
2,
23. lim f x exists for all c 3. In particular, lim f x 2.
24. lim f x exists for all c 2, 0. In particular, lim f x 2.
25.
26.
x→c
x→2
y
x→c
x→4
y
6
2
5
4
1
3
f
2
1
−2 −1
−1
1
2
3
4
π
2
−π
2
x
5
π
x
−1
−2
lim f x exists for all values of c 4.
x→c
lim f x exists for all values of c .
x→c
Section 2.2
y
27. One possible answer is
Finding Limits Graphically and Numerically
y
28. One possible answer is:
6
4
5
3
4
f
2
2
1
1
−2 −1
−1
75
−3
x
1
2
3
4
−2
x
−1
1
2
−1
5
29. Ct 0.75 0.50 t 1
(a)
(b)
3
3
3.3
3.4
3.5
3.6
3.7
4
C 1.75
2.25
2.25
2.25
2.25
2.25
2.25
lim Ct 2.25
5
0
t
t→3.5
0
(c)
t
2
2.5
2.9
3
3.1
3.5
4
C
1.25
1.75
1.75
1.75
2.25
2.25
2.25
lim Ct does not exist. The values of C jump from 1.75 to 2.25 at t 3.
t→3
30. Ct 0.35 0.12 t 1
(a)
(b)
1
t
Ct
0
3
3.3
3.4
3.5
3.6
3.7
4
0.59
0.71
0.71
0.71
0.71
0.71
0.71
lim Ct 0.71
5
0
t→3.5
(c)
t
Ct
2
2.5
2.9
3
3.1
3.5
4
0.47
0.59
0.59
0.59
0.71
0.71
0.71
lim Ct does not exist. The values of C jump from 0.59 to 0.71 at t 3.
t→3
31. We need f x 3 x 1 3 x 2 < 0.4. Hence, take 0.4. If 0 < x 2 < 0.4, then
x 2 x 1 3 f x 3 < 0.4, as desired.
76
Chapter 2
Limits and Their Properties
32. We need f x 1 1
2x
1
1
1 < 0.01. Let . If 0 < x 2 <
, then
x1
x1
101
101
1
1
1
1
< x2 <
⇒ 1
< x1 < 1
101
101
101
101
⇒
100
102
< x1 <
101
101
⇒ x1 >
and we have
f x 1 100
101
1
1
2x
1101
1 <
0.01.
x1
x1
100101 100
33. You need to find such that 0 < x 1 < implies
1
f x 1 1 < 0.1. That is,
x
0.1 <
1
1 < 0.1
x
1 0.1 <
1
x
< 1 0.1
9
<
10
1
x
<
10
>
9
So take x
1
. Then 0 < x 1 < implies
11
1
1
< x1 <
11
11
1
1
< x1 < .
11
9
Using the first series of equivalent inequalities, you obtain
11
10
f x 1 10
>
11
1
1 < 0.1.
x
10
10
1 > x 1 >
1
9
11
1
1
> x 1 > .
9
11
34. You need to find such that 0 < x 2 < implies
f x 3 x2 1 3 x2 4 < 0.2. That is,
0.2
4 0.2
3.8
3.8
3.8 2
< x2 4
< x2
< x2
<
x
< x2
<
<
<
<
<
0.2
4 0.2
4.2
4.2
4.2 2
So take 4.2 2 0.0494.
Then 0 < x 2 < implies
4.2 2 < x 2 < 4.2 2
3.8 2 < x 2 < 4.2 2.
Using the first series of equivalent inequalities, you obtain
f x 3 x2 4 < 0.2.
35. lim 3x 2 8 L
x→2
3x 2 8 < 0.01
3x 6 < 0.01
3x 2 < 0.01
0.01
0.0033 3
0.01
Hence, if 0 < x 2 < , you have
3
0 < x2 <
3x 6 < 0.01
3x 2 8 < 0.01
f x L < 0.01.
3 x 2 < 0.01
Section 2.2
36. lim 4 x→4
x
2L
2
x→2
x2 3 1 < 0.01
x2 4 < 0.01
x 2x 2 < 0.01
x 2 x 2 < 0.01
x
< 0.01
2
1
x 4 < 0.01
2
0.01
Hence, if 0 < x 4 < 0.02, you have
x 2 < x 2
0 < x 4 < 0.02 1
x 4 < 0.01
2
2
4
If we assume 1 < x < 3, then 0.015 0.002.
1
x 2x 2 < 0.01
x2 4 < 0.01
x2 3 1 < 0.01
f x L < 0.01.
f x L < 0.01.
38. lim x2 4 29 L
39. lim x 3 5
x→5
x→2
x2 4 29 < 0.01
Given > 0:
x 3 5 < x 2 < x2 25 < 0.01
x 5x 5 < 0.01
Hence, let .
0.01
x 5 < x 5
0.01
, you have
11
x 5 <
0.01
1
<
0.01
11
x5
x 5x 5 < 0.01
x2 25 < 0.01
x2 4 29 < 0.01
f x L < 0.01.
Hence, if 0 < x 2 < , you have
If we assume 4 < x < 6, then 0.0111 0.0009.
Hence, if 0 < x 5 < x 3 5 < f x L < .
x2 < Hence, if 0 < x 3 < , you have
2
40. lim 2x 5 1
x→3
Given > 0:
2x 5 1 < 2x 6 < 2x 3 < 1
x 2 < 0.002 50.01 < x 20.01
x
2 < 0.01
2
x 3 < 2 Hence, let 2.
Hence, if 0 < x 2 < 0.002, you have
x
< 0.01
2
77
37. lim x2 3 1 L
x
4
2 < 0.01
2
2
Finding Limits Graphically and Numerically
x 3 < 2
2x 6 < 2x 5 1 < f x L < .
78
Chapter 2
41. lim
x→4
Limits and Their Properties
12 x 1 12 4 1 3
42. lim 23 x 9 23 1 9 29
3
x→1
Given > 0:
x 1 3
x 2
1
2
1
2
1
2
Given > 0:
x 9 x 2
3
< 2
3
< x 4 < x 4 < 2
2
3
1 3
1
2x
2 < < f x L < .
x→6
2
3x
2
3x
29
3
< f x L < .
x→2
Hence, any > 0 will work.
Hence, any > 0 will work.
Hence, for any > 0, you have
Hence, for any > 0, you have
0 < 0 < 1 1 < f x L < .
33 < f x L < .
3
45. lim x0
46. lim x 4 2
x→0
Given > 0:
x
x→4
Given > 0:
x0 < 3
3
x 2
<
x 2
Hence for 0 < x 2 < , you have
Given > 0:
Hence, let .
<
⇒ x 2 < .
47. lim x 2 2 2 4
x 2 4 < x 2 4 < x 2 x 2 x 2 < 0 < x 4 < 3 ⇒ x 4 < x 2
Hence for 0 < x 0 < 3, you have
x < 3
3 x < 3 x 0 < f x L < .
x 2 x 4 x 2 < Assuming 1 < x < 9, you can choose 3. Then,
Hence, let 3.
x 2 < x < 3 23 < 9 33 < x→2
Given > 0: 1 1 < Given > 0:
x 1 < 32
44. lim 1 1
43. lim 3 3
x 1 < Hence, if 0 < x 1 < 32, you have
x 4 < 2
1
2x
< Hence, let 32.
Hence, if 0 < x 4 < 2, you have
< 2
3
x 1 < 32
Hence, let 2.
29
3
x 2 < 0
x 2 < x 2 < x 2 4 < x 2 4 < f x L < .
(because x 2 < 0)
Section 2.2
Finding Limits Graphically and Numerically
48. lim x 3 0
49. lim x2 1 2
Given > 0:
Given > 0:
x→3
x→1
x 3 0 < x 3 < x2 1 2 < x2 1 < x 1x 1 < Hence, let .
79
Hence for 0 < x 3 < , you have
x 1 < x 1
x 3 < x 3 0 < f x L < .
If we assume 0 < x < 2, then 3.
Hence for 0 < x 1 < , you have
3
1
1
x 1 < 3 < x 1
x2 1 < x2 1 2 < f x 2 < .
Hence for 0 < x 3 < , you have
4
50. lim x2 3x 0
x→3
Given > 0:
1
xx 3 < xx 3 < 3x 0 < f x L < .
x2
x 3 < x
If we assume 4 < x < 2, then 4.
x 5 3
x→4
0.5
52. f x x4
lim f x 1
6
1
x 3 < 4 < x
x2 3x 0 < 51. f x −6
6
x2
x3
4x 3
lim f x x→3
4
−3
1
2
5
−0.1667
−4
The domain is 5, 4 4, .
The graphing utility does not show the hole at 4, 16 .
53. f x x9
x 3
The domain is all x 1, 3. The graphing utility does not
1
show the hole at 3, 2 .
54. f x 10
lim f x 6
e x2 1
x
lim f x x→9
x→0
0
10
55. lim f x 25 means that the
x→8
values of f approach 25 as x gets
closer and closer to 8.
−2
2
−1
0
The domain is all x ≥ 0 except x 9. The graphing
utility does not show the hole at 9, 6.
1
2
2
The domain is all x 0. The graphing utility does not
show the hole at 0, 12 .
56. No. The fact that f 2 4 has no
bearing on the existence of the
limit of f x as x approaches 2.
57. No. The fact that lim f x 4 has
x→2
no bearing on the value of f at 2.
80
Chapter 2
Limits and Their Properties
58. (i) The values of f approach different
numbers as x approaches c from
different sides of c:
(ii) The values of f increase without bound as x approaches c:
(iii) The values of f oscillate
between two fixed numbers as
x approaches c:
y
y
y
6
4
5
3
4
2
3
4
3
2
1
1
x
−4 −3 −2 −1
−1
1
2
3
4
−3 −2 −1
−1
3
4
−3
−4
−4
4
60. V r 3, V 2.48
3
59. (a) C 2 r
r
2
5
−2
−3
x
−4 −3 −2
x
2
C
6
3
0.9549 cm
2 2 (b) If C 5.5, r 5.5
0.87535 cm.
2
If C 6.5, r 6.5
1.03451 cm.
2
4
(a) 2.48 r 3
3
r3 1.86
r 0.8397 in.
(b)
Thus 0.87535 < r < 1.03451.
(c) lim 2 r 6; 0.5; 0.0796
2.45 ≤
2.45 ≤
r →3
≤ 2.51
V
4 3
r ≤ 2.51
3
0.5849 ≤
r 3 ≤ 0.5992
0.8363 ≤
r
≤ 0.8431
(c) For 2.51 2.48 0.03, 0.003.
61. f x 1 x1x
lim 1 x1x e 2.71828
x→0
y
7
3
(0, 2.7183)
2
1
f x
x
f x
0.1
2.867972
0.1
2.593742
0.01
2.731999
0.01
2.704814
0.001
2.719642
0.001
2.716942
0.0001
2.718418
0.0001
2.718146
0.00001
2.718295
0.00001
2.718268
0.000001
2.718283
0.000001
2.718280
x
−3 −2 −1
−1
62. f x x
1
2
3
4
5
x 1 x 1
y
x
3
x
f x
1
0.5
0.1
0
0.1
0.5
1.0
2
2
2
Undef.
2
2
2
1
−2
lim f x 2
1
−1
x→0
Note that for 1 < x < 1, x 0, f x x
−1
x 1 x 1
2.
x
2
3
4
Section 2.2
63.
64.
0.002
Finding Limits Graphically and Numerically
0.005
(1.999, 0.001)
(2.001, 0.001)
1.998
0
2.99
2.002
3.01
0
Using the zoom and trace feature, 0.001. That is, for
0 < x 2 < 0.001,
From the graph, 0.001. Thus
3 , 3 2.999, 3.001.
x2 4
4 < 0.001.
x2
Note:
65. False; f x sin xx is undefined when x 0.
From Exercise 7, we have
lim
x→0
x2 3x
x for x 3.
x3
66. True.
sin x
1.
x
67. False; let
f x 68. False; let
x10, 4x,
2
x4
.
x4
f x x10, 4x,
2
x4
x4
.
lim f x lim x2 4x 0 and f 4 10 0
f 4 10
x→4
x→4
lim f x lim x2 4x 0 10
x→4
x→4
69. f x x
70. The value of f at c has no bearing on the limit as x
approaches c.
(a) lim x 0.5 is true.
x→0.25
As x approaches 0.25 14, f x x approaches
1
2
0.5.
(b) lim x 0 is false.
x→0
f x x is not defined on an open interval
containing 0 because the domain of f is x ≥ 0.
71. If lim f x L1 and lim f x L2, then for every > 0, there exists 1 > 0 and 2 > 0 such that
x→c
x→c
x c < 1 ⇒ f x L1 < and x c < 2 ⇒ f x L2 < . Let equal the smaller of 1 and 2.
Then for x c < , we have L1 L2 L1 f x f x L2 ≤ L1 f x f x L2 < .
Therefore, L1 L2 < 2. Since > 0 is arbitrary, it follows that L1 L2.
72. f x mx b, m 0. Let > 0 be given. Take If 0 < x c < m
, then
mx c < mx mc < mx b mc b < which shows that lim mx b mc b.
x→c
m
.
73. lim f x L 0 means that for every > 0 there
x→c
exists > 0 such that if
0 < x c < ,
then
f x L 0 < .
This means the same as f x L < 0 < x c < .
Thus, lim f x L.
x→c
when
81
82
Chapter 2
Limits and Their Properties
74. (a) 3x 13x 1x2 0.01 9x2 1x2 1
100
1
(b) We are given lim gx L > 0. Let 2L. There
x→c
1
10x2 190x2 1
100
Thus, 3x 13x 1x2 0.01 > 0 if
10x2 1 < 0 and 90x2 1 < 0.
Let a, b 1
,
1
90 90
exists > 0 such that 0 < x c < implies that
L
gx L < . That is,
2
1
9x4 x2 100
L
L
< gx L <
2
2
L
<
2
.
gx
<
3L
2
For x in the interval c , c , x c, we have
L
gx > > 0, as desired.
2
For all x 0 in a, b, the graph is positive. You can
verify this with a graphing utility.
75. Answers will vary.
x2 x 12
7
x→4
x4
4 0.1n
n
76. lim
f 4 0.1n
n
4 0.1n
f 4 0.1n
1
4.1
7.1
1
3.9
6.9
2
4.01
7.01
2
3.99
6.99
3
4.001
7.001
3
3.999
6.999
4
4.0001
7.0001
4
3.9999
6.9999
h
h
77. The radius OP has a length equal to the altitude z of the triangle plus . Thus, z 1 .
2
2
1
h
Area triangle b 1 2
2
P
h
O
Area rectangle bh
b
h
1
Since these are equal, b 1 bh
2
2
1
h
2h
2
5
h1
2
2
h .
5
78. Consider a cross section of the cone, where EF is a diagonal of the inscribed cube.
AD 3, BC 2. Let x be the length of a side of the cube. Then EF x2.
A
By similar triangles,
E
EF
AG
BC AD
x2 3 x
2
3
Solving for x,
B
32x 6 2x
32 2x 6
x
92 6
6
0.96.
7
32 2
G
D
F
C
Section 2.3
Section 2.3
1.
Evaluating Limits Analytically
Evaluating Limits Analytically
(a) lim hx 0
7
2.
x→5
−8
(a) lim gx 2.4
10
x→4
(b) lim gx 4
(b) lim hx 6
x→0
x→1
13
0
−7
−5
hx x 5x
2
3.
10
gx (a) lim f x 0
4
4.
x→0
(b) lim f x 0.524
x→ 3
π
−π
(a) lim f t 0
10
t→4
(b) lim f t 5
−5
6
t→1
10
−4
12 x 3
x9
− 10
f t t t 4
f x x cos x
5. lim x4 24 16
6. lim x5 25 32
7. lim 2x 1 20 1 1
8. lim 3x 2 33 2 7
x→2
x→2
x→3
x→0
9. lim 2x2 4x 1 232 43 1 18 12 1 7
x→3
10. lim 3x3 4x2 3 313 412 3 2
11. lim
1 1
x
2
13. lim
x3
13
2
2
x2 4 12 4
5
5
x→2
x→1
2
2
2
x 2 3 2
12. lim
x→3
14. lim
x→3
16. lim
x→3
2x 5 23 5
1
x3
33
6
x 1
x4
3 1
19. lim sin x sin
x→ 2
22. lim sin
x→1
25.
34
28. lim sec
15. lim
x→7
2
1
2
5x
x 2
3
35
9
35
3
20. lim tan x tan 0
21. lim cos
23. lim sec 2x sec 0 1
24. lim cos 5x cos 5 1
26.
6x sec 76 23
57
7 2
3 x 23 3 4 23 3
18. lim x→3
x→0
5 1
6
2
17. lim x 1 3 1 2
x→ x
sin 1
2
2
lim sin x sin
x→56
x→7
x→1
lim cos x cos
x→53
5 1
3
2
29. lim ex cos 2x e0 cos 0 1
x→0
x→4
x→2
x
2
1
cos
3
3
2
x→ 27. lim tan
x→3
4x tan 34 1
30. lim ex sin x e0 sin 0 0
x→0
83
84
Chapter 2
Limits and Their Properties
ex ln1e ln e
31. lim ln 3x ex ln 3 e
32. lim ln
33. (a) lim f x 5 1 4
34. (a) lim f x 3 7 4
x→1
x→1
x→1
1
x
1
x→3
(b) lim gx 43 64
(b) lim gx 42 16
(c) lim g f x g f 1 g4 64
(c) lim g f x g4 16
x→4
x→4
x→1
x→3
36. (a) lim f x 242 34 1 21
35. (a) lim f x 4 1 3
x→4
x→1
(b) lim gx 3 1 2
3 21 6 3
(b) lim gx (c) lim g f x g3 2
(c) lim g f x g21 3
x→21
x→3
x→4
x→1
37. (a) lim 5gx
5 lim gx 53 15
x→c
38. (a) lim 4f x
4 lim f x 4
x→c
x→c
(b) lim f x gx
lim f x lim gx 2 3 5
x→c
x→c
x→c
(b) lim f x gx
lim f x lim gx x→c
x→c
(c) lim f xgx
lim f x
lim gx
23 6
x→c
x→c
32 6
x→c
x→c
x→c
(c) lim f xgx
lim f x
lim gx
lim f x 2
f x
x→c
(d) lim
x→c gx
lim gx 3
x→c
x→c
x→c
3 1
2
2 2
3212 43
lim f x 32
f x
x→c
3
x→c gx
lim gx 12
x→c
(d) lim
x→c
39. (a) lim f x
3 lim f x
3 43 64
x→c
x→c
3
3
3 lim f x 27 3
f x 40. (a) lim x→c
(b) lim f x lim f x 4 2
x→c
x→c
lim f x 27 3
f x x→c
(b) lim
x→c 18
lim 18
18 2
x→c
x→c
(c) lim 3 f x
3 lim f x 34 12
x→c
(c) lim f x
lim f x
2 272 729
2
x→c
(d) lim f x
32
x→c
x→c
lim f x
32
x→c
4
32
8
x→c
(d) lim f x
23 lim f x
23 2723 9
x→c
41. f x 2x 1 and gx x 0.
2x2 x
agree except at
x
x→c
42. f x x 3 and hx x2 3x
agree except at x 0.
x
(a) lim gx lim f x 1
(a) lim hx lim f x 5
(b) lim gx lim f x 3
(b) lim hx lim f x 3
x→0
x→2
x→0
x→1
x→0
x→1
43. f x xx 1 and gx x3 x
agree except at x 1.
x1
44. gx x→2
x→0
1
x
and f x 2
agree except at x 0.
x1
x x
(a) lim gx lim f x 2
(a) lim f x does not exist.
(b) lim gx lim f x 0
(b) lim f x 1
x→1
x→1
x→1
45. f x x→1
x→1
x→0
x2 1
and gx x 1 agree except at x 1.
x1
lim f x lim gx 2
x→1
3
−3
4
x→1
−4
Section 2.3
2x2 x 3
and gx 2x 3 agree except at
x1
x 1.
47. f x 46. f x x 2.
x3 8
and gx x2 2x 4 agree except at
x2
lim f x lim gx 12
lim f x lim gx 5
x→1
Evaluating Limits Analytically
x→2
x→1
x→2
12
4
−8
4
−9
9
0
−8
x 4lnx 6
lnx 6
and gx x2 16
x4
ln 2
0.0866
lim f x lim gx x→4
x→4
8
x3 1
and gx x2 x 1 agree except at
x1
x 1.
48. f x 49. f x lim f x lim gx 3
x→1
x→1
1
7
−7
−4
3
4
−2
−1
50. f x e 2x 1
and gx e x 1
ex 1
x5
x5
lim
x2 25 x→5 x 5x 5
51. lim
x→5
lim f x lim gx e0 1 2
x→0
lim
x→0
x→5
3
−2
1
1
x 5 10
2
−1
3x
x 3
lim
x→3 x2 9
x→3 x 3x 3
x2 x 6
x 3x 2
lim
x→3
x→3 x 3x 3
x2 9
52. lim
lim
x→3
54. lim
x→3
55. lim
x→0
1
1
x3
6
x 5 5
x
lim
x→0
x→0
x→0
lim
x→3
x2 x 6
x 3x 2
x2 5
lim
5
lim
x2 5x 6 x→3 x 3x 2 x→3 x 2 1
lim
56. lim
53. lim
3 x 3
x
lim
x→0
lim
x→0
x 5 5
x
x 5 5
x 5 5
5
x 5 5
1
1
lim
xx 5 5 x→0 x 5 5 25
10
3 x 3
x
3x3
3 x 3
3 x 3
3 x 3 x
lim
x→0
1
3 x 3
3
1
9
33
x 2 5 5
x 3 6 6
85
86
Chapter 2
57. lim
Limits and Their Properties
x 5 3
x4
x→4
lim
lim
x→4
58. lim
x 1 2
x3
x→3
x 5 3
x4
x→4
lim
x 5 3
x 5 3
x 5 9
1
1
1
lim
x→4
3
6
x 5 3
9
x 4x 5 3
x 1 2
x3
x→3
x 1 2
x 1 2
lim
x→3
1
x3
1
lim
x 3x 1 2
x→3 x 1 2 4
1
1
3 3 x
3x 3
33 x
1
1
59. lim
lim
lim
x→0
x→0
x→0 33 x
x
x
9
1
1
4 x 4
1
1
x4 4
4x 4
lim
lim
60. lim
x→0
x→0
x→0 4x 4
x
x
16
61. lim
x→0
2x x 2x
2x 2x 2x
lim
lim 2 2
x→0
x→0
x
x
x x2 x2
x2 2xx x2 x2
x2x x
lim
lim
lim 2x x 2x
x→0
x→0
x→0
x→0
x
x
x
62. lim
x x2 2x x 1 x2 2x 1
x2 2xx x2 2x 2x 1 x2 2x 1
lim
x→0
x→0
x
x
63. lim
lim 2x x 2 2x 2
x→0
x x3 x3
x3 3x2x 3xx2 x3 x3
lim
x→0
x→0
x
x
64. lim
lim
x→0
65. lim
x 2 2
x→0
x
f x
x
0.1
0.358
Analytically, lim
x→0
x3x2 3xx x2
lim 3x2 3xx x2 3x2
x→0
x
0.354
0.01
2
0.001
0
0.001
0.01
0.1
0.345
?
0.354
0.353
0.349
0.354
x 2 2
x
lim
x→0
lim
x→0
x 2 2
x
−3
3
−2
x 2 2
x 2 2
2
x22
1
1
0.354.
lim
x→0
4
x 2
2
xx 2 2
22
Section 2.3
66. f x Evaluating Limits Analytically
4 x
x 16
1
0
x
15.9
15.99
15.999
16
16.001
16.01
16.1
f x
0.1252
0.125
0.125
?
0.125
0.125
0.1248
Analytically, lim
x→16
x→16
1
x 4
20
−1
It appears that the limit is 0.125.
4 x
4 x
lim
x→16 x 4 x 4
x 16
lim
1
.
8
1
1
1
2x 2
67. lim
x→0
x
4
3
−5
x
0.1
0.01
0.001
0
0.001
f x
0.263
0.251
0.250
?
0.250
1
1
2x 2
2 2 x
lim
Analytically, lim
x→0
x→0
x
22 x
68. lim
x→2
0.01
0.1
0.249
0.238
x
1
1
−2
1
1
1
lim
.
x x→0
lim
22 x x x→0 22 x
4
x5 32
80
x2
x
f x
100
1.9
1.99
72.39
79.20
Analytically, lim
x→2
1.999
79.92
1.9999
2.0
79.99
?
2.0001
80.01
2.001
80.08
2.01
2.1
80.80
88.41
−4
3
−25
x5 32
x 2x4 2x3 4x2 8x 16
lim
x→2
x2
x2
lim x4 2x3 4x2 8x 16 80.
x→2
(Hint: Use long division to factor x5 32.)
sinx x15 115 51
69. lim
sin x
lim
x→0
5x
71. lim
sin x1 cos x
1
lim
x→0 2
2x2
x→0
x→0
sin x
x
70. lim
51 cos x
1 cos x
lim 5
x→0
x
x
72. lim
cos tan sin lim
1
→0 74. lim
2 tan2 x
2 sin2 x
sin x
lim
2 lim
x→0 x cos2 x
x→0
x
x
x→0
1 cos x
x
→0
50 0
1
10 0
2
sin x
sin2 x
lim
sin x 1 sin 0 0
x→0
x→0
x
x
73. lim
x→0
210 0
1 cos h2
1 cos h
lim
1 cos h
h→0
h→0
h
h
75. lim
87
00 0
76. lim sec 1 →
cos2 x
sin x
88
Chapter 2
77. lim
x→ 2
Limits and Their Properties
cos x
lim sin x 1
x→ 2
cot x
78. lim
x→ 4
1 tan x
cos x sin x
lim
sin x cos x x→4 sin x cos x cos2 x
sin x cos x
lim
x→ 4 cos xsin x cos x
1
lim
x→ 4 cos x
lim sec x
x→ 4
2
4e 2x 1
4e x 1e x 1
lim
x
x→0
x→0
e 1
ex 1
1 ex
1 ex ex
lim x
ex
x
x→0 e 1
x→0 e 1
1 exex
lim
lim ex 1
x→0
x→0
1 ex
80. lim
79. lim
81. lim
t→0
82. lim
x→0
sin 3t
sin 3t
lim
t→0
2t
3t
13sin3x3x 21131 32
sin 3t
t
4
0.1
t
x→0
32 132 23
sin 2x
sin 2x
lim 2
sin 3x x→0
2x
83. f t lim 4e x 1 42 8
f t
0.01
2.96
0.001
2.9996
3
0
0.001
0.01
0.1
?
3
2.9996
2.96
84. From the graph, lim
x→0
2
−1
The limit appears to equal 3.
sin 3t
sin 3t
lim 3
31 3.
t→0
t→0
t
3t
Analytically, lim
− 2
cos x 1
0.25
2x2
1
−
1
0.1
0.01
0.01
0.1
1
0.2298
0.2498
0.25
0.25
0.2498
0.2298
x
f x
lim
x→0
cos x 1
0.25
2x2
Analytically,
cos x 1
2x2
cos x 1
cos2 x 1
cos x 1 2x2cos x 1
x→0
sin2 x
x2
1
2cos x 1
sinx x 2cos1x 1 114 14 0.25
2
lim
sin2 x
cos x 1
2x2
2
−1
Section 2.3
85. f x Evaluating Limits Analytically
sin x2
x
1
x
0.1
f x
0.099998 0.01 0.001 ? 0.001 0.01 0.099998
− 2
0.01 0.001 0 0.001 0.01 0.1
2
−1
sin x2
sin x2
lim x
01 0.
x→0
x→0
x
x2
Analytically, lim
86. f x sin x
3 x
2
−3
0.1
x
f x
0.215
0.01
0.001
0
0.001
0.01
0.1
0.01
?
0.01
0.0464
0.215
0.0464
87. f x ln x
x1
4
x
0.5
0.9
0.99
1.01
1.1
1.5
f x
1.3863
1.0536
1.0050
0.9950
0.9531
0.8109
lim
x→1
−2
The limit appears to equal 0.
sin x
3 2 sin x
lim x
01 0.
3
x→0 x→0
x
x
Analytically, lim
3
−1
6
−1
ln x
1
x1
88. f x e3x 8
e2x 4
5
x
0.5
0.6
0.69
0.70
0.8
0.9
f x
2.7450
2.8687
2.9953
3.0103
3.1722
3.3565
−1
2
0
ex 2e2x 2ex 4
e3x 8
e2x 2ex 4 4 4 4
lim
3
lim
2x
x
x
x→ln 2 e 4
x→ln 2
x→ln 2
e 2e 2
ex 2
22
lim
89. lim
h→0
90. lim
h→0
f x h f x
2x h 3 2x 3
2x 2h 3 2x 3
2h
lim
lim
lim
2
h→0
h→0
h→0 h
h
h
h
x h x
x h x
f x h f x
lim
lim
h→0
h→0
h
h
h
lim
h→0
x h x
x h x
xhx
1
1
lim
h x h x h→0 x h x 2x
4
4
f x h f x
xh
x
4x 4x h
4
4
91. lim
lim
lim
lim
2
h→0
h→0
h→0
h→0 x hx
h
h
x hxh
x
89
90
Chapter 2
Limits and Their Properties
x h2 4x h x2 4x
f x h f x
x2 2xh h2 4x 4h x2 4x
lim
lim
h→0
h→0
h→0
h
h
h
92. lim
lim
h→0
h2x h 4
lim 2x h 4 2x 4
h→0
h
93. lim 4 x2 ≤ lim f x ≤ lim 4 x2
x→0
x→0
x→0
x→a
Therefore, lim f x b.
Therefore, lim f x 4.
x→a
x→0
95. f x x cos x
96. f x x sin x
4
6
3
2
− 2
2
−4
−2
lim x sin x 0
lim x cos x 0
x→0
x→0
98. f x x cos x
97. f x x sin x
6
6
− 2
2
− 2
2
−6
−6
lim x sin x 0
99. f x x sin
lim x cos x 0
x→0
1
x
100. hx x cos
0.5
−0.5
0.5
lim x sin
1
x
0.5
− 0.5
0.5
−0.5
x→0
x→a
x→a
x→0
x→0
x→a
b ≤ lim f x ≤ b
4 ≤ lim f x ≤ 4
− 3
2
94. lim b x a ≤ lim f x ≤ lim b x a 1
0
x
− 0.5
lim x cos
x→0
1
0
x
101. We say that two functions f and g agree at all but one point (on an open interval) if f x gx for all x
in the interval except for x c, where c is in the interval.
x2 1
and gx x 1 agree at all points
x1
except x 1.
102. f x 103. An indeterminant form is obtained when evaluating a
limit using direct substitution produces a meaningless
fractional expression such as 00. That is,
lim
x→c
f x
gx
for which lim f x lim gx 0.
x→c
x→c
Section 2.3
Evaluating Limits Analytically
91
104. If a function f is squeezed between two functions h and g,hx ≤ f x ≤ gx, and h and g have the same limit L as x → c,
then lim f x exists and equals L.
x→c
105. f x x, gx sin x, hx sin x
x
106. f x x, gx sin2 x, hx 3
sin2 x
x
2
f
g
h
g
−5
5
−3
3
h
f
−3
−2
When you are “close to” 0 the magnitude of f is
approximately equal to the magnitude of g.
Thus, g f 1 when x is “close to” 0.
When you are “close to” 0 the magnitude of g is “smaller”
than the magnitude of f and the magnitude of g is
approaching zero “faster” than the magnitude of f.
Thus, g f 0 when x is “close to” 0.
107. st 16t2 1000
lim
t→5
s5 st
600 16t2 1000
16t 5t 5
lim
lim
lim 16t 5 160 ftsec
t→5
t→5
t→5
5t
5t
t 5
Speed 160 ftsec
108. st 16t 2 1000 0 when t 5 210 st s
lim
t→5102
510 t
2
5 10
seconds
1000
16
2
0 16t 2 1000
510
t→5102
t
2
lim
16 t 2 lim
t→5102
lim
t→5102
125
2
510 t
2
16 t 16 t lim
t→5102
510
2
t 5 210 t 5 10
2
s3 st
4.932 150 4.9t2 150
4.99 t2
lim
lim
t→3
t→3
t→3
3t
3t
3t
lim
t→3
4.93 t3 t
lim 4.93 t 29.4 msec
t→3
3t
110. 4.9t2 150 0 when t 1500
5.53 seconds.
150
4.9
49
The velocity at time t a is
lim
t→a
510
8010 ftsec 253 ftsec
2
2
109. st 4.9t 150
lim
sa st
4.9a ta t
4.9a2 150 4.9t2 150
lim
lim
t→a
t→a
at
at
at
lim 4.9a t 2a4.9 9.8a msec.
t→a
Hence, if a 150049, the velocity is 9.8150049 54.2 msec.
92
Chapter 2
Limits and Their Properties
111. Let f x 1x and gx 1x. lim f x and lim gx do not exist.
x→0
x→0
lim 0
0
1
1
lim f x gx
lim
x→0
x→0 x
x
x→0
112. Suppose, on the contrary, that lim gx exists. Then, since lim f x exists, so would lim f x gx
, which is a
x→c
x→c
contradiction. Hence, lim gx does not exist.
x→c
x→c
113. Given f x b, show that for every > 0 there exists a > 0 such that f x b < whenever x c < .
Since f x b b b 0 < for any > 0, then any value of > 0 will work.
114. Given f x x n, n is a positive integer, then
lim x n lim x x n1 lim x
lim x n1
x→c
x→c
x→c
x→c
c lim x x n2
clim x
lim x n2
x→c
x→c
c cx→c
lim x→c
. . . c n.
x x n3
115. If b 0, then the property is true because both sides are equal to 0. If b 0, let > 0 be given.
Since lim f x L, there exists > 0 such that f x L < b whenever 0 < x c < .
x→c
Hence, wherever 0 < x c < , we have
b f x L
< bf x bL
or
< which implies that lim bf x
bL.
x→c
116. Given lim f x 0:
x→c
Now f x 0 f x f x 0 < for x c
. Therefore, lim f x 0.
x→c
For every > 0, there exists > 0 such that f x 0 < whenever 0 < x c < .
M f x ≤ f xgx ≤ M f x
117.
<
lim M f x ≤ lim f xgx ≤ lim M f x x→c
x→c
x→c
M0 ≤ lim f xgx ≤ M0
x→c
0 ≤ lim f xgx ≤ 0
x→c
Therefore, lim f xgx 0.
x →c
f x ≤ f x ≤ f x
lim f x
≤ lim f x ≤ lim f x
x→c
x→c
x→c
118. (a) If lim f x 0, then lim f x 0.
x→c
x→c
0 ≤ lim f x ≤ 0
x→c
Therefore, lim f x 0.
x→c
(b) Given lim f x L:
x→c
For every > 0, there exists > 0 such that f x L < whenever 0 < x c < .
Since f x L ≤ f x L < for x c < , then lim f x L .
x→c
Section 2.3
119. False. As x approaches 0 from the left,
x 1.
x
120. False. lim
x→ Evaluating Limits Analytically
93
0
sin x
0
x
2
−3
3
−2
122. False. Let f x 121. True.
3, x 1
x, x 1
, c 1.
Then lim f x 1 but f 1 1.
x→1
123. False. The limit does not exist.
4
124. False. Let f x 12 x2 and gx x2. Then f x < gx
for all x 0. But lim f x lim gx 0.
x→0
−3
x→0
6
−2
125. Let
126. lim
x→0
f x 4,4,
if x ≥ 0
if x < 0
lim f x lim 4 4.
x→0
x→0
lim f x does not exist since for x < 0, f x 4
and for x ≥ 0, f x 4.
x→0
1 cos x
1 cos x
lim
x→0
x
x
1 cos2 x
sin2 x
lim
x→0 x1 cos x
x →0 x1 cos x
lim
lim
x→0
sin x
x
lim
x→0
sin x
1 cos x
sin x
x
lim 1 cos x
sin x
x→0
10 0
rational
0,1, ifif xx isis irrational
0, if x is rational
g x x, if x is irrational
127. f x lim f x does not exist.
x→0
No matter how “close to” 0 x is, there are still an infinite number of rational and irrational numbers
so that lim f x does not exist.
x→0
lim gx 0.
x→0
When x is “close to” 0, both parts of the function are “close to” 0.
1 cos x
1 cos x
94
Chapter 2
128. f x Limits and Their Properties
sec x 1
x2
129. (a) lim
x→0
1 cos x
1 cos x
lim
x→0
x2
x2
(a) The domain of f is all x 0, 2 n.
(b)
lim
x→0
2
− 3
2
sin2 x
x→0
x2
1
−2
The domain is not obvious. The hole at x 0 is not
apparent.
(c) lim f x x→0
(d)
1 cos2 x
x21 cos x
1
1 cos x
lim
3
2
(b) Thus,
1 cos x
1 cos x
12 21
1
1 cos x 1
⇒ 1 cos x x2
x2
2
2
1
2
1
⇒ cos x 1 x2 for x 0.
2
sec x 1 sec x 1
x2
x2
sec x 1
sec2 x 1
sec x 1 x2sec x 1
1
(c) cos0.1 1 0.12 0.995
2
(d) cos0.1 0.9950, which agrees with part (c).
tan2 x
1
1 sin2 x
sec x 1 cos2 x x2 sec x 1
x2
sec x 1
1 sin2 x
1
lim
2
x→0
x→0 cos2 x
x
x2 sec x 1
Hence, lim
11
12 21.
130. The calculator was set in degree mode, instead of radian mode.
Section 2.4
Continuity and One-Sided Limits
1. (a) lim f x 1
2. (a)
(b) lim f x 1
(b)
(c) lim f x 1
(c) lim f x 2
(c) lim f x 0
The function is continuous at
x 3.
The function is continuous at
x 2.
The function is NOT continuous at
x 3.
x→3
x→3
x→3
4. (a)
3. (a) lim f x 0
lim f x 2
(b) lim f x 0
x→3
x→2
x→3
x→2
lim f x 2
x→3
5. (a) lim f x 2
x→2
6. (a)
x→4
lim f x 2
lim f x 0
x→1
(b) lim f x 2
(b)
(c) lim f x 2
(c) lim f x does not exist
(c) lim f x does not exist.
The function is NOT continuous at
x 2.
The function is NOT continuous
at x 4.
The function is NOT continuous at
x 1.
(b)
x→2
x→4
x→2
7. lim
x→5
9.
lim f x 2
x→2
lim
x→4
x5
1
1
lim
x2 25 x→5 x 5 10
x→3 x
x2 9
does not exist because
without bound as x → 3
.
8. lim
x→2
x
x2 9
grows
10. lim
x→4
lim f x 2
x→1
x→1
2x
1
1
lim x2 4 x→2 x 2
4
x 2
x4
lim
x→4
lim
x→4
lim
x→4
x 2
x4
x 2
x 2
x4
x 4x 2
1
x 2
1
4
Section 2.4
11. lim
x→0
x x
lim
x→0 x
1
x
12. lim
x→3
1
1
x x
x
x x x
lim 13. lim x→0
x→0
x
xx x
14. lim x→0
x
1
Continuity and One-Sided Limits
x 3 x3
lim
x→3
32
1
32
1
1
95
1
1
lim
2
x x→0
lim
xx x x x→0 xx x
xx 0
x
x2 2xx x2 x x x2 x
x x2 x x x2 x
lim x→0
x
x
lim x→0
2xx x2 x
x
lim 2x x 1
x→0
2x 0 1 2x 1
15. lim f x lim
x→3
x→3
x2 5
2
2
16. lim f x lim x2 4x 2 2
x→2
x→2
lim f x lim x2 4x 6 2
x→2
x→2
lim f x 2
x→2
17. lim f x lim x 1 2
x→1
x→1
18. lim f x lim 1 x 0
x→1
x→1
lim f x lim x 1 2
3
x→1
19. lim cot x does not exist since
x→ lim cot x and lim cot x
x→ x→1
x→ do not exist.
lim f x 2
x→1
20. lim sec x does not exist since
x→ 2
lim
x→ 2
sec x and
lim
x→ 2
sec x
21. lim 3x
5 33 5 4
x→4
x
3 for 3 < x < 4
22. lim 3x x
33 3 6
x→3
do not exist.
23. lim 2 x
does not exist
x→3
2x 1 1 2
24. lim 1 x→1
because
25. lim lnx 3 ln 0
x→3
does not exist.
lim 2 x
2 3 5
x→3
and
lim 2 x
2 4 6.
x→3
26. lim ln6 x ln 0
x→6
27. lim lnx23 x ln41 ln 4
x→2
28. lim ln
x→5
x
x 4
ln
5
ln 5
1
does not exist.
29. f x x2
1
4
has discontinuities at x 2 and
x 2 since f 2 and f 2 are not
defined.
30. f x x2 1
x1
has a discontinuity at x 1
since f 1 is not defined.
1
31. f x x
x
2
has discontinuities at each integer
k since lim f x lim f x.
x→k
x→k
96
Chapter 2
Limits and Their Properties
x,
32. f x 2,
2x 1,
x < 1
x 1 has discontinuity at x 1 since f 1 2 lim f x 1.
x→1
x > 1
33. gx 25 x2 is continuous on 5, 5.
34. f t 2 9 t2 is continuous on 2, 2.
35. lim f x 3 lim f x. f is continuous on 1, 4.
36. g2 is not defined. g is continuous on 1, 2.
37. f x x2 2x 1 is continuous for all real x.
38. f x 39. f x 3x cos x is continuous for all real x.
40. f x cos
x→0
x→0
41. f x x
is not continuous at x 0, 1. Since
x2 x
1
x
for x 0, x 0 is a removable
x2 x x 1
discontinuity, whereas x 1 is a nonremovable
discontinuity.
43. f x x
is continuous for all real x.
x2 1
1
is continuous for all real x.
x2 1
x
is continuous for all real x.
4
x
has nonremovable discontinuities at
x2 1
x 1 and x 1 since lim f x and lim f x
42. f x x→1
x→1
do not exist.
x3
has a nonremovable discontinuity at
x2 9
x 3 since lim f x does not exist, and has a
44. f x x→3
removable discontinuity at x 3 since
lim f x lim
x→3
45. f x x2
x 2x 5
46. f x has a nonremovable discontinuity at x 5 since lim f x
x→5
does not exist, and has a removable discontinuity at
x 2 since
lim f x lim
x→2
47. f x x 2
x→2
1
1
.
x5
7
x2
has a nonremovable discontinuity at x 2 since lim f x
x→2
does not exist.
x→3
1
1
.
x3 6
x1
x 2x 1
has a nonremovable discontinuity at x 2 since
lim f x does not exist, and has a removable
x→2
discontinuity at x 1 since
lim f x lim
x→1
48. f x x→1
x 3
1
1
.
x2 3
x3
has a nonremovable discontinuity at x 3 since lim f x
x→3
does not exist.
Section 2.4
49. f x x,x ,
x ≤ 1
x > 1
2
50. f x Continuity and One-Sided Limits
3,
2x
x,
2
x < 1
x ≥ 1
has a possible discontinuity at x 1.
has a possible discontinuity at x 1.
1. f 1 1
1. f 1 12 1
2.
lim f x lim x 1
lim f x 1
lim f x lim x 1
x→1
x→1
x→1
x→1
2.
x→1
2
lim f x lim x2 1
x→1
lim f x 1
x→1
x→1
x→1
f is continuous at x 1, therefore, f is
continuous for all real x.
f is continuous at x 1, therefore, f is
continuous for all real x.
x→1
3. f 1 lim f x
3. f 1 lim f x
1
x 1,
51. f x 2
3 x,
lim f x lim 2x 3 1
x→1
x→1
x ≤ 2
has a possible discontinuity at x 2.
x > 2
2
12
2
1. f 2 2. lim f x lim
x→2
x→2
12 x 1 2
lim f x lim 3 x 1
x→2
x→2
lim f x does not exist.
x→2
Therefore, f has a nonremovable discontinuity at x 2.
52. f x 2x,
x 4x 1,
2
x ≤ 2
has a possible discontinuity at x 2.
x > 2
1. f 2 22 4
2.
lim f x lim 2x 4
x→2
x→2
x→2
x→2
lim f x does not exist.
x→2
lim f x lim x2 4x 1 3
Therefore, f has a nonremovable discontinuity at x 2.
53. f x x
tan 4 ,
x,
x
x
x
< 1
tan 4 ,
≥ 1
x,
1. f 1 1
1 < x < 1
has possible discontinuities at x 1, x 1.
x ≤ 1 or x ≥ 1
f 1 1
2. lim f x 1
lim f x 1
x→1
x→1
3. f 1 lim f x
x→1
f 1 lim f x
x→1
f is continuous at x ± 1, therefore, f is continuous for all real x.
54. f x csc x ,
6
2,
1. f 1 csc
2
6
2. lim f x 2
x→1
3. f 1 lim f x
x→1
x 3
x 3
≤ 2
> 2
csc x ,
6
2,
f 5 csc
1 ≤ x ≤ 5
x < 1 or x > 5
has possible discontinuities at x 1, x 5.
5
2
6
lim f x 2
x→5
f 5 lim f x
x→5
f is continuous at x 1 and x 5, therefore, f is continuous for all real x.
97
98
Chapter 2
55. f x Limits and Their Properties
ln1 x x ,1,
2
x ≥ 0
x < 0
56. f x has a possible discontinuity at x 0.
lim f x 1 0 1;
5x
3
5
, x > 5
x ≤ 5
has a possible discontinuity at x 5.
f 0 ln0 1 ln 1 0
x→0
1010 3ex,
lim f x 10 3e55 7
x→5 lim f x 0
lim f x 10 355 7
x→0 x→5
Therefore, f has a nonremovable discontinuity at x 0.
f is continuous for all x.
x
has nonremovable discontinuities at each
4
4k 2, k is an integer.
57. f x csc 2x has nonremovable discontinuities at integer
multiples of 2.
58. f x tan
59. f x x 1
has nonremovable discontinuities at each
integer k.
60. f x 3 x
has nonremovable discontinuities at each
integer k.
61. lim f x 0
62. lim f x 0
50
20
x→0
x→0
lim fx 0
lim f x 0
x→0
x→0
f is not continuous at x 2.
−8
f is not continuous at x 4
8
−8
−10
−10
64. lim g(x lim
63. f 2 8
x→0
Find a so that lim ax2 8 ⇒ a x→2
8
2.
22
x→0
4 sin x
4
x
lim gx lim a 2x a
x→0
x→0
Let a 4.
65. Find a and b such that lim ax b a b 2 and lim ax b 3a b 2.
x→1
x→3
a b 2
3a b 2
4
4a
a 1
b
2,
f x x 1,
2,
x ≤ 1
1 < x < 3
x ≥ 3
2 1 1
x2 a2
x→a x a
66. lim gx lim
x→a
lim x a 2a
67. f gx x 12
Continuous for all real x
x→a
Find a such that 2a 8 ⇒ a 4.
68. f gx 1
x 1
Nonremovable discontinuity at x 1. Continuous for all
x > 1. Because f g is not defined for x < 1, it is better
to say that f g is discontinuous from the right at x 1.
70. f gx sin x2
Continuous for all real x.
8
69. f gx 1
1
x2 5 6 x2 1
Nonremovable discontinuities at x ± 1
Section 2.4
72. hx 71. y x
x
Nonremovable discontinuity at each integer
Continuity and One-Sided Limits
99
1
x 1x 2
Nonremovable discontinuity at x 1 and x 2
0.5
2
−3
3
−3
4
−1.5
−2
73. f x 2xx 2x,4,
2
x ≤ 3
x > 3
74. f x cos x 1 , x < 0
5x, x
x ≥ 0
−7
f 0 50 0
Nonremovable discontinuity at x 3
5
lim f x lim
x→0
−5
3
7
x→0
2
cos x 1
0
x
−3
lim f x lim 5x 0
x→0
x→0
Therefore, lim f x 0 f 0 and f is continuous
−5
x→0
on the entire real line.
(x 0 was the only possible discontinuity.)
75. f x x
x2 1
76. f x xx 3
Continuous on 3, Continuous on , 77. f x sec
x
4
78. f x Continuous on 0, Continuous on:
. . . , 6, 2, 2, 2, 2, 6, 6, 10, . . .
79. f x sin x
x
3
−4
x1
x
80. f x x3 8
x2
4
−4
−2
lnx2 1
x
3
−4
4
0
The graph appears to be continuous on the interval
4, 4. Since f 0 is not defined, we know that f has
a discontinuity at x 0. This discontinuity is removable
so it does not show up on the graph.
81. f x 14
The graph appears to be continuous on the interval
4, 4. Since f 2 is not defined, we know that f has
a discontinuity at x 2. This discontinuity is removable
so it does not show up on the graph.
82. f x ex 1
ex 1
5
4
−4
−3
The graph appears to be continuous on the interval
4, 4. Since f 0 is not defined, it is discontinuous
there. Removable discontinuities do not always show up.
4
−2
The graph appears to be continuous on the interval
4, 4. Since f 0 is not defined, it is discontinuous
there. Removable discontinuities do not always show up.
100
Chapter 2
Limits and Their Properties
83. f x x2 4x 3
84. f x x3 3x 2
f x is continuous on 2, 4.
f x is continuous on 0, 1.
f 2 1 and f 4 3
f 0 2 and f 1 2
By the Intermediate Value Theorem, f c 0 for at least
one value of c between 2 and 4.
By the Intermediate Value Theorem, f c 0 for at least
one value of c between 0 and 1.
2 .
85. h is continuous on 0,
h0 2 < 0 and h
0.91 > 0.
2
By the Intermediate Value Theorem, h c 0 for at least
one value of c between 0 and .
2
87. f x x3 x 1
86. g is continuous on 0, 1.
g 0 2.77 < 0 and g 1 1.61 > 0.
By the Intermediate Value Theorem, g c 0 for at least
one value of c between 0 and 1.
88. f x x3 3x 3
f x is continuous on 0, 1.
f x is continuous on 0, 1.
f 0 1 and f 1 1
f 0 3 and f 1 1
By the Intermediate Value Theorem, f x 0 for at least
one value of c between 0 and 1. Using a graphing utility,
we find that x 0.6823.
By the Intermediate Value Theorem, f x 0 for at least
one value of c between 0 and 1. Using a graphing utility,
we find that x 0.8177.
89. gt 2 cos t 3t
90. h
1 3 tan g is continuous on 0, 1.
h is continuous on 0, 1.
g0 2 > 0 and g1 1.9 < 0.
h0 1 > 0 and h1 2.67 < 0.
By the Intermediate Value Theorem, gt 0 for at least
one value c between 0 and 1. Using a graphing utility, we
find that t 0.5636.
By the Intermediate Value Theorem, h
0 for at least
one value between 0 and 1. Using a graphing utility, we
find that 0.4503.
91. f x x2 x 1
92. f x x2 6x 8
f is continuous on 0, 5.
f is continuous on 0, 3.
f 0 1 and f 5 29
f 0 8 and f 3 1
1 < 11 < 29
The Intermediate Value Theorem applies.
x2
x 1 11
1 < 0 < 8
The Intermediate Value Theorem applies.
x2 6x 8 0
x2 x 12 0
x 2x 4 0
x 4x 3 0
x 2 or x 4
x 4 or x 3
c 3 (x 4 is not in the interval.)
Thus, f 3 11.
c 2 (x 4 is not in the interval.)
Thus, f 2 0.
Section 2.4
93. f x x3 x2 x 2
f is continuous on 0, 3.
f 0 2 and f 3 19
2 < 4 < 19
94. f x x3 x2 x 2 4
x3 x2 x 6 0
x 2x2 x 3 0
x2 x
x1
5
52 356 and f 4 203
20
35
< 6 <
6
3
The Intermediate Value Theorem applies.
x2 x
6
x1
x2
(x2 x 3 has no real solution.)
c2
Thus, f 2 4.
101
f is continuous on 2 , 4. The nonremovable discontinuity,
x 1, lies outside the interval.
f
The Intermediate Value Theorem applies.
Continuity and One-Sided Limits
x2 x 6x 6
x2 5x 6 0
x 2x 3 0
x 2 or x 3
c 3 (x 2 is not in the interval.)
Thus, f 3 6.
95. (a) The limit does not exist at x c.
(b) The function is not defined at x c.
(c) The limit exists at x c, but it is not equal to the
value of the function at x c.
(d) The limit does not exist at x c.
96. A discontinuity at x c is removable if you can define
(or redefine) the function at x c in such a way that the
new function is continuous at x c. Answers will vary.
(a) f x x 2
x2
sinx 2
(b) f x x2
1,
0,
(c) f x 1,
0,
97.
y
5
4
3
2
1
−2 −1
x
1
y
3
2
1
if x ≥ 2
if 2 < x < 2
if x 2
if x < 2
−3
−2
−1
x
−1
1
2
3
−2
−3
98. If f and g are continuous for all real x, then so is f g
(Theorem 1.11, part 2). However, fg might not be
continuous if gx 0. For example, let f x x and
gx x2 1. Then f and g are continuous for all real x,
but fg is not continuous at x ± 1.
3 4 5 6 7
−2
−3
The function is not continuous at x 3 because
lim f x 1 0 lim f x.
x→3
x→3
100. True; if f x gx, x c, then lim f x lim gx (if
99. True.
x→c
x→c
1. f c L is defined.
they exist) and at least one of these limits then does not
2. lim f x L exists.
equal the corresponding function value at x c.
x→c
3. f c lim f x
x→c
All of the conditions for continuity are met.
102
Chapter 2
Limits and Their Properties
101. False; a rational function can be written as PxQx
where P and Q are polynomials of degree m and n,
respectively. It can have, at most, n discontinuities.
102. False; f 1 is not defined and lim f x does not exist.
103. lim f t 28
104. The functions agree for integer values of x:
t→4
lim f t 56
t→4
At the end of day 3, the amount of chlorine in the pool
has decreased to about 28 oz. At the beginning of
day 4, more chlorine was added, and the amount was
about 56 oz.
x→1
gx 3 x
3 x 3 x
for x an integer
f x 3 x
3 x
However, for non-integer values of x, the functions differ
by 1.
f x 3 x
gx 1 2 x
.
1
1
For example, f 2 3 0 3, g2 3 1 4.
1.04,
105. C 1.04 0.36t 1
,
1.04 0.36t 2,
0 < t ≤ 2
t > 2, t is not an integer
t > 2, t is an integer
Nonremovable discontinuity at each integer greater than
or equal to 2.
You can also write C as
C
1.04,
1.04 0.362 t
,
0 < t ≤ 2
.
t > 2
t 2 2 t
106. Nt 25 2
t
0
1
1.8
2
3
3.8
Nt
50
25
5
50
25
5
Discontinuous at every positive even integer. The
company replenishes its inventory every two months.
N
C
50
Number of units
4
3
2
40
30
20
10
1
t
2
t
1
2
3
4
6
8
10 12
Time (in months)
4
107. Let st be the position function for the run up to the campsite. s0 0 (t 0 corresponds to 8:00 A.M.,
s20 k (distance to campsite)). Let rt be the position function for the run back down the mountain:
r0 k, r10 0. Let f t st rt.
When t 0 (8:00 A.M.), f 0 s0 r0 0 k < 0.
When t 10 (8:10 A.M.), f 10 s10 r10 > 0.
Since f 0 < 0 and f 10 > 0, then there must be a value t in the interval 0, 10 such that f t 0.
If f t 0, then st rt 0, which gives us st rt. Therefore, at some time t, where 0 ≤ t ≤ 10,
the position functions for the run up and the run down are equal.
108. Let V 43 r 3 be the volume of a sphere of radius r. V is
continuous on 1, 5.
V1 43 4.19
V5 43 53 523.6
Since 4.19 < 275 < 523.6, the Intermediate Value
Theorem implies that there is at least one value r between
1 and 5 such that Vr 275. (In fact, r 4.0341.)
109. Suppose there exists x1 in a, b such that f x1 > 0 and
there exists x2 in a, b such that f x2 < 0. Then by the
Intermediate Value Theorem, f x must equal zero for
some value of x in x1, x2 (or x2, x1 if x2 < x1). Thus,
f would have a zero in a, b, which is a contradiction.
Therefore, f x > 0 for all x in a, b or f x < 0 for all
x in a, b.
Section 2.4
110. Let c be any real number. Then lim f x does not exist
x→c
since there are both rational and irrational numbers
arbitrarily close to c. Therefore, f is not continuous at c.
Continuity and One-Sided Limits
103
111. If x 0, then f 0 0 and lim f x 0. Hence, f is
x→0
continuous at x 0.
If x 0, then lim f t 0 for x rational, whereas
t→x
lim f t lim kt kx 0 for x irrational. Hence,
t →x
t →x
f is not continuous for all x 0.
y
1, if x < 0
112. sgnx 0, if x 0
1, if x > 0
4
113. (a)
3
S
2
60
1
(a) lim sgnx 1
−4 −3 −2 −1
x→0
(b) lim sgnx 1
x→0
50
x
1
2
3
4
40
−2
30
−3
20
−4
10
(c) lim sgnx does not exist.
t
5
x→0
10
15 20 25 30
(b) There appears to be a limiting speed and a possible
cause is air resistance.
0 ≤ x < b
b < x ≤ 2b
0
114. (a) f x b
(b) gx y
x
2
0 ≤ x ≤ b
b
x
2
b < x ≤ 2b
y
2b
2b
b
b
x
b
2b
x
b
NOT continuous at x b.
115. f x 1 xx,,
2
Continuous on 0, 2b.
x ≤ c
x > c
f is continuous for x < c and for x > c. At x c, you
need 1 c2 c. Solving c2 c 1, you obtain
c
117. f x 1 ± 1 4 1 ± 5
.
2
2
x c2 c
x
116. Let y be a real number. If y 0, then x 0. If y > 0,
then let 0 < x0 < 2 such that M tan x0 > y (this is
possible since the tangent function increases without
bound on 0, 2). By the Intermediate Value Theorem,
f x tan x is continuous on 0, x0 and 0 < y < M ,
which implies that there exists x between 0 and x0 such
that tan x y. The argument is similar if y < 0.
, c > 0
Domain: x c2 ≥ 0 ⇒ x ≥ c2 and x 0, c2, 0 0, lim
x→0
x c2 c
x
lim
x→0
2b
x c2 c
x
x c2 c
x c2 c
1
x c2 c2
1
lim
x→0 xx c2 c
x→0 x c2 c
2c
lim
Define f 0 12c to make f continuous at x 0.
104
Chapter 2
Limits and Their Properties
118. 1. f c is defined.
119. hx xx
2. lim f x lim f c x f c exists.
x→c
15
x→0
Let x c x. As x → c, x → 0
3. lim f x f c.
x→c
Therefore, f is continuous at x c.
−3
3
−3
h has nonremovable discontinuities at
x ± 1, ± 2, ± 3, . . . .
120. (a) Define f x f2x f1x. Since f1 and f2 are continuous on a, b, so is f.
f a f2a f1a > 0 and f b f2b f1b < 0
By the Intermediate Value Theorem, there exists c in a, b such that f c 0.
f c f2c f1c 0 ⇒ f1c f2c
(b) Let f1x x and f2x cos x, continuous on 0, 2, f10 < f20 and f12 > f22.
Hence by part (a), there exists c in 0, 2 such that c cosc.
Using a graphing utility, c 0.739.
4
1 24x
(a) Domain: all x 0
121. f x (b)
(c) lim f x 4
x→0
lim f x 0
x→0 6
(d) For x near 0 and negative,
−10
10
For x near 0 and positive,
4
4
4.
1 24x 1 0
4
0.
1 24x
−6
122. The statement is true.
If y ≥ 0 and y ≤ 1, then y y 1 ≤ 0 ≤ x2, as desired. So assume y > 1. There are now two cases.
1
Case 1: If x ≤ y 2, then 2x 1 ≤ 2y and
Case 2: If x ≥ y 12
y y 1 y y 1 2y
x2 ≥ y 12 2
≤ x 12 2y
y2 y 14
x2 2x 1 2y
> y2 y
≤ x2 2y 2y
y y 1
x
2
In both cases, y y 1 ≤ x2.
123. P1 P02 1 P02 1 1
P2 P12 1 P12 1 2
P5 P22 1 P22 1 5
Continuing this pattern, we see that Px x for infinitely many values of x.
Hence, the finite degree polynomial must be constant: Px x for all x.
Section 2.5
Section 2.5
1.
x
x2 4
lim 2
x
x 4
x→2
5. f x 2.
lim
1
x2
lim
1
x2 x→2
x→2
2
1
x2 9
2.99
2.9
2.5
f x
0.308
1.639
16.64
166.6
166.7
16.69
1.695
0.364
x
lim sec
x
4
lim f x x→3
3.5
f x 1.077
x
4
x→2
2.999
3.1
3.01
3.001
2.999
2.99
2.9
2.5
5.082 50.08
500.1
499.9
49.92
4.915
0.9091
lim f x x→3
x
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
f x
3.769
15.75
150.8
1501
1499
149.3
14.25
2.273
2
lim f x 8. f x sec
x
6
x
3.5
f x 3.864
lim f x x→3
3.1
3.01
19.11
191.0
lim f x x→3
x→0
x
4
lim sec
x→2
3.001
x→3
9. lim
lim tan
x→2
4.
3.01
lim f x x2
x 9
x
4
3.1
x→3
7. f x lim tan
x→2
3.5
lim f x x
x2 9
3.
x
x→3
6. f x 3.001 2.999 2.99 2.9
2.5
1910
3.864
1910
191.0
19.11
lim f x x→3
1
1
lim
x2 x→0 x2
Therefore, x 0 is a vertical asymptote.
10. lim
x→2
lim
x→2
4
x 23 4
x 23
Therefore, x 2 is a vertical asymptote.
11. lim
x→2
lim
x→2
x2 2
x 2x 1 x2 2
x 2x 1
Therefore, x 2 is a vertical asymptote.
lim x2 2
x 2x 1 lim x2 2
x 2x 1
x→1
x→1
105
Infinite Limits
lim 2
x→2
Infinite Limits
Therefore, x 1 is a vertical asymptote.
12. lim
x→0
2x
2x
lim 2
x→0
1 x
x 1 x
x2
Therefore, x 0 is a vertical asymptote.
lim
2x
x21 x lim
2x
x21 x
x→1
x→1
Therefore, x 1 is a vertical asymptote.
106
13.
Chapter 2
lim x→2
Limits and Their Properties
x2
x2
and lim 2
x→2 x 4
4
x2
14. No vertical asymptote since the denominator is never zero.
Therefore, x 2 is a vertical asymptote.
lim
x→2
x2
x2
and lim 2
x→2 x 4
x2 4
Therefore, x 2 is a vertical asymptote.
15. No vertical asymptote since the denominator is never zero.
16.
lim hs and lim hs .
s→5
s→5
Therefore, s 5 is a vertical asymptote.
lim hs and lim hs .
s→5
s→5
Therefore, s 5 is a vertical asymptote.
17. f x tan 2x x
18. f x sec x 2n 1 n
, n any integer.
4
4
2
19. lim 1 t→0
sin 2x
has vertical asymptotes at
cos 2x
4
4
lim 1 2
t→0
t2
t
x
20. gx Therefore, t 0 is a vertical asymptote.
1
has vertical asymptotes at
cos x
2n 1
, n any integer.
2
12x3 x2 4x 1 xx2 2x 8
3x2 6x 24
6 x2 2x 8
1
x,
6
x 2, 4
No vertical asymptotes. The graph has holes at x 2
and x 4.
21.
lim
x
x 2x 1 lim
x
x 2x 1
x→2
x→2
Therefore, x 2 is a vertical asymptote.
lim
x
x 2x 1 lim
x
x 2x 1
x→1
x→1
22. f x 4x2 x 6
4x 3x 2
xx3 2x2 9x 18 xx 2x2 9
4
, x 3, 2
xx 3
Vertical asymptotes at x 0 and x 3. The graph has
holes at x 3 and x 2.
Therefore, x 1 is a vertical asymptote.
x3 1 x 1x2 x 1
x1
x1
has no vertical asymptote since
23. gx lim gx lim x2 x 1 3.
x→1
x→1
24. hx has no vertical asymptote since
lim hx lim
x→2
e2x
x1
x 1 is a vertical asymptote.
25. f x x2 4
x 2x 2
x3 2x2 x 2 x 2x2 1
x→2
4
x2
.
x2 1 5
26. g x xe2x
No vertical asymptotes.
Section 2.5
lnt2 1
t2
27. ht z ± 2 are vertical asymptotes.
1
ex 1
30. f x lnx 3
x 3 is a vertical asymptote.
x 0 is a vertical asymptote.
t
has vertical asymptotes at t n,
sin t
n a nonzero integer. There is no vertical asymptote at
t 0 since
31. st lim
t→0
t
1.
sin t
32. g 2n 1 n, n any integer.
2
2
There is no vertical asymptote at 0 since
lim
x2 1
lim x 1 2
x→1 x 1
x→1
tan sin has vertical asymptotes at
cos →0
tan 1.
x2 6x 7
lim x 7 8
x→1
x→1
x1
33. lim
34. lim
2
2
−3
−3
3
3
−12
−5
Removable discontinuity at x 1
35.
lim x2 1
x1
lim x2 1
x1
x→1
x→1
f x Removable discontinuity at x 1
36. lim
8
x→1
−3
Vertical asymptote at
x 1
37.
107
28. f z lnz2 4 lnz 2 lnz 2
t 2 is a vertical asymptote.
29. f x Infinite Limits
3
sinx 1
1
x1
Removable discontinuity at
x 1
38. f x ex1 1ex1 1
ex1,
ex1 1
x 1
lnx2 1
x1
Vertical asymptote
at x 1
3
−5
3
−5
2
−1
Removable discontinuity at x 1
39. lim
x3
x2
40. lim
x2 x
1x
41. lim
x2
x 3x 3 42. lim
x2
x→3
3
−2
7
x→2
−3
−8
e2x1 1
ex1 1
−3
2
x→1
x→4
x2
1
16 2
108
43.
Chapter 2
Limits and Their Properties
x2 2x 3
x1 4
lim x→3 x 2
x2 x 6
5
lim x→3
44.
x2 x
x
1
lim 2
x→1 x 1x 1
x→1 x 1
2
45. lim
x→0
49. lim
x→0
51.
53.
46. lim
2
47. lim 1 x→3
1
x
lim
ln cos x ln cos
lim
x sec x and
x→ 12
x2 1
x2
9
x→0
50.
6x2 x 1
3x 1 5
lim
4x2 4x 3 x→12 2x 3 8
48. lim x2 2
sin x x→ 2
lim
x→12
ln 0 2
lim
x→ 12
lim
x→ 2
2
cos x 52. lim e0.5x sin x 10 0
x→0
x sec x 54.
lim
x→ 12
x2 tan x and
lim f x lim
x→1
x→1
3
1
x1 56. f x −4
5
x2
x3 1
x1
4
lim f x lim x 1 0
x→1
−8
x2
1
25
−4
0.3
−8
lim f x 8
x→1
−3
57. f x x2 tan x .
x→ 12
x→ 12
x2 x 1
x3 1
lim
x→ 12
Therefore, lim x2 tan x does not exist.
Therefore, lim x sec x does not exist.
55. f x 2
x
58. f x sec
8
x→5
x
6
lim f x 6
−9
9
x→3
−0.3
59. A limit in which f x increases or decreases without
bound as x approaches c is called an infinite limit. is
not a number. Rather, the symbol
−6
60. The line x c is a vertical asymptote if the graph of f
approaches ± as x approaches c.
lim f x x→c
says how the limit fails to exist.
61. One answer is f x 63.
x3
x3
.
x 6x 2 x2 4x 12
y
62. No. For example, f x 64. P 3
2
lim
V→0
1
−2
x
−1
1
−1
−2
3
1
has no vertical asymptote.
x2 1
k
V
k
k (In this case we know that k > 0.)
V
Section 2.5
200
ftsec
6
3
(b) r 50 sec2 200 ftsec
3
65. (a) r 50 sec2
(c)
lim
→ 2
66. C (b) C50 $528 million
50 sec2 (c) C75 $1584 million
m0
lim m lim
v→c
lim
x→100
68. (a) r 1 v2c2
v→c
528x
, 0 ≤ x < 100
100 x
(a) C25 $176 million
(d)
67. m m0
1 v2c2
(b) r 27
Total distance
Total time
50 2d
dx dy
50 2xy
yx
(b)
215
60
y
150
66.667
50
42.857
x→25
25x
x 25 As x gets close to 25 mph, y becomes larger and larger.
Domain: x > 25
1
0.5
0.2
0.1
0.01
0.001
0.0001
0.1585
0.0411
0.0067
0.0017
0
0
0
0.5
lim
x→0
x sin x
0
x
1.5
− 0.25
x
f x
1
0.5
0.2
0.1
0.01
0.001
0.0001
0.1585
0.0823
0.0333
0.0167
0.0017
0
0
0.25
− 1.5
1.5
− 0.25
—CONTINUED—
50
25x
y
x 25
(b)
3
ftsec
2
40
50x 2yx 25
− 1.5
30
50x 2xy 50y
f x
2x
625 x2
7
ftsec
12
x
(c) lim
50y 50x 2xy
x
625 225
x→25
69. (a) Average speed 528
Thus, it is not possible.
100 x 625 49
(c) lim
70. (a)
Infinite Limits
lim
x→0
x sin x
0
x2
109
110
Chapter 2
Limits and Their Properties
70. ––CONTINUED––
(c)
x
f x
1
0.5
0.2
0.1
0.1585
0.1646
0.1663
0.1666
0.01
0.001
0.0001
0.1667 0.1667
0.1667
0.25
− 1.5
lim
1.5
x→0
x sin x
0.1167 16
x3
− 0.25
(d)
x
f x
1
0.5
0.2
0.1
0.1585
0.3292
0.8317
1.6658
0.01
0.001
0.0001
16.67
166.7
1667.0
1.5
− 1.5
1.5
lim
x→0
x sin x
x4
− 1.5
For n ≥ 3, lim
x→0
x sin x
.
xn
1
1
1
1
71. (a) A bh r2 1010 tan 102
2
2
2
2
(b)
f 50 tan 50 Domain:
(c)
(d)
0, 2 0.3
0.6
0.9
1.2
1.5
0.47
4.21
18.0
68.6
630.1
100
lim A → 2
lim A → 2
0
1.5
0
72. (a) Because the circumference of the motor is
half that of the saw arbor, the saw makes
17002 850 revolutions per minute.
(b) The direction of rotation is reversed.
(d)
(c) 220 cot 210 cot : straight sections.
The angle subtended in each circle is
2 2
2
(e)
0.3
0.6
0.9
1.2
1.5
L
306.2
217.9
195.9
189.6
188.5
450
2.
Thus, the length of the belt around the pulleys is
20 2 10 2 30 2.
Total length 60 cot 30 2
Domain: 0,
2
2
0
0
(f)
lim
→ 2
L 60 188.5
(All the belts are around pulleys.)
(g) lim L →0
Section 2.5
73. False; for instance, let
74. False; for instance, let
x2 1
.
x1
f x The graph of f has a hole at 1, 2,
not a vertical asymptote.
f x x2 1
or
x1
gx x
.
x2 1
1,
f x x
3,
x0
x 0.
1
1
and gx 4, and c 0.
x2
x
lim
1
1
and lim 4 , but
x→0 x
x2 lim
x1 x1 lim x x 1 0.
x→0
The graph of f has a vertical asymptote at x 0, but
f 0 3.
111
75. True.
77. Let f x 76. False; let
Infinite Limits
2
x→0
2
4
x→0
4
78. Given lim f x and lim gx L:
x→c
x→c
(2) Product:
If L > 0, then for L2 > 0 there exists 1 > 0 such that gx L < L2 whenever 0 < x c < 1. Thus,
L2 < gx < 3L2. Since lim f x then for M > 0, there exists 2 > 0 such that f x > M2L whenever
x→c
x c < 2. Let be the smaller of 1 and 2. Then for 0 < x c < , we have f xgx > M2LL2 M.
Therefore lim f xgx . The proof is similar for L < 0.
x→c
(3) Quotient: Let > 0 be given.
There exists 1 > 0 such that f x > 3L2 whenever 0 < x c < 1 and there exists 2 > 0 such that gx L < L2
whenever 0 < x c < 2. This inequality gives us L2 < gx < 3L2. Let be the smaller of 1 and 2. Then for
0 < x c < , we have
gx
3L2
<
.
f x
3L2
Therefore, lim
x→c
gx
0.
f x
79. Given lim f x , let g x 1. Then lim
x →c
80. Given lim
x→c
x →c
gx
0 by Theorem 1.15.
f x
1
0. Suppose lim f x exists and equals L.
x→c
f x
lim 1
1
1
x→c
0.
x→c f x
lim f x L
Then, lim
x→c
1
is defined for all x > 3. Let M > 0 be
x3
1
given. We need > 0 such that f x > M
x3
whenever 3 < x < 3 .
81. f x Equivalently, x 3 <
This is not possible. Thus, lim f x does not exist.
x→c
1
whenever x 3 < , x > 3.
M
1
So take . Then for x > 3 and x 3 < ,
M
1
1
> M and hence f x > M.
x3 8
1
1
< N whenever 4 < x < 4.
is defined for all x < 4. Let N < 0 be given. We need > 0 such that f x x4
x4
1
1
1
< whenever x 4 < , x < 4. So take
Equivalently, x 4 > whenever x 4 < , x < 4. Equivalently,
N
x4
N
1
1
1
1
1
. Note that > 0 because N < 0. For x 4 < and x < 4,
> N, and
< N.
N
x4
x4
x4
82. f x 112
Chapter 2
Limits and Their Properties
Review Exercises for Chapter 2
2. Precalculus. L 9 12 3 12 8.25
1. Calculus required. Using a graphing utility, you can
estimate the length to be 8.3. Or, the length is slightly
longer than the distance between the two points, 8.25.
11
−9
9
−1
3.
x
0.1
0.01
0.001
0.001
0.01
f x
1.0526
1.0050
1.005
0.9995
0.9950
0.01
0.001
0.001
0.01
0.1
1.4140
1.4125
1.3970
0.1
0.9524
lim f x 1
x→0
4.
0.1
x
f x
1.4323
1.6569
exact: 2
lim f x 1.414
x→0
5.
0.1
x
f x
1.4144
0.01
0.8867
0.001
0.0988
0.0100
0.001
0.01
0.1
0.0100
0.1013
1.1394
lim f x 0
x→0
6.
0.1
x
f x
0.01
0.001
0.2
0.2
0.202
0.001
0.01
0.1
0.2
0.2
0.198
lim f x 0.2
x→0
7. hx x2 2x
x
8. gx 3x
x2
9. (a) lim f t does not exist.
t →0
(a) lim hx 2
(a) lim gx does not exist.
(b) lim hx 3
(b) lim gx 0
x→0
x→1
10. (a) lim gx 0
x→0
(b) lim gx 0
x→2
(b) lim f t 0
t →1
x→2
x→0
11. lim 3 x 3 1 2
x→1
Let > 0 be given. Choose . Then for
0 < x 1 < , you have
x 1 < 1 x < 3
x 2 < f x L < .
Review Exercises for Chapter 2
Assuming 4 < x < 16, you can choose 5.
12. lim x 9 3.
x→9
Hence, for 0 < x 9 < 5, you have
Let > 0 be given. We need
x 9 < 5 < x 3 x 3 < ⇒ x 3x 3 < x 3
x 9 < x 3 .
x 3 < f x L < .
13. lim x2 3 1
x→2
1
x 2.
Let > 0 be given. We need x2 3 1 < ⇒ x2 4 x 2x 2 < ⇒ x 2 <
Assuming, 1 < x < 3, you can choose 5. Hence, for 0 < x 2 < 5 you have
1
x 2 < 5 < x 2
x 2x 2 < x2 4 < x2 3 1 < f x L < .
14. lim 9 9. Let > 0 be given. can be any positive
x→5
15. lim t 2 4 2 6 2.45
t→4
number. Hence, for 0 < x 5 < , you have
9 9 < f x L < .
16. lim 3 y 1 3 4 1 9
17. lim
t2 9
lim t 3 6
t→3
t3
19. lim
y→4
18. lim
t→3
t→2
x→4
t2
1
1
lim
t2 4 t→2 t 2
4
x 2
x4
lim
x→4
lim
x→4
20. lim
x→0
4 x 2
x
lim
lim
x→0
22. lim
s→0
4 x 2
x
x→0
1
4 x 2
4 x 2
4 x 2
21. lim
1
4
11 s 1 lim 11 s 1 11 s 1
s→0
s
s
11 s 1
lim
s→0
x→0
x 2
x 2x 2
1
x 2
1
4 2
1x 1 1
1 x 1
lim
x→0
x
xx 1
1
1
lim
x→0 x 1
11 s 1
1
1
lim
s 11 s 1 s→0 1 s 11 s 1 2
1
4
113
114
Chapter 2
Limits and Their Properties
x3 125
x 5x2 5x 25
lim
x→5 x 5
x→5
x5
23. lim
24. lim
x→2
x 2x 2
x2 4
lim
x3 8 x→2 x 2x2 2x 4
lim x2 5x 25
lim
x→5
x→2
75
1 cos x
x
lim
x→0 sin x
sin x
25. lim
x→0
27. lim
x→0
1 xcos x 10 0
x→0
0
x→0
lim
x→ 4
1
4
12
3
4x
44
tan x
1
sin6 x 12
sin6 cos x cos6 sin x 12
lim
x→0
x
x
lim
28. lim
26.
x2
x2 2x 4
1
2
3
2
3 sin x
cos x 1
lim
x
x→0 2
x
1 3
2
cos x 1
cos cos x sin sin x 1
lim
x→0
x
x
lim
x→0
cos xx 1
lim sin x→0
sin x
x
0 01 0
29. lim e x1 sin
x→1
lnx 12
2 lnx 1
lim
lim 2 2
x→2 lnx 1
x→2 lnx 1
x→2
x
e0 sin 1
2
2
30. lim
4323 21
3
2
7
32. lim f x 2gx 2
x→c
4
3
12
31. lim f x gx x→c
33. f x (a)
2x 1 3
x1
(b)
x
1.1
1.01
1.001
1.0001
f x
0.5680
0.5764
0.5773
0.5773
lim
x→1
(c) lim
x→1
2x 1 3
x1
2x 1 3
x1
2
−1
0.577
lim
x→1
lim
x→1
lim
x→1
Actual limit is 33.
2x 1 3
x1
2x 1 3
2x 1 3
2x 1 3
x 12x 1 3 2
2x 1 3
2
1
3
23 3
3
2
0
Review Exercises for Chapter 2
34. f x (a)
3
1 x
x1
x
f x
lim
x→1
(c) lim
x→1
115
1.1
1.01
1.001
0.3228
0.3322
0.3332
3
1 x
0.333
x1
0.3333
3
−3
3
3
1 x x
2
3
3
1 x x
1x
3 x x 11 3 x2
lim
1
3 x 3 x2
1 −3
2
lim
x→1
2
Actual limit is 13 .
3
3
1 x
1 x
lim
x→1
x1
x1
x→1
(b)
1.0001
1
3
sa st
4.942 200 4.9t2 200
lim
t→a
t→4
at
4t
35. lim
lim
t→4
4.9t 4t 4
4t
lim 4.9t 4 39.2 msec
t→4
36. st 0 ⇒ 4.9t2 200 0 ⇒ t2 40.816 ⇒ t 6.39 sec
When t 6.39, the velocity is approximately
lim
t→a
sa st
lim 4.9a t
t→a
at
lim 4.96.39 6.39 62.6 msec.
t→6.39
37. lim
x→3
x 3 x3
lim
x→3 x 3
1
x3
x→1
x→2
41. lim ht does not exist because lim ht 1 1 2 and
t→1
lim ht t→1
x→4
at x 4.
40. lim gx 1 1 2
39. lim f x 0
t→1
1
2 1
38. lim x 1 does not exist. The graph jumps from 2 to 3
42. lim f s 2
s→2
1 1.
43. f x x 3
44. f x lim x 3 k 3 where k is an integer.
x→k
lim x 3 k 2 where k is an integer.
x→k
Nonremovable discontinuity at each integer k
Continuous on k, k 1 for all integers k
3x2 x 2 ,
x1
0,
lim f x lim
x→1
x→1
x
1
x1
3x x 2
x1
2
lim 3x 2 5 0
x→1
Removable discontinuity at x 1
Continuous on , 1 1, 116
Chapter 2
45. f x Limits and Their Properties
3x2 x 2 3x 2x 1
x1
x1
46. f x lim f x lim 3x 2 5
x→1
lim 5 x 3
Removable discontinuity at x 1
Continuous on , 1 1, lim
x→2
lim 2x 3 1
x→2
Nonremovable discontinuity at x 2
Continuous on , 2 2, x x 1 1 1x
1
lim 1 x
1
x 22
48. f x 1
x 22 x→0
Domain: , 1, 0, Nonremovable discontinuity at x 2
Continuous on , 2 2, 49. f x Nonremovable discontinuity at x 0
Continuous on , 1 0, 3
x1
50. f x lim f x x→1
lim
x→1
lim f x x→1
x1
2x 2
x1
1
2x 1 2
Removable discontinuity at x 1
Continuous on , 1 1, Nonremovable discontinuity at x 1
Continuous on , 1 1, 51. f x csc
x ≤ 2
x > 2
x→2
x→1
47. f x 52xx,3,
x
2
52. f x tan 2x
Nonremovable discontinuities when
Nonremovable discontinuities at each even integer.
Continuous on 2k, 2k 2 for all integers k.
x
2n 1
4
Continuous on
2n 4 1, 2n 4 1
for all integers n.
54. hx 5 ln x 3 is continuous
on , 3 and 3, . There
is a nonremovable discontinuity
at x 3.
53. gx 2e x4 is continuous on
all intervals n, n 1, where n
is an integer. g has nonremovable
discontinuities at each n.
55. f 2 5
Find c so that lim cx 6 5.
x→2
c2 6 5
2c 1
c 12
56. lim x 1 2
x→1
lim x 1 4
x→3
Find b and c so that lim x2 bx c 2 and lim x2 bx c 4.
x→1
Consequently we get
Solving simultaneously,
x→3
1bc2
b
and 9 3b c 4.
3 and
c 4.
Review Exercises for Chapter 2
57. f is continuous on 1, 2. f 1 1 < 0 and
f 2 13 > 0. Therefore by the Intermediate Value
Theorem, there is at least one value c in 1, 2 such
that 2c3 3 0.
58. C 9.80 2.50x 1, x > 0
9.80 2.50x 1
C has a nonremovable discontinuity at each integer.
30
0
5
0
59. A 50001.062t
60. f x x 1x
(a) Domain: , 0 1, Nonremovable discontinuity every 6 months
(b) lim f x 0
9000
x→0
(c) lim f x 0
x→1
5
0
4000
61. gx 1 2
x
62. hx Vertical asymptote at x 0
63. f x 4x
4 x2
Vertical asymptotes at x 2 and x 2
8
x 102
64. f x csc x
Vertical asymptote at every integer k
Vertical asymptote at x 10
65. gx ln9 x2 ln3 x ln3 x
66. f x 10e2x
Vertical asymptote at x 0
Vertical asymptotes at x ± 3
lim f x x→0
67.
69.
lim
2x2 x 1
x2
68.
lim
x1
1
1
lim
x3 1 x→1 x2 x 1 3
70.
x→2
x→1
71. lim
x2 2x 1
x1
73. lim
sin 4x
4 sin 4x
lim
x→0
5x
5 4x
75. lim
csc 2x
1
lim
x →0 x sin 2x
x
x→1
x→0
x→0
77. lim lnsin x x→0
72.
54
lim
x→ 12
x
2x 1 lim
x1
1
1
lim
x4 1 x→1 x2 1x 1
4
lim x2 2x 1
x1
x→1
x→1
74. lim
sec x
x
76. lim
cos2 x
x
x→0
x→0
78. lim 12e2x x→0
117
118
Chapter 2
79. f x (a)
tan 2x
x
0.1
x
f x
lim
Limits and Their Properties
x→0
0.01
2.0271
0.001
2.0003
2.0000
0.001
0.01
0.1
2.0000
2.0003
2.0271
tan 2x
2
x
Analytically,
tan 2x sin 2x
x
2x
2
cos 2x
→ 2.
(b) Yes, define
f x 2, x
tan 2x ,
x0
x0
.
Now f x is continuous at x 0.
Problem Solving for Chapter 2
1. (a) Perimeter PAO x2 y 12 x2 y2 1
x2 x2 12 x2 x4 1
Perimeter PBO x 12 y2 x2 y2 1
x 12 x4 x2 x4 1
(b) rx x2 x2 12 x2 x4 1
(c) lim r x x 12 x4 x2 x4 1
x
x→0
4
2
1
0.1
0.01
Perimeter PAO
33.02
9.08
3.41
2.10
2.01
Perimeter PBO
33.77
9.60
3.41
2.00
2.00
rx
0.98
0.95
1
1.05
1.005
1
1
x
2. (a) Area PAO bh 1x 2
2
2
1
1
y
x2
Area PBO bh 1y 2
2
2
2
(b) ax Area PBO x22
x
Area PAO
x2
x
4
2
1
0.1
0.01
Area PAO
2
1
12
120
1200
Area PBO
8
2
12
1200
120,000
ax
4
2
1
110
1100
(c) lim ax lim x 0
x→0
x→0
101 2
1
101 2
Problem Solving for Chapter 2
3. (a) There are six triangles, each with a central angle of
60 3. Hence,
4. (a) Slope 3
25
y x
4
4
33
2.598.
2
(c) Let Q x, y x, 25 x2
h = sin θ
h = sin 60°
mx 1
1
θ
60°
40 4
30 3
3
3
(b) Slope , Tangent line: y 4 x 3
4
4
12bh 6 121 sin 3 Area hexagon 6
25 x2 4
x3
(d) lim mx lim
x→3
(b) There are n triangles, each with central angle of
2n. Hence,
12bh n 121 sin 2n n sin 22n.
An
24
48
96
2.598
3
3.106
3.133
3.139
25 x2 4
25 x2 4
lim
3 x3 x
x 325 x2 4
lim
x→3
12
25 x2 16
x 325 x2 4
x→3
An n
6
x3
lim
x→3
n
25 x2 4
x→3
33
Error: 0.5435.
2
(c)
119
3 x
25 x2 4
3
6
44
4
This is the slope of the tangent line at P.
(d) As n gets larger and larger, 2n approaches 0.
Letting x 2n,
An sin2n sin2n
sin x
2n
2n
x
which approaches 1 .
5. (a) Slope 12
5
6.
5
(b) Slope of tangent line is .
12
5
y 12 x 5
12
y
(c) Q x, y x, 169 mx x
a bx 3
x
12 169 x2
x→5
x5
(d) lim mx lim
lim
3 bx 3
x
x→0
12 169 x2
12 169 x2
lim
x2 25
x 512 169 x2
lim
x 5
12 169 x2
lim
x→0
bx
x 3 bx 3
b
.
lim
x→0 3 bx 3
144 169 x2
lim
x→5 x 5 12 169 x2 x→5
a bx 3
Thus,
169 12
x5
x→5
a bx 3
a bx 3
xa bx 3
x2
x→5
Letting a 3 simplifies the numerator.
5
169
x
, Tangent line
12
12
x2
a bx 3
10
5
12 12 12
This is the same slope as part (b).
Setting
b
3 3
3, you obtain b 6.
Thus, a 3 and b 6.
120
Chapter 2
Limits and Their Properties
7. (a) 3 x13 ≥ 0
(d) lim f x lim
x→1
x13 ≥ 3
lim
x→1
0.5
lim
x→1
− 30
12
− 0.1
3 x13 2
x13 1x23
x13 1
x13 13 x13 2
1
x23
x13
13 x13 2
1
1
1 1 12 2 12
3 2713 2
lim f x 27 1
x→27
1
2
0.0714
28 14
8. lim f x lim a2 2 a2 2
x→0
3 x13 2
3 x13 4
x→1 x 1 3 x13 2 Domain: x ≥ 27, x 1
(c)
x1
x→1
lim
x ≥ 27
(b)
3 x13 2
9. (a) lim f x 3: g1, g4
x→0
x→2
ax
tan x
lim f x lim
a because lim
1
x→0
x→0 tan x
x→0
x
(b) f continuous at 2: g1
(c) lim f x 3: g1, g3, g4
x→2
Thus,
a2 2 a
a2 a 2 0
a 2a 1 0
a 1, 2.
10. f x 1 x1
x
4
Near x 0, f x 2.718.
There is no y-intercept because f 0 is not defined.
−2
0.1
x
f x
0.01
2.8680
0.001
2.7320
2.7196
0
0.001
0.01
0.1
undef.
2.7169
2.7048
2.5937
lim f x .
x→1 11.
y
(a)
4
f 0 0
3
f 12 0 1 1
2
1
−4 −3 −2 −1
−2
−3
−4
f 1 1 1 1 1 0
x
1
2
3
f 2.7 3 2 1
4
(b)
lim f x 1
x→1
lim f x 1
x→1
lim f x 1
x→12
(c) f is continuous for all real numbers except
x 0, ± 1, ± 2, ± 3, . . .
4
0
Problem Solving for Chapter 2
v2 12. (a)
192,000
v02 48
r
13. (a)
192,000
v 2 v02 48
r
r
lim r v→0
2
192,000
v v02 48
1
x
192,000
48 v02
a
v2 x→a
(ii) lim Pa, bx 0
1920
v02 2.17
r
x→a
(iii) lim Pa, bx 0
x→b
(iv) lim Pa, bx 1
1920
v 2 v02 2.17
r
r
lim r v→0
x→b
(c) Pa, b is continuous for all positive real numbers
except x a, b.
1920
v2 v02 2.17
(d) The area under the graph of u, and above the
x-axis, is 1.
1920
2.17 v02
Let v0 2.17 misec 1.47 mi
sec.
r
(c)
lim r v→0
10,600
v 2 v02 6.99
10,600
6.99 v02
Let v0 6.99 2.64 misec.
Since this is smaller than the escape velocity for earth,
the mass is less.
14. Let a 0 and let > 0 be given. There exists 1 > 0
such that if 0 < x 0 < , then f x L < .
Let 1 a . Then for 0 < x 0 < 1 a ,
you have
x < a1
ax < 1
f ax L < .
As a counterexample, let f x Then lim f x 1 L,
x→0
but lim f ax lim f 0 2.
x→0
x→0
b
(b) (i) lim Pa, bx 1
Let v0 48 43 feetsec.
(b)
y
12
x0
.
x0
121