C H A P T E R 1
Limits and Their Properties
Section 1.1
A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 26
Section 1.2
Finding Limits Graphically and Numerically . . . . . . . . 26
Section 1.3
Evaluating Limits Analytically
Section 1.4
Continuity and One-Sided Limits . . . . . . . . . . . . . . 36
Section 1.5
Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 41
. . . . . . . . . . . . . . . 31
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
C H A P T E R 1
Limits and Their Properties
Section 1.1
A Preview of Calculus
3. Calculus required: slope of tangent line at x 2 is rate of
change, and equals about 0.16.
1. Precalculus: 20 ftsec15 seconds 300 feet
5. Precalculus: Area 12 bh 12 53 15
2 sq. units
7. f x 4x x2
(a)
(b) slope m y
4
P
3
4x x2 3
x1
x 13 x
3 x, x 1
x1
x 2: m 3 2 1
x 1.5: m 3 1.5 1.5
x
1
2
3
x 0.5: m 3 0.5 2.5
(c) At P1, 3 the slope is 2.
You can improve your approximation of the slope at
x 1 by considering x-values very close to 1.
5 5 5
10.417
2 3 4
5
5
5
5
5
5
5
1
9.145
Area 5 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.
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 1.2
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
26
Finding Limits Graphically and Numerically
x2
0.3333
x2 x 2
Actual limit is 13 .
Section 1.2
3.
0.1
x
f x
0.01
0.2911
lim
5.
x
x
2.9
f x
0.0641
lim
x→3
7.
0.2889
x 3 3
x→0
f x
lim
0.001
0.01
0.1
0.2887
0.2884
0.2863
2.999
3.001
3.01
0.0627
0.0625
0.0625
0.0623
0.01
0.99998
sin x
1.0000
x
0.001
1.0000
0.01
0.1
1.0000
0.99998
0.9983
(Actual limit is 1.) (Make sure you use radian mode.)
11. lim f x lim 4 x 2
x→2
x 5 does not exist. For values of x to the left of 5,
x5
x 5 x 5 equals 1, whereas for values of x to the
right of 5, x 5 x 5 equals 1.
0.0610
0.001
x→3
x→5
3.1
Actual limit is 161 .
9. lim 4 x 1
13. lim
27
Actual limit is 123 .
0.2887
2.99
0.9983
x→0
0.2887
1x 1 14
0.0625
x3
0.1
x
0.001
Finding Limits Graphically and Numerically
x→2
15. lim sin x 0
x→1
17. lim cos1x does not exist since the function oscillates between 1 and 1 as x approaches 0.
x→0
19. (a) f 1 exists. The black dot at 1, 2 indicates that
f 1 2.
(b) lim f x does not exist. As x approaches 1 from the
x→1
left, f x approaches 2.5, whereas as x approaches 1
from the right, f x approaches 1.
(c) f 4 does not exist. The hollow circle at 4, 2
indicates that f is not defined at 4.
(d) lim f x exists. As x approaches 4, f x approaches 2:
x→4
lim f x 2.
x→4
21. lim f x exists for all c 3. In particular, lim f x 2.
x→c
x→2
23.
25. One possible answer is
y
6
y
5
6
4
3
5
f
2
4
f
1
−2 −1
−1
x
1
2
3
4
5
−2
lim f x exists for all values of c 4.
x→c
2
1
−2 −1
−1
x
1
2
3
4
5
28
Chapter 1
Limits and Their Properties
27. Ct 0.75 0.50 t 1
(a)
(b)
t
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
3
lim Ct 2.25
t→3.5
5
0
(c)
0
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
29. 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.
31. You need to find such that 0 < x 1 < implies
1
f x 1 1 < 0.1. That is,
x
So take 1
1 < 0.1
x
0.1 <
1 0.1 <
1
x
< 1 0.1
9
<
10
1
x
<
11
10
>
10
11
10
>
9
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
f x 1 1
1 < 0.1.
x
10
10
1 > x 1 >
1
9
11
1
1
> x 1 > .
9
11
33. 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
, you have
Hence, if 0 < x 2 < 3
0 < x2 <
3x
6
< 0.01
3x 2 8 < 0.01
f x L < 0.01
3 x 2 < 0.01
35. lim x2 3 1 L
x→2
x2 3 1 < 0.01
x2 4 < 0.01
x 2x 2 < 0.01
x 2 x 2 < 0.01
0.01
x 2 < x 2
If we assume 1 < x < 3, then 0.015 0.002.
Hence, if 0 < x 2 < 0.002, you have
1
1
x 2 < 0.002 50.01 < x 20.01
x 2x 2 < 0.01
x2 4 < 0.01
x2 3 1 < 0.01
f x L < 0.01
Section 1.2
37. lim x 3 5
Finding Limits Graphically and Numerically
39. lim
x→4
x→2
12 x 1 12 4 1 3
Given > 0:
Given > 0:
x 1 3
x 2
x 3 5 < x 2 < 1
2
1
2
Hence, let .
1
2
Hence, if 0 < x 2 < , you have
x 2 < x 3 5 < f x L < .
< < x 4 < x 4 < 2
Hence, let 2.
Hence, if 0 < x 4 < 2, you have
x 4 < 2
x 2
x 1 3
1
2
1
2
< < f x L < .
41. lim 3 3
3
x0
43. lim x→6
x→0
Given > 0:
3 3 < 0 < Hence, let 3.
Hence, for any > 0, you have
x < 3
3 x < 3 x 0 < f x L < .
Hence for 0 x 0 3,
3 3 < f x L < .
3 x < x < 3 Hence, any > 0 will work.
you have
47. lim x2 1 2
45. lim x 2 2 2 4
x→2
3 x 0 < Given > 0: x→1
Given > 0:
Given > 0:
x 2 4 < x 2 4 < x 2 x 2 x 2 < x 2 < 0
Hence, let .
Hence for 0 < x 2 < , you have
x 2 < x 2 < x 2 4 < x 2 4 < f x L < .
x2 1 2 < x2 1 < x 1x 1 < x 1 < x 1
If we assume 0 < x < 2, then 3.
Hence for 0 < x 1 < , you have
3
(because x 2 < 0)
1
1
x 1 < 3 < x 1
x2 1 < x2 1 2 < f x 2 < .
29
30
Chapter 1
49. f x Limits and Their Properties
x 5 3
51. f x x4
1
6
lim f x x→4
x9
0.5
−6
10
x 3
lim f x 6
6
x→9
−0.1667
0
10
0
The domain is 5, 4 4, . The graphing utility does
not show the hole at 4, 16 .
53. lim f x 25 means that the values of f approach 25 as
x→8
x gets closer and closer to 8.
57. (a) C 2 r
r
The domain is all x ≥ 0 except x 9. The graphing
utility does not show the hole at 9, 6.
55. No. The fact that lim f x 4 has no bearing on the value
x→2
of f at 2.
(b) If C 5.5, r 5.5
0.87535 cm
2
If C 6.5, r 6.5
1.03451 cm
2
6
3
C
0.9549 cm
2 2 Thus 0.87535 < r < 1.03451
(c) lim 2 r 6; 0.5; 0.0796
r →3
59. 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
61.
x
1
2
3
4
5
63. False; f x sin xx is undefined when x 0.
From Exercise 7, we have
0.002
(1.999, 0.001)
(2.001, 0.001)
lim
x→0
1.998
0
sin x
1.
x
2.002
Using the zoom and trace feature, 0.001. That is, for
0 < x 2 < 0.001,
67. f x x
65. False; let
f x x10, 4x,
2
x4
.
x4
f 4 10
(a) lim x 0.5 is true.
x→0.25
As x approaches 0.25 14, f x x approaches
1
2
lim f x lim x→4
x2 4
4 < 0.001.
x2
x→4
x2
4x 0 10
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.
Section 1.3
Evaluating Limits Analytically
69. 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.
71. lim f x L 0 means that for every > 0 there
x→c
exists > 0 such that if
73. Answers will vary.
0 < x c < ,
then
f x L 0 < .
This means the same as f x L < 0 < x c < .
when
Thus, lim f x L.
x→c
75. The radius OP has a length equal to the altitude z of the
h
h
triangle plus . Thus, z 1 .
2
2
1
h
Area triangle b 1 2
2
P
h
O
Area rectangle bh
b
1
h
bh
Since these are equal, b 1 2
2
1
h
2h
2
5
h1
2
h
Section 1.3
1.
2
5
Evaluating Limits Analytically
(a) lim hx 0
7
x→5
−8
13
(b) lim hx 6
x→1
3.
x→0
5. lim x4 24 16
x→2
9. lim x2 3x 32 33 9 9 0
x→3
π
−π
−7
hx x2 5x
(a) lim f x 0
4
(b) lim f x 0.524
−4
f x x cos x
7. lim 2x 1 20 1 1
x→0
x→ 3
6 31
32
Chapter 1
Limits and Their Properties
11. lim 2x2 4x 1 232 43 1 18 12 1 7
x→3
x3
13
2
2
x2 4 12 4
5
5
13. lim
1 1
x
2
15. lim
17. lim
5x
57
35
35
x 2
7 2
9
3
19. lim x 1 3 1 2
x→2
x→7
x→1
x→3
21. lim x 32 4 32 1
23. (a) lim f x 5 1 4
x→4
x→1
(b) lim gx 43 64
x→4
(c) lim g f x g f 1 g4 64
x→1
25. (a) lim f x 4 1 3
27. lim sin x sin
x→ 2
x→1
1
2
(b) lim gx 3 1 2
x→3
(c) lim g f x g3 2
x→1
x
2
1
cos
3
3
2
29. lim cos
x→2
35. lim tan
x→3
31. lim sec 2x sec 0 1
lim sin x sin
33.
x→0
x→56
4x tan 34 1
37. (a) lim 5gx 5 lim gx 53 15
x→c
x→c
(b) lim f x gx lim f x lim gx 2 3 5
x→c
x→c
x→c
(c) lim f xgx lim f x lim gx 23 6
x→c
x→c
x→c
lim f x
f x
2
x→c
(d) lim
x→c gx
lim gx 3
41. f x 2x 1 and gx x 0.
39. (a) lim f x3 lim f x3 43 64
x→c
x→c
(b) lim f x lim f x 4 2
x→c
x→c
(c) lim 3 f x 3 lim f x 34 12
x→c
x→c
(d) lim f x32 lim f x32 432 8
x→c
x→c
2x2 x
agree except at
x
x→c
43. f x xx 1 and gx x3 x
agree except at x 1.
x1
(a) lim gx lim f x 1
(a) lim gx lim f x 2
(b) lim gx lim f x 3
(b) lim gx lim f x 0
x→0
x→1
x→0
x→1
45. f x 5 1
6
2
x→1
x→1
x2 1
and gx x 1 agree except at x 1.
x1
lim f x lim gx 2
x→1
x→1
47. f x x 2.
x→1
x3 8
and gx x2 2x 4 agree except at
x2
lim f x lim gx 12
x→1
x→2
x→2
3
12
−3
4
−4
−9
9
0
Section 1.3
49. lim
x→5
x5
x5
lim
x2 25 x→5 x 5x 5
lim
x→5
53. lim
x
lim
x→0
x 5 3
x4
x→4
x2 x 6
x 3x 2
lim
x→3 x 3x 3
x2 9
lim
x→3
x 5 5
x
x→0
lim
55. lim
x→3
1
1
x 5 10
x 5 5
x→0
51. lim
Evaluating Limits Analytically
lim
x→4
x 5 5
x 5 5
5
x 5 5
1
1
lim
x
x 5 5 x→0 x 5 5 2
5
10
x 5 3
x4
x→4
lim
x 2 5 5
x 3 6 6
x 5 3
x 5 3
1
1
x 5 9
1
lim
x 4
x 5 3 x→4 x 5 3 9 3 6
1
1
3x 3
3 3 x
x
1
1
lim
lim
57. lim
lim
x→0
x→0 3 x3x
x→0 3 x3x
x→0 3 x3
x
9
59. lim
x→0
2x x 2x
2x 2x 2x
lim
x→0
x
x
lim 2 2
x→0
x x2 2x x 1 x2 2x 1
x2 2xx x2 2x 2x 1 x2 2x 1
lim
x→0
x→0
x
x
61. lim
lim 2x x 2 2x 2
x→0
63. lim
x 2 2
x→0
x
f x
x
0.1
0.354
0.01
0.358
Analytically, lim
2
0
0.001
0.01
0.1
0.354
?
0.354
0.353
0.349
0.354
x 2 2
x
x→0
−3
0.001
lim
x 2 2
x
x→0
lim
x→0
3
−2
x 2 2
x 2 2
2
x22
1
1
lim
0.354.
4
x
x 2 2 x→0 x 2 2 2
2
1
1
2x 2
1
65. 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
Analytically, lim
lim
x→0
x→0
x
22 x
1
0.01
0.1
0.249
0.238
x
1
1
−2
1
1
lim
.
x x→0
lim
22 x x x→0 22 x
4
33
34
Chapter 1
67. lim
x→0
Limits and Their Properties
sin x
lim
x→0
5x
sinx x15 115 5
1
69. lim
x→0
1
sin x1 cos x
lim
x→0 2
2x2
sin x
x
1 cos x
x
1
10 0
2
1 cos h2
1 cos h
lim
1 cos h
h→0
h→0
h
h
sin2 x
sin x
lim
sin x 1 sin 0 0
x→0
x→0
x
x
71. lim
73. lim
00 0
cos x
lim sin x 1
x→ 2
cot x
75. lim
x→ 2
79. f t f t
t→0
32 132 23
4
0.01
2.96
0.001
2.9996
3
0
0.001
0.01
0.1
?
3
2.9996
2.96
− 2
2
−1
The limit appears to equal 3.
sin 3t
sin 3t
Analytically, lim
lim 3
31 3.
t→0
t→0
t
3t
81. f x sin 3t
sin 3t
lim
t→0
2t
3t
sin 3t
t
0.1
t
77. lim
sin x2
x
1
− 2
x
0.1
0.01
0.001
0
0.001
0.01
0.1
f x
0.099998
0.01
0.001
?
0.001
0.01
0.099998
2
−1
sin x2
sin x2
lim x
01 0.
x→0
x→0
x
x2
Analytically, lim
83. lim
x →0
f x x f x
2x x 3 2x 3
2x 2 x 3 2x 3
2 x
lim
lim
2
lim
x →0
x →0
x →0 x
x
x
x
4
4
4x 4x x
4
4
f x x f x
x x
x
lim
2
85. lim
lim
lim
x→0 x xx x
x→0 x xx
x→0
x→0
x
x
x
87. lim 4 x2 ≤ lim f x ≤ lim 4 x2
x→0
x→0
x→0
4 ≤ lim f x ≤ 4
x→0
Therefore, lim f x 4.
x→0
89. f x x cos x
91. f x x sin x
4
93. f x x sin
1
x
6
0.5
− 3
2
3
2
− 2
−4
lim x cos x 0
x→0
2
−0.5
0.5
−6
x→0
−0.5
lim x sin x 0
lim x sin
x→0
1
0
x
Section 1.3
95. 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.
Evaluating Limits Analytically
35
97. 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
99. f x x, gx sin x, hx x→c
sin x
x
3
f
g
h
−5
5
−3
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.
101. 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
103. st 4.9t2 150
lim
t→3
s3 st
4.932 150 4.9t2 150
4.99 t2
lim
lim
t→3
t→3
3t
3t
3t
lim
t→3
4.93 t3 t
lim 4.93 t 29.4 msec
t→3
3t
105. 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
107. 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.
109. 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
x→c
f x L < b whenever 0 < x c < . Hence,
wherever 0 < x c < , we have
b f x L < or bf x bL < which implies that lim bf x bL.
x→c
M f x ≤ f xgx ≤ M f x
111.
113. False. As x approaches 0 from the left,
lim M f x ≤ lim f xgx ≤ lim M f x x→c
x→c
x→c
2
M0 ≤ lim f xgx ≤ M0
x→c
0 ≤ lim f xgx ≤ 0
−3
3
x→c
Therefore, lim f xgx 0.
x →c
−2
x 1.
x
36
Chapter 1
Limits and Their Properties
115. True
117. False. The limit does not exist.
119. Let
4
f x −3
4,4,
if x ≥ 0
if x < 0
lim f x lim 4 4.
6
x→0
x→0
lim f x does not exist since for x < 0, f x 4 and for
x ≥ 0, f x 4.
−2
x→0
rational
0,1, ifif xx isis irrational
0, if x is rational
g x x, if x is irrational
121. 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.
123. (a) lim
x→0
1 cos x
1 cos x
lim
x→0
x2
x2
1 cos x
1 cos x
(b) Thus,
1 cos x 1
1
⇒ 1 cos x x2
x2
2
2
1 cos2 x
x→0 x 1 cos x
lim
sin2 x
x→0
x2
lim
1
Section 1.4
1
⇒ cos x 1 x2 for x 0.
2
2
1
(c) cos0.1 1 0.12 0.995
2
1
1 cos x
(d) cos0.1 0.9950, which agrees with part (c).
12 21
Continuity and One-Sided Limits
lim f x 2
1. (a) lim f x 1
3. (a) lim f x 0
(b) lim f x 1
(b) lim f x 0
(b) lim f x 2
(c) lim f x 0
(c) lim f x does not exist
x→3
x→3
(c) lim f x 1
x→3
x→5
11. lim
x→0
x5
1
1
lim
x2 25 x→5 x 5 10
x x
lim
x→0 x
1
x
x→4 x→4
x→3
x→4
x→3
The function is continuous at
x 3.
7. lim
5. (a)
x→3
The function is NOT continuous
at x 4.
The function is NOT continuous
at x 3.
9.
lim
x→3 x
x2 9
does not exist because
decreases without bound as x → 3 .
x
x2 9
Section 1.4
1
1
x x x
x x
x
13. lim lim x→0
x→0
x
xx x
x
1
x→3
1
x→0
x→3
37
lim
x x→0
xx x x
lim 15. lim f x lim
Continuity and One-Sided Limits
1
xx x
1
1
2
xx 0
x
x2 5
2
2
17. lim f x lim x 1 2
x→1
x→1
lim f x lim x3 1 2
x→1
x→1
lim f x 2
x→1
19. lim cot x does not exist since
x→ lim cot x and lim cot x do not
x→ 21. lim 3x 5 33 5 4
x→4
x 3 for 3 ≤ x < 4
23. lim 2 x does not exist
x→3
because
x→ lim 2 x 2 3 5
x→3
exist.
and
lim 2 x 2 4 6.
x→3
25. f x 1
x2 4
27. f x has discontinuities at x 2 and
x 2 since f 2 and f 2 are not
defined.
31. lim f x 3 lim f x.
x→0
x→0
f is continuous on 1, 4
.
x
x
2
29. gx 25 x2 is continuous
on 5, 5
.
has discontinuities at each integer
k since lim f x lim f x.
x→k
x→k
33. f x x2 2x 1 is continuous
for all real x.
35. f x 3x cos x is continuous
for all real x.
x
1
x
is not continuous at x 0, 1. Since 2
for x 0, x 0 is a removable discontinuity,
x2 x
x x x1
whereas x 1 is a nonremovable discontinuity.
37. f x 39. f x x
is continuous for all real x.
x2 1
41. f x x2
x 2x 5
43. 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
x→2
1
1
.
x5
7
x 2
x2
has a nonremovable discontinuity at x 2 since
lim f x does not exist.
x→2
38
Chapter 1
x
1,
47. f x 2
3 x,
x ≤ 1
x > 1
x,
x2,
45. f x Limits and Their Properties
1. f 1 1
lim f x lim x 1
x→1
x→1
x→1
x→1
1. f 2 lim f x 1
lim f x lim x2 1
x→2
2.
f is continuous at x 1, therefore, f is continuous for
all real x.
x→2
2x 1 2
lim f x lim 3 x 1
x→2
x→1
2
12
2
lim f x lim
x→1
3. f 1 lim f x
49. f x x > 2
has a possible discontinuity at x 2.
has a possible discontinuity at x 1.
2.
x ≤ 2
x→2
Therefore, f has a nonremovable discontinuity at x 2.
x < 1
x ≥ 1
x
tan 4 ,
x,
x
tan 4 ,
x,
1 < x < 1
x ≤ 1 or x ≥ 1
has possible discontinuities at x 1, x 1.
1. f 1 1
f 1 1
2. lim f x 1
lim f x 1
x→1
x→1
3. f 1 lim f x
f 1 lim f x
x→1
x→1
f is continuous at x ± 1, therefore, f is continuous for all
real x.
51. f x csc 2x has nonremovable discontinuities at integer
multiples of 2.
53. f x x 1 has nonremovable discontinuities at each
integer k.
55. lim f x 0
57. f 2 8
50
x→0
lim f x 0
Find a so that lim ax2 8 ⇒ a x→0
x→2
f is not continuous at x 2.
−8
8
2.
22
8
−10
59. 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
4a
4
a 1
b
2,
f x x 1,
2,
x ≤ 1
1 < x < 3
x ≥ 3
2 1 1
61. f gx x 12
Continuous for all real x.
63. f gx 1
1
x2 5 6 x2 1
Nonremovable discontinuities at x ± 1
Section 1.4
67. f x 65. y x x
Nonremovable discontinuity at each integer
0.5
Continuity and One-Sided Limits
2xx 2x,4,
2
39
x ≤ 3
x > 3
Nonremovable discontinuity at x 3
5
−3
3
−5
7
−1.5
−5
69. f x x
x2 1
71. f x sec
Continuous on , 73. f x x
4
Continuous on:
. . . , 6, 2, 2, 2, 2, 6, 6, 10, . . .
sin x
x
3
−4
The graph appears to be continuous on the interval 4, 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.
−2
1
75. f x 16x4 x3 3 is continuous on 1, 2
.
f 1 16 and f 2 4. By the Intermediate Value
Theorem, f c 0 for at least one value of c between
1 and 2.
33
79. f x x3 x 1
77. f x x2 2 cos x is continuous on 0, .
f 0 3 and f 2 1 > 0. By the Intermediate
Value Theorem, f c 0 for the least one value of c
between 0 and .
81. gt 2 cos t 3t
f x is continuous on 0, 1
.
g is continuous on 0, 1
.
f 0 1 and f 1 1
g0 2 > 0 and g1 1.9 < 0.
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, gt 0 for at least
one value c between 0 and 1. Using a graphing utility, we
find that t 0.5636.
83. f x x2 x 1
85. f x x3 x2 x 2
f is continuous on 0, 5
.
f is continuous on 0, 3
.
f 0 1 and f 5 29
f 0 2 and f 3 19
1 < 11 < 29
The Intermediate Value Theorem applies.
2 < 4 < 19
The Intermediate Value Theorem applies.
x3 x2 x 2 4
x2 x 1 11
x2 x 12 0
x 4x 3 0
x3 x2 x 6 0
x 2x2 x 3 0
x2
x 4 or x 3
c 3 (x 4 is not in the interval.)
Thus, f 3 11.
(x2
x 3 has no real solution.)
c2
Thus, f 2 4.
40
Chapter 1
Limits and Their Properties
87. (a) The limit does not exist at x c.
(c) The limit exists at x c, but it is not equal to the
value of the function at x c.
(b) The function is not defined at x c.
89.
(d) The limit does not exist at x c.
91. True
y
5
4
3
2
1
1. f c L is defined.
2. lim f x L exists.
x→c
x
−2 −1
1
3. f c lim f x
3 4 5 6 7
−2
−3
x→c
All of the conditions for continuity are met.
The function is not continuous at x 3 because
lim f x 1 0 lim f x.
x→3
x→3
95. lim f t 28
93. 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.
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.
1.04,
97. 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
C
1.04,
1.04 0.362 t,
0 < t ≤ 2
.
t > 2
4
3
2
1
t
1
2
3
4
99. 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.
101. 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
.
103. 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, f is
t →x
t →x
not continuous for all x 0.
Section 1.5
105. (a)
107. f x S
60
1 xx,,
2
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
50
40
30
c
20
10
1 ± 1 4 1 ± 5
.
2
2
t
5
10
15 20 25 30
(b) There appears to be a limiting speed and a possible
cause is air resistance.
109. f x x c2 c
x
, c > 0
Domain: x c2 ≥ 0 ⇒ x ≥ c2 and x 0, c2, 0 0, lim
x c2 c
x
x→0
lim
x c2 c
x
x→0
lim
x→0
x c2 c
x c2 c
x c2 c2
1
1
lim
xx c2 c x→0 x c2 c 2c
Define f 0 12c to make f continuous at x 0.
111. hx xx
15
−3
3
−3
h has nonremovable discontinuities at
x ± 1, ± 2, ± 3, . . . .
113. 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 2: If x ≥ y 2
1
Case 1: If x ≤ y 2, then 2x 1 ≤ 2y and
x2 ≥ y 12 2
y y 1 y y 1 2y
≤ x 12 2y
y2 y 14
x2 2x 1 2y
> y2 y
≤ x2 2y 2y
y y 1
x2
In both cases, y y 1 ≤ x2.
Section 1.5
1.
Infinite Limits
lim 2
x
x2 4
lim 2
x
x 4
x→ 2
x→2
Infinite Limits
2
3.
lim tan
x→2
lim tan
x→2
x
4
x
4
41
42
Chapter 1
Limits and Their Properties
5. f x 1
x2 9
x
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
f x
0.308
1.639
16.64
166.6
166.7
16.69
1.695
0.364
lim f x x→3
lim f x x→3
7. f x x2
x 9
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 x→3
lim f x x→3
9. lim
x→0
1
1
lim
x2 x→0 x2
Therefore, x 0 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
13.
lim
x→2
x2
x2
and lim 2
x→2 x 4
x2 4 Therefore, x 2 is a vertical asymptote.
lim
x→2
x2
x2
x2
and lim 2
x→2 x 4
4
Therefore, x 2 is a vertical asymptote.
Therefore, x 1 is a vertical asymptote.
15. No vertical asymptote since the denominator is never zero.
17. f x tan 2x x
sin 2x
has vertical asymptotes at
cos 2x
2n 1 n
, n any integer.
4
4
2
19. lim 1 t→0
4
4
lim 1 2
t→0
t2
t
Therefore, t 0 is a vertical asymptote.
Section 1.5
21.
lim
x
x 2x 1 lim
x
x 2x 1
x→2
x→2
lim
x→1
lim
x→1
x 2x 1
43
x3 1 x 1x2 x 1
x1
x1
23. f x has no vertical asymptote since
lim f x lim x2 x 1 3.
x→1
Therefore, x 2 is a vertical asymptote.
x
Infinite Limits
x→1
The graph has a hole at x 1.
x
x 2x 1
Therefore, x 1 is a vertical asymptote.
25. f x x 5x 3
x3
,x5
x 5x2 1 x2 1
27. st No vertical asymptotes. The graph has a hole at x 5.
t
has vertical asymptotes at t n, n
sin t
a nonzero integer. There is no vertical asymptote at
t 0 since
lim
t→0
x2 1
lim x 1 2
x→1 x 1
x→1
29. lim
31.
lim x2 1
x1
lim x2 1
x1
x→1
2
−3
x→1
3
t
1.
sin t
Vertical asymptote at x 1
−5
Removable discontinuity at x 1
33. lim
x→2
37.
x3
x2
lim
x→3
35. lim
x→3
x2 2x 3
x 1x 3
x1 4
lim lim x→3 x 2x 3
x→3 x 2
x2 x 6
5
x2 x
x
1
lim 2
x→1 x 1x 1
x→1 x 1
2
39. lim
43. lim
x→0
47.
x2
x 3x 3 x→0
2
sin x lim
x→ 12
x sec x and
45. lim
x→ lim
x→ 12
41. lim 1 2
x
lim x sin x 0
csc x x→
x sec x .
Therefore, lim x sec x does not exist.
x→ 12
x2 x 1
x2 x 1
3
x 1
x 1x2 x 1
1
lim f x lim
x→1
x→1 x 1
3
49. f x −4
5
−3
1
x
8
−3
3
−8
44
Chapter 1
51. f x Limits and Their Properties
0.3
1
x2 25
−8
lim f x 8
x→5
53. 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
lim f x x→c
−0.3
says how the limit fails to exist.
55. One answer is f x x3
x3
.
x 6x 2 x2 4x 12
200
ftsec
6
3
(b) r 50 sec2 200 ftsec
3
59. (a) r 50 sec2
y
57.
3
2
1
−2
(c)
x
−1
1
3
lim
→ 2
50 sec2 −1
−2
m0
61. m 1 v2c2
lim m lim
v→c
v→c
m0
1 v2c2
63. (a) Average speed Total distance
Total time
50 2d
dx dy
50 2xy
yx
(b)
x
30
40
50
60
y
150
6.667
50
42.857
(c) lim
x→25
25x
x 25 As x gets close to 25 mph, y becomes larger and larger.
50y 50x 2xy
50x 2xy 50y
50x 2yx 25
25x
y
x 25
Domain: x > 25
1
1
1
1
65. (a) A bh r 2 1010 tan 102
2
2
2
2
(b)
50 tan 50 0.3
0.6
0.9
1.2
1.5
f 0.47
4.21
18.0
68.6
630.1
Domain: 0,
2
(c)
100
0
(d)
1.5
0
lim A → 2
Review Exercises for Chapter 1
67. False; for instance, let
f x 69. True
x2 1
.
x1
The graph of f has a hole at 1, 2, not a
vertical asymptote.
71. Let f x 1
1
and gx 4, and c 0.
x2
x
73. Given lim f x , let g x 1. then lim
x →c
x →c
by Theorem 1.15.
1
1
2 and lim 4 , but
x→0 x
x→0 x
lim
x1 x1 lim x x 1 0.
2
lim
x→0
2
75. f x 4
4
x→0
1
is defined for all x > 3. Let M > 0 be given.
x3
We need > 0 such that f x Equivalently, x 3 <
So take 1
> M whenever 3 < x < 3 x3
1
whenever x 3 < , x > 3.
M
1
1
1
. Then for x > 3 and x 3 < ,
> M and hence f x > M.
M
x3 8
Review Exercises for Chapter 1
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, approximately 8.25.
4
2
x2
3. f x x
1
−3
x
0.1
0.01
0.001
0.001
0.01
0.1
f x
1.0526
1.0050
1.0005
0.9995
0.9950
0.9524
lim f x 1.0
x→0
5. hx x2 2x
x
(a) lim hx 2
x→0
(b) lim hx 3
x→1
7. 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 < .
3
−5
gx
0
fx
45
46
Chapter 1
Limits and Their Properties
9. 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 < .
11. lim t 2 4 2 6 2.45
t→4
15. lim
x 2
x4
x→4
lim
x→4
lim
x→4
17. lim
x→0
13. lim
1
t2
1
lim
t2 4 t→2 t 2
4
21. lim
1 cos x
x
lim
x→0 sin x
sin x
t→2
x 2
x 2x 2
1
x 2
1
4 2
1
4
1 x 1
1
1x 1
1
lim
lim
1
x→0
x→0 x 1
x
xx 1
x3 125
x 5x2 5x 25
lim
x→5 x 5
x→5
x5
19. lim
x→0
1 xcos x 10 0
lim x2 5x 25 75
x→5
23. lim
x→0
sin
6 x
12
sin
6 cos x cos
6 sin x 12
lim
x→0
x
x
lim
x→0
1
2
3 sin x
3
3
cos x 1
lim
x 0 2 1 2
x→0 2
x
25. lim f x gx
4 3 2
3
2
1
x→c
27. f x (a)
2x 1 3
x1
x
f x
lim
x→1
1.1
1.01
1.001
1.0001
0.5680
0.5764
0.5773
0.5773
2x 1 3
x1
—CONTINUED—
0.577
Actual limit is 33.
(b)
2
−1
2
0
Review Exercises for Chapter 1
27. —CONTINUED—
(c) lim
2x 1 3
x1
x→1
x→1
lim
x→1
t→a
x1
x→1
lim
29. lim
2x 1 3
lim
2x 1 3
2x 1 3
2x 1 3
x 1 2x 1 3 2
2x 1 3
2
1
3
23 3
3
sa st
4.942 200 4.9t2 200
lim
t→4
at
4t
lim
t→4
31. lim
x→3
x 3 x3
lim
x→3 x 3
x3
1
4.9t 4t 4
4t
lim 4.9t 4 39.2 msec
t→4
33. lim f x 0
35. lim ht does not exist because
x→2
t→1
lim ht 1 1 2 and
t→1
1
lim ht 21 1 1.
t→1
39. f x 37. f x x 3
lim x 3 k 3 where k is an integer.
x→k
lim f x lim 3x 2 5
lim x 3 k 2 where k is an integer.
x→1
x→k
lim
x→2
1
x 22
43. f x x→1
lim f x x→1
Nonremovable discontinuity at x 2
Continuous on , 2 2, x
2
Nonremovable discontinuity at x 1
Continuous on , 1 1, 47. f 2 5
Find c so that lim cx 6 5.
Nonremovable discontinuities at
each even integer.
Continuous on
x→2
c2 6 5
2c 1
2k, 2k 2
c
for all integers k.
51. f x (a) lim f x 4
x→2
1
2
x2 4
x2
x 2
x2
x2
3
x1
lim f x 1
x 22 45. f x csc
x→1
Removable discontinuity at x 1
Continuous on , 1 1, Nonremovable discontinuity at each integer k
Continuous on k, k 1 for all integers k
41. f x 3x2 x 2 3x 2x 1
x1
x1
(b) lim f x 4
x→2
(c) lim f x does not exist.
x→2
49. 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.
47
48
Chapter 1
53. gx 1 Limits and Their Properties
2
x
55. f x lim x→1
x1
1
1
lim
x3 1 x→1 x2 x 1 3
63. lim x x→0
67. lim
x→0
57.
2x2 x 1
x2
lim
x→2
Vertical asymptote at x 10
Vertical asymptote at x 0
59.
8
x 102
61. lim
x2 2x 1
x1
65. lim
sin 4x
4 sin 4x
lim
x→0
5x
5 4x
69. C 80,000p
, 0 ≤ p < 100
100 p
x→1
1
x3
x→0
csc 2x
1
lim
x →0 x sin 2x
x
54
(a) C15 $14,117.65
(b) C50 $80.000
(c) C90 $720,000
(d)
lim
p→100
80,000p
100 p Problem Solving for Chapter 1
1. (a) Perimeter PAO x2 y 12 x2 y2 1
x2 x2 12 x2 x4 1
(b) rx x 12 x4 x2 x4 1
x2 x2 12 x2 x4 1
x
Perimeter PBO x 12 y2 x2 y2 1
x 12 x4 x2 x4 1
(c) lim r x x→0
101 2
1
101 2
3. (a) There are 6 triangles, each with a central angle of
60 3. Hence,
12bh 6 121 sin 3 (c)
1
60°
33
0.5435
Error: 2
θ
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
r x
0.98
0.95
1
1.05
1.005
12bh n 121 sin n h = sin θ
1
1
2
An n
33
2.598.
2
h = sin 60°
2
(b) There are n triangles, each with central angle of
2n. Hence,
Area hexagon 6
4
n sin2n
.
2
n
6
12
24
48
96
An
2.598
3
3.106
3.133
3.139
(d) As n gets larger and larger, 2n approaches 0.
Letting x 2n,
An sin2n sin2n
sin x
2n
2n
x
which approaches 1 .
Problem Solving for Chapter 1
5. (a) Slope 12
5
(d) lim mx lim
x→5
(b) Slope of tangent line is
y 12 y
x→5
5
.
12
lim
x→5
5
x 5
12
lim
5
169
Tangent line
x
12
12
lim
12 169 x2
x5
x→5
144 169 x2
x 512 169 x2
5
x 5
10
12 169 x2 12 12 12
This is the same slope as part (b).
169 x2 12
mx x5
7. (a) 3 x13 ≥ 0
(b)
lim f x (c)
0.5
x→27
x13 ≥ 3
x ≥ 27
− 30
Domain: x ≥ 27, x 1
x→1
3 x13 2
x1
x→1
12 169 x2
12 169 x2
x2 25
x→5 x 5 12 169 x2 (c) Q x, y x, 169 x2 (d) lim f x lim
12
3 2713 2
27 1
1
2
0.0714
28 14
− 0.1
3 x13 2
3 x13 2
3 x13 4
x→1 x 1 3 x13 2 lim
lim
x→1
lim
x→1
x13
x13 1
1x23 x13 13 x13 2
1
x23 x13 13 x13 2
1
1
1 1 12 2 12
9. (a) lim f x 3: g1, g4
x→2
11.
(b) f continuous at 2: g1
(c) lim f x 3: g1, g3, g4
x→2
13. (a)
y
y
4
2
3
2
1
−4 −3 −2 −1
x
1
2
3
1
4
−2
−3
x
a
−4
(a)
f 1 1 1 1 1 0
f 0 0
f 12 0 1 1
f 2.7 3 2 1
(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, . . .
49
b
(b) (i) lim Pa, bx 1
x→a
(ii) lim Pa, bx 0
x→a
(iii) lim Pa, bx 0
x→b
(iv) lim Pa, bx 1
x→b
(c) Pa, b is continuous for all positive real numbers
except x a, b.
(d) The area under the graph of U, and above the
x-axis, is 1.