C H A P T E R 3
Limits and Their Properties
Section 3.1
A Preview of Calculus . . . . . . . . . . . . . . . . . . . 256
Section 3.2
Finding Limits Graphically and Numerically . . . . . . . 257
Section 3.3
Evaluating Limits Analytically . . . . . . . . . . . . . . . 265
Section 3.4
Continuity and One-Sided Limits
Section 3.5
Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . .280
Review Exercises
. . . . . . . . . . . . . 272
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291
C H A P T E R 3
Limits and Their Properties
Section 3.1
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. Precalculus: Area 2
2
2
8. f x x
7. f x 4x x2
(a)
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 0.5: m 3 0.5 2.5
x 3: m (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.
x 5: m x 2
x4
x 2
x 2x 2
1
x 2
, x4
1
1
3
1 2
1
3 2
1
5 2
(c) At P4, 2 the slope is
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. Area 5 256
Section 3.2
10. For the graph on the left,
Finding Limits Graphically and Numerically
For the graph on the right,
Area 12 1.75 3 3.75 4 3.75 3 1.75 0
Area 3 4 3 0 10.
12 21 10.5
Note:
The exact area is 10 23.
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 3.2
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.
3.
x2
0.3333
x2 x 2
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
x→2
lim
x2
0.25
x2 4
x
0.1
lim
0.01
x
0.001
0.2889
x 3 3
0.2887
0.2887
0.001
0.01
0.1
0.2887
0.2884
0.2863
Actual limit is 1 23.
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
x
lim
x→3
x3
0.0641
Actual limit is 14 .
0.25
2.9
f x
6.
Actual limit is 14 .
0.2911
x→0
5.
Actual limit is 13 .
x
f x
4.
Finding Limits Graphically and Numerically
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 .
x
3.9
3.99
3.999
4.001
4.01
4.1
f x
0.0408
0.0401
0.0400
0.0400
0.0399
0.0392
lim
x→4
xx 1 45
0.04
x4
Actual limit is 251 .
257
258
Chapter 3
Limits and Their Properties
7. lim 4 x 1
8. lim x2 2 3
x→3
x→1
9. lim f x lim 4 x 2
x→2
11. lim
x→5
10. lim f x lim x2 2 3
x→2
x→1
x 5 does not exist. For values of x to the left of 5,
1
does not exist since the function increases and
x3
decreases without bound as x approaches 3.
12. lim
x5
x→3
x 5 x 5 equals 1, whereas for values of x to the
right of 5, x 5 x 5 equals 1.
13. Ct 0.75 0.50
t 1
(a)
x→1
(b)
3
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
lim Ct 2.25
t→3.5
(c)
5
0
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
t→3
at t 3.
14. Ct 0.35 0.12
t 1
(a)
(b)
1
t
Ct
0
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
(c)
3
t→3.5
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
15. 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
<
11
10
>
10
11
10
>
9
x
10
10
1 > x 1 >
1
9
11
1
1
> x 1 > .
9
11
So take 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
Section 3.2
16. 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
Finding Limits Graphically and Numerically
17. lim 3x 2 8 L
x→2
3x 2 8
3x 6
3x 2
x→4
x
4
2 < 0.01
2
2
x
< 0.01
2
1
x 4 < 0.01
2
Hence, if 0 < x 4 < 0.02, you have
0 < x 4 < 0.02 1
x 4 < 0.01
2
2
4
x
< 0.01
2
x
2 < 0.01
2
f x L < 0.01.
3x 6 < 0.01
3x 2 8 < 0.01
f x L < 0.01.
3 x 2 < 0.01
f x 3 x2 4 < 0.2.
< 0.01
Using the first series of equivalent inequalities, you obtain
x
2
2
< 0.01
0.01
0.0033 3
0.01
Hence, if 0 < x 2 < , you have
3
3.8 2 < x 2 < 4.2 2.
< 0.01
0 < x2 <
4.2 2 < x 2 < 4.2 2
18. lim 4 259
19. lim x2 3 1 L
x→2
x2 3 1
x2 4
x 2x 2
x 2 x 2
< 0.01
< 0.01
< 0.01
< 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.
260
Chapter 3
Limits and Their Properties
20. lim x2 4 29
x→5
2
x
2
x 25
x 5x 5
21. lim x 3 5
x→2
4 29 < 0.01
Given > 0:
x 3 5
x 2
< 0.01
< 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 < 22. lim 2x 5 1
23. lim
x→4
x→3
12 x 1 12 4 1 3
Given > 0:
Given > 0:
2x 5 1 < 2x 6 < 2x 3 < x 3
<
2
Hence, let 2.
Hence, if 0 < x 3 < , you have
2
x 3 < 2
2x 6 < 2x 5 1 < f x L < .
x 1 3 < x 2 < 1
2
1
2
1
2
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 < .
Section 3.2
24. lim 23 x 9 23 1 9 29
3
25. lim 3 3
x→1
x→6
Given > 0:
Given > 0:
x 9 < x < 2
3
Finding Limits Graphically and Numerically
3 3
29
3
2
3
2
3
x 1
2
3
x 1
0 < < Hence, any > 0 will work.
3
2
<
Hence, for any > 0, you have
3 3 < f x L < .
Hence, let 32.
Hence, if 0 < x 1 < x 1 <
x x 9 2
3
2
3
< 3
2 ,
you have
3
2
2
3
< 29
3
< f x L < .
26. lim 1 1
x→2
3
x0
27. lim x→0
Given > 0: 1 1 < 3 x 0 < Given > 0: 0 < 3 x < x
Hence, any > 0 will work.
< 3 Hence, let 3.
Hence, for any > 0, you have
1 1 < f x L < .
Hence for 0 < x 0 < 3, you have
3
x 0 < f x L < .
x < 3
3 x < 28. lim x 4 2
x→4
Given > 0:
x
2
x 2 < x 4 < x→2
x 2 < Given > 0:
x 2
x 2 4
x 2 4
< x 2 x 2 x 2
< x 2 < 0
0 < x 4 < 3 ⇒ x 4 < x 2
x 2
Assuming 1 < x < 9, you can choose 3. Then,
29. lim x 2 2 2 4
⇒ x 2 < .
< Hence, .
Hence for 0 < x 2 < , you have
x 2 < x 2 < x 2 4 < x 2 4 < f x L < .
(because x 2 < 0)
261
262
Chapter 3
Limits and Their Properties
30. lim x 3 0
31. lim x2 1 2
Given > 0:
Given > 0:
x→3
x 3 0
x 3
x→1
< x2 1 2 < x2 1 < x 1x 1 < x 1 < x 1
< Hence, let .
Hence for 0 < x 3 < , you have
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 < .
33. f x 32. lim x2 3x 0
x→3
Given > 0:
x2 3x 0
xx 3
x 5 3
x4
lim f x < x→4
0.5
1
6
−6
6
−0.1667
< The domain is 5, 4 4, .
The graphing utility does not show the hole at 4, 16 .
x 3 < x
If we assume 4 < x < 2, then 4.
Hence for 0 < x 3 < , you have
4
1
1
x 3 < 4 < x
xx 3 < x2 3x 0 < f x L < .
34. f x x3
x 4x 3
lim f x x→3
4
35. f x 2
1
2
−3
5
x9
8
x 3
lim f x 6
x→9
−4
0
12
0
The domain is all x 1, 3. The graphing utility does not
1
show the hole at 3, 2 .
The domain is all x ≥ 0 except x 9. The graphing
utility does not show the hole at 9, 6.
Section 3.2
36. f x x3
x2 9
lim f x x→3
−9
263
37. lim f x 25 means that the values of f approach 25 as x
x→8
gets closer and closer to 8.
3
1
6
Finding Limits Graphically and Numerically
3
−3
The domain is all x ± 3. The graphing utility does not
1
show the hole at 3, 6 .
38. (a) No. The fact that f 2 4 has no bearing on the existence of the limit of f x as x approaches 2.
39. (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:
y
(iii) The values of f oscillate
between two fixed numbers as
x approaches c:
y
y
6
4
5
3
4
2
3
4
3
2
1
−4 −3 −2 −1
−1
(b) No. The fact that lim f x 4 has no bearing on the
x→2
value of f at 2.
1
x
1
2
3
4
−3 −2 −1
−1
3
4
2
−4
40. Let px be the atmospheric pressure in a plane at
altitude x (in feet).
lim px 14.7 lbin
41. False. For example, let f x exist, but lim f x x→2
2
x2
. Then f 2 does not
x2 4
1
(See Exercise 2, Section 3.2.)
4
x→0 43. False. Let f x 42. True
4 0,x,
x2
.
x2
lim f x 2, but f 2 0 (See Exercise 9.)
x→2
45. (a) C 2 r
44. False. Let
x10, 4x,
2
x4
x4
.
lim f x lim x 4x 0 and f 4 10 0
2
x→4
x→4
4
−3
−4
f x 3
5
−2
−3
x
−4 −3 −2
x
2
r
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
Thus, 0.87535 < r < 1.03451.
(c) lim 2 r 6; 0.5; 0.0796
r →3
264
Chapter 3
Limits and Their Properties
4
46. V r 3, V 2.48
3
4
(a) 2.48 r 3
3
(b)
r 0.8397 in.
47. f x 1 x1x
lim 1 x
1x
x→0
e 2.71828
y
7
3
(0, 2.7183)
2
1
−3 −2 −1
−1
48. Let f x ≤ 2.51
V
2.45 ≤
1.86
r3 2.45 ≤
(c) For 2.51 2.48 0.03,
0.003.
4 3
r ≤ 2.51
3
0.5849 ≤
r 3 ≤ 0.5992
0.8363 ≤
r
≤ 0.8431
x
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
2.716924
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
1
2
3
4
5
1
1
and gx .
x
x
49. f x x
(a) lim x 0.5 is true.
Then lim f x and lim gx do not exist.
x→0
x→0.25
x→0
But, lim f x gx lim
x→0
x→0
1
As x approaches 0.25 4, f x x approaches
1x 1x 0 does exist.
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.
50. The value of f at c has no bearing on the limit as x approaches c.
51. 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.
52. 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
53. lim f x L 0 means that for every > 0 there
x→c
exists > 0 such that if
f x L 0
This means the same as f x L < when
0 < x c < .
0 < x c < , then
Thus, lim f x L.
x→c
< .
Section 3.3
54. (a) 3x 13x 1x2 0.01 9x2 1x2 9x4 x2 1
100
1
,
1
90 90
x→c
1.
1
100
1
10x2 190x2 1
100
L
L
< gx L <
2
2
L
< gx
2
.
(a) lim hx 0
x→5
<
3L
.
2
For x in the interval c , c , x c, we have
L
gx > > 0, as desired.
2
2.
(a) lim gx 2.4
10
x→4
(b) lim gx 4
(b) lim hx 6
x→0
x→1
13
0
−7
10
−5
hx x 5x
2
3.
Evaluating Limits Analytically
7
−8
exists > 0 such that 0 < x c < implies that
L
gx L < . That is,
2
For all x 0 in a, b, the graph is positive. You can
verify this with a graphing utility.
Section 3.3
gx (a) lim f x 0
6
x→0
−8
10
12 x 3
x9
4.
(b) lim f x 4
(a) lim f t 0
10
t→4
(b) lim f t 5
−5
x→2
t→1
10
−6
− 10
f t t t 4
5. lim x4 24 16
6. lim x3 23 8
7. lim 2x 1 20 1 1
8. lim 3x 2 33 2 7
x→2
x→0
9. lim x2 3x 32 33 9 9 0
x→3
x→2
x→3
10. lim x2 1 12 1 0
x→1
11. lim 2x2 4x 1 232 43 1 18 12 1 7
x→3
12. lim 3x3 2x2 4 313 212 4 5
x→1
14. lim
x→3
16. lim
x→3
2
2
2
x 2 3 2
2x 3 23 3 3
x5
35
8
265
1
(b) We are given lim gx L > 0. Let 2L. There
Thus, 3x 13x 1x2 0.01 > 0 if
10x2 1 < 0 and 90x2 1 < 0.
Let a, b Evaluating Limits Analytically
13. lim
1 1
x
2
15. lim
x3
13
2
2
x2 4 12 4
5
5
x→2
x→1
17. lim
x→7
5x
x 2
57
7 2
35
9
35
3
266
Chapter 3
18. lim
x→3
x 1
x4
Limits and Their Properties
3 1
34
19. lim x 1 3 1 2
2
x→3
3
3
20. lim x 4 442
21. lim x 32 4 32 1
22. lim 2x 13 20 13 1
23. (a) lim f x 5 1 4
x→4
x→4
x→1
x→0
(b) lim gx 43 64
x→4
(c) lim g f x g f 1 g4 64
x→1
24. (a) lim f x 3 7 4
25. (a) lim f x 4 1 3
x→3
x→1
(b) lim gx 42 16
(b) lim gx 3 1 2
(c) lim g f x g4 16
(c) lim g f x g3 2
x→4
x→3
x→3
x→1
27. (a) lim 5gx 5 lim gx 53 15
26. (a) lim f x 242 34 1 21
x→c
x→4
3 21 6 3
(b) lim gx x→c
(b) lim f x gx lim f x lim gx 2 3 5
x→21
x→c
(c) lim g f x g21 3
x→4
x→c
x→c
(c) lim f xgx lim f x lim gx 23 6
x→c
(d) lim
x→c
x→c
x→c
lim f x
f x
2
x→c
gx
lim gx 3
x→c
28. (a) lim 4f x 4 lim f x 4
x→c
x→c
29. (a) lim f x3 lim f x3 43 64
3
6
2
x→c
(b) lim f x gx lim f x lim gx x→c
x→c
x→c
(c) lim f xgx lim f x lim gx x→c
x→c
x→c
3 1
2
2 2
3
2
1
3
2
4
lim f x
f x
32
x→c
(d) lim
3
x→c gx
lim gx 12
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
x→c
3
3
3 lim f x 27 3
30. (a) lim f x x→c
x→c
lim f x 27 3
f x x→c
(b) lim
x→c 18
lim 18
18 2
x→c
(c) lim f x2 lim f x2 272 729
x→c
x→c
31. f x 2x 1 and gx x 0.
2x2 x
agree except at
x
(a) lim gx lim f x 1
x→0
x→0
(b) lim gx lim f x 3
x→1
x→1
(d) lim f x23 lim f x23 2723 9
x→c
x→c
32. f x x 3 and hx x2 3x
agree except at x 0.
x
33. f x xx 1 and gx x3 x
agree except at x 1.
x1
(a) lim hx lim f x 5
(a) lim gx lim f x 2
(b) lim hx lim f x 3
(b) lim gx lim f x 0
x→2
x→0
x→2
x→0
x→1
x→1
x→1
x→1
Section 3.3
34. gx 1
x
and f x 2
agree except at x 0.
x1
x x
35. f x x→1
x→1
(b) lim f x lim gx 1
x→1
3
x→0
−4
5
−3
2x2 x 3
and gx 2x 3 agree except at
x1
x 1.
36. f x x3 8
and gx x2 2x 4 agree except
x2
at x 2.
37. f x lim f x lim gx 12
lim f x lim gx 5
x→1
x→2
x→1
x→2
12
4
−8
4
−9
x3 1
and gx x2 x 1 agree except at
x1
x 1.
38. f x 39. lim
x→5
x5
x5
lim
x2 25 x→5 x 5x 5
lim
lim f x lim gx 3
x→1
9
0
−8
x→5
x→1
1
1
x 5 10
7
−4
4
−1
x2 x 6
x 3x 2
lim
x→3
x→3 x 3x 3
x2 9
2x
x 2
lim
x→2 x2 4
x→2 x 2x 2
41. lim
40. lim
lim
x→2
1
1
x2
4
lim
x→3
x 4x 1
x2 5x 4
lim
x→4 x2 2x 8
x→4 x 4x 2
42. lim
lim
x→4
43. lim
x→0
x 5 5
x
x 1 3 1
x 2 6 2
lim
x→0
lim
x→0
x 5 5
x
267
x2 1
and gx x 1 agree except at x 1.
x1
lim f x lim gx 2
(a) lim f x does not exist.
x→0
Evaluating Limits Analytically
x 5 5
x 5 5
5
x 5 5
1
1
lim
xx 5 5 x→0 x 5 5 25
10
x 2 5 5
x 3 6 6
268
Chapter 3
44. lim
Limits and Their Properties
2 x 2
x
x→0
lim
lim
x→0
45. lim
x 5 3
x4
x→4
lim
x→4
x 1 2
x3
x→3
x
lim
2x2
2 x 2
2 x 2
2 x 2 x
x 5 3
x4
x→4
lim
46. lim
2 x 2
x→0
lim
x→0
1
2 x 2
2
1
4
22
x 5 3
x 5 3
1
x 5 9
1
1
lim
x 4x 5 3 x→4 x 5 3 9 3 6
x 1 2
x3
x→3
x 1 2
x 1 2
lim
x→3
x3
1
1
lim
x 3x 1 2 x→3 x 1 2 4
1
1
3x 3
3 3 x
1
1
47. lim
lim
lim
x→0
x→0 33 xx
x→0 33 x
x
9
1
4 x 4
1
1
1
x4 4
4x 4
48. lim
lim
lim
x→0
x→0
x→0 4x 4
x
x
16
49. lim
x→0
2x x 2x
2x 2x 2x
lim
lim 2 2
x→0
x→0
x
x
x x2 x2
x2 2x x x2 x2
x2x x
lim
lim
lim 2x x 2x
x→0
x→0
x→0
x→0
x
x
x
50. lim
51. lim
x→0
x x2 2x x 1 x2 2x 1
x2 2xx x2 2x 2x 1 x2 2x 1
lim
x→0
x
x
lim 2x x 2 2x 2
x→0
x x3 x3
x3 3x2 x 3xx2 x3 x3
lim
x→0
x→0
x
x
52. lim
lim
x→0
53. lim
x 2 2
x→0
x
f x
x
0.1
0.358
x3x2 3x x x2
lim 3x2 3x x x2 3x2
x→0
x
0.354
0.01
2
0.001
0
0.001
0.01
0.1
0.354
?
0.354
0.353
0.349
0.354
−3
3
−2
Analytically, lim
x→0
x 2 2
x
lim
x→0
lim
x→0
x 2 2
x
x 2 2
x 2 2
2
x22
1
1
0.354.
lim
4
xx 2 2 x→0 x 2 2 22
Section 3.3
54. f x Evaluating Limits Analytically
4 x
x 16
269
1
0
x
15.9
15.99
15.999
16
16.001
16.01
16.1
f x
.1252
.125
.125
?
.125
.125
.1248
Analytically, lim
x→16
−1
4 x
4 x lim
x→16 x 4 x 4 x 16
lim
x→16
1
x 4
20
It appears that the limit is
0.125.
1
.
8
1
1
1
2x 2
55. lim
x→0
x
4
2
−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
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
x→2 x 2
100
56. lim
x
f x
1.9
1.99
72.39
79.20
1.999
79.92
1.9999
79.99
2.0
?
2.0001
80.01
2.001
80.08
2.01
2.1
80.80
88.41
−4
x5 32
x 2x4 2x3 4x2 8x 16
lim
x→2 x 2
x→2
x2
Analytically, lim
lim x4 2x3 4x2 8x 16 80.
x→2
(Hint: Use long division to factor x5 32.)
57. f x 2x 3
lim
x→0
f x x f x
2x x 3 2x 3
2x 2 x 2x
2 x
lim
lim
lim
lim 2 2
x→0
x→0
x→0 x
x→0
x
x
x
x x x
x x x x x x
f x x f x
lim
lim
x→0
x→0
x→0
x
x
x
x x x
1
1
x x x
lim
lim x x x 2x
x→0 x x x x x→0
58. lim
59. f x 4
x
4
4
f x x f x
x x
x
4x 4x x
lim
lim
lim
x→0
x→0
x→0 x xx x
x
x
lim
x→0
4 x
4
4
2
lim
x→0 x xx
x xx x
x
3
−25
270
Chapter 3
60. lim
x→0
Limits and Their Properties
f x x f x
x2 2x x x2 4x 4 x x2 4x
x x2 4x x x2 4x
lim
lim
x→0
x→0
x
x
x
x2x x 4
lim 2x x 4 2x 4
lim
x→0
x→0
x
61. lim 4 x2 4
lim 4 x→0
62. lim b x a ≤ lim f x ≤ lim b x a x→0
x→a
4
x2
x→a
x→a
b ≤ lim f x ≤ b
x→a
Since 4 x2 ≤ f x ≤ 4 x2 the Squeeze Theorem says
lim f x 4.
Therefore, lim f x b.
x→a
x→0
x2 1
and gx x 1 agree at all points
x1
except x 1.
63. 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.
64. f x 65. An indeterminant form is obtained when evaluating a limit
using direct substitution produces a meaningless fractional
expression such as 00. That is,
66. 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
f x
lim
x→c gx
for which lim f x lim gx 0.
x→c
x→c
67. st 16t2 1000
s5 st
600 16t2 1000
16t 5t 5
lim 16t 5 160 ftsec.
lim
lim
t→5
t→5
t→5
5t
5t
t 5
Speed 160 ftsec
lim
t→5
68. st 16t2 1000 0 when t 5 210 st s
lim
t→5102
5 10
seconds
1000
16
2
510 t
2
0 16t2 1000
t→5102
510
t
2
lim
16 t2 lim
t→5102
lim
t→5102
125
2
510 t
2
16 t lim
t→5102
510
2
t 5 210
t 5 10
2
510
8010 ftsec 253 ftsec
2
69. st 4.9t2 150
s3 st
4.932 150 4.9t2 150
lim
lim
t→3
t→3
3t
3t
4.99 t2
lim
t→3
3t
4.93 t3 t
lim
t→3
3t
lim 4.93 t 29.4 msec
t→3
16 t Section 3.3
70. 4.9t2 150 0 when t Evaluating Limits Analytically
271
1500
5.53 seconds.
150
4.9
49
The velocity at time t a is
sa st
4.9a2 150 4.9t2 150
4.9a ta t
lim
lim
t→a
t→a
t→a
at
at
at
lim
lim 4.9a t 2a4.9 9.8a msec.
t→a
Hence, if a 150049, the velocity is 9.8150049 54.2 msec.
71. False. As x approaches 0 from the left,
x 1.
72. True. lim x3 03 0
x
x→0
2
−3
3
−2
74. False. Let f x 73. True
3x xx 11
, c 1.
Then lim f x 1 but f 1 1.
x→1
75. False. The limit does not exist.
76. False. Let f x 12 x2 and gx x2. Then f x < gx for
all x 0. But lim f x lim gx 0.
4
x→0
−3
x→0
6
−2
77. Let f x 1x and gx 1x. lim f x and lim gx
x→0
x→0
78. Suppose, on the contrary, that lim gx exists. Then,
x→c
since lim f x exists, so would lim f x gx, which
do not exist.
x→c
lim f x gx lim
x→0
x→0
x→c
is a contradiction. Hence, lim gx does not exist.
1x 1x x→c
lim 1 1
x→0
79. 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.
80. lim x n lim x x
x→c
x→c
.
. . x
lim x lim x . . . lim by Property 3, Theorem 3.2
cc. . . c
by Property 2, Theorem 3.1
x→c
cn
x→c
x→c
272
Chapter 3
Limits and Their Properties
81. 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,
x→c
there exists > 0 such that 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
82. f x 0,1,
if x is rational
if x is irrational
g x 0,x,
if x is rational
if x is irrational
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.
Section 3.4
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)
(b)
lim f x 2
5. (a)
lim f x 2
x
does not exist because
without bound as x → 3 .
6. (a)
x→4 (b)
8. lim
x→2
x
x2 9
grows
10. lim
x→4
lim f x 0
x→1
lim f x 2
x→1
(c) lim f x does not exist.
x→4
x5
1
1
lim
x2 25 x→5 x 5 10
x2 9
x→3
x→1
The function is NOT continuous
at x 4.
The function is NOT continuous at
x 2.
x→3 x→3
x→4
x→2
lim
(b) lim f x 0
x→2
(c) lim f x does not exist
(c) lim f x 2
9.
lim f x 2
x→3
(b) lim f x 2
lim f x 2
x→2
x→5
3. (a) lim f x 0
x→2
x→2
7. lim
lim f x 2
x→2
The function is NOT continuous at
x 1.
1
2x
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 4 x 2
1
x 2
1
4
Section 3.4
11. lim
x→0
x x
lim
x→0 x
1
x
12. lim
x→2
1
1
x x
x
x x x
13. lim lim x→0
x→0
x
xx x
x
1
x2
lim
x→2
x2
1
x2
1
x→0
x→0
x 2 273
lim
x x→0
xx x x
lim 14. lim Continuity and One-Sided Limits
1
xx x
1
1
2
xx 0
x
x x2 x x x2 x
x2 2xx 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
19. lim 3x 5 33 5 4
18. lim f x lim 1 x 0
x→1
x→4
x→1
x 3 for 3 < x < 4
lim f x lim x 1 2
3
x→1
x→1
lim f x 2
x→1
20. lim 2x x 22 2 2
x→2
21. lim 2 x does not exist
x→3
because
lim 2 x 2 3 5
x→3
and
lim 2 x 2 4 6.
x→3
2x 1 1 2
22. lim 1 x→1
23. f x 1
x2 4
has discontinuities at x 2 and x 2 since f 2 and
f 2 are not defined.
24. f x x2 1
has a discontinuity at x 1 since f 1
x1
is not defined.
25. f x x
x
2
has discontinuities at each integer k since
lim f x lim f x.
x→k
x→k
274
Chapter 3
Limits and Their Properties
x,
26. 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
27. gx 25 x2 is continuous
on 5, 5.
28. f t 3 9 t2
is continuous on 3, 3.
29. lim f x 3 lim f x.
30. g2 is not defined.
g is continuous on 1, 2.
31. f x x2 2x 1 is continuous
for all real x.
32. f x x
is not continuous at x 0, 1. Since
x2 x
33. f x x
1
for x 0, x 0 is a removable
x2 x x 1
discontinuity, whereas x 1 is a nonremovable
discontinuity.
35. f x x→0
x→0
f is continuous on 1, 4.
1
x2 1
is continuous for all real x.
x
has nonremovable discontinuities at x 1
x2 1
and x 1 since lim f x and lim f x do not exist.
34. f x x→1
x→1
x
is continuous for all real x.
x2 1
x3
has a nonremovable discontinuity at x 3 since lim f x does not exist,
x→3
x2 9
and has a removable discontinuity at x 3 since
36. f x lim f x lim
x→3
37. f x x→3
1
1
.
x3 6
x2
x 2x 5
38. 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
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
1
1
.
x→2 x 5
7
lim f x lim
x→2
39. f x x 2 has a nonremovable discontinuity at
x2
x 2 since lim f x does not exist.
x→2
41. f x x,x ,
2
x ≤ 1
x > 1
x1
x 2x 1
x→1
40. f x x→1
1
1
.
x2 3
x 3
x3
has a nonremovable discontinuity at x 3 since lim f x
x→3
does not exist.
42. f x 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
x→1
x→1
lim f x lim
x→1
x→1
x2
lim f x 1
1
x→1
3. f 1 lim f x
x→1
f is continuous at x 1, therefore, f is continuous for
all real x.
2.
lim f x lim 2x 3 1
x→1
lim f x 1
x→1
lim f x lim x2 1
x→1
x→1
x→1
3. f 1 lim f x
x→1
f is continuous at x 1, therefore, f is continuous
for all real x.
Section 3.4
x
1,
43. f x 2
3 x,
x ≤ 2
has a possible discontinuity
275
x ≤ 2
x > 2
2
has a possible discontinuity at x 2.
1. f 2 1. f 2 22 4
2
12
2
lim f x lim
x→2
2.
2x,
x 4x 1,
44. f x x > 2
at x 2.
Continuity and One-Sided Limits
x→2
2x 1 2
lim f x lim 3 x 1
x→2
x→2
lim f x does not exist.
2.
lim f x lim 2x 4
x→2
x→2
x→2
x→2
x→2
lim f x
lim f x lim x2 4x 1 3
x→2
does not exist.
Therefore, f has a nonremovable discontinuity at x 2.
Therefore, f has a nonremovable discontinuity at x 2.
45. f x x 1 has nonremovable discontinuities at each
integer k.
46. f x 3 x has nonremovable discontinuities at each
integer k.
47. lim f x 0
48. f x x→0
lim f x 0
x2 4xx 2
x4
20
x→0
50
−6
−8
−10
8
−10
4
lim f x 0 and lim f x 0
x→0 f is not continuous at x 2.
x→a
lim ax 2 4a
lim x a 2a
x→2
x→2
x2 a2
x→a x a
50. lim gx lim
49. Find a such that
lim
x→0
The function is not continuous at x 4.
x3
8.
4a 8
x→a
Find a such that 2a 8 ⇒ a 4.
a2
52. hx f gx f x2 x2
51. f gx x 12
Continuous for all real x.
53. hx f g x f
1x 11
x x, x 0
Continuous on 0, .
55. f gx 1
1
x2 5 6 x2 1
Nonremovable discontinuities at x ± 1.
Continuous for all real x.
54. hx f gx f x 1 1
x 1
.
Continuous on 1, .
56. hx f gx f
x 1 1 1
x 1, x 1
1
x1
Continuous for all x 1. Removable discontinuity
at x 1.
276
Chapter 3
Limits and Their Properties
58. hx 57. f x x x
Nonremovable discontinuity at each integer.
0.5
1
x 1x 2
Nonremovable discontinuity at x 1 and x 2.
2
−3
3
−3
4
−1.5
−2
59. gx 2xx 2x,4,
2
x ≤ 3
x > 3
60. f x Nonremovable discontinuity at x 3.
xx2x6x2, 6,
2
2
x < 2
x ≥ 2
10
5
−8
−5
12
10
−10
−5
61. f x Continuous for all real x.
x
x2 1
62. f x xx 3
Continuous on , .
63. f x x2
x2
36
64. f x Continuous on: , 6, 6, 6, 6, .
65. f x Continuous on 3, .
x2 x 2
x1
Continuous on 0, .
66. f x 2
−4
x1
x
x3 8
x2
14
4
−6
The graph appears to be continuous on the interval
4, 4. Since f 1 is not defined, we know that f has a
discontinuity at x 1. This discontinuity is removable
so it does not show up on the graph.
1 4
33
67. f x 16
x 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.
−4
4
0
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.
68. f x x3 3x 2 is continuous on 0, 1.
f 0 2 and f 1 2.
By the Intermediate Value Theorem, f x 0 for at least
one value of c between 0 and 1.
Section 3.4
69. f x x3 x 1
Continuity and One-Sided Limits
277
70. f x x3 3x 2
f x is continuous on 0, 1.
f x is continuous on 0, 1.
f 0 1 and f 1 1.
f 0 2 and f 1 2.
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 0.5961.
71. gt 3t 2 1 4. g is continuous on 0, 1.
g0 3 4 1 and g1 32 4 0.2426.
By the Intermediate Value Theorem, gt 0 for at least
one value of c between 0 and 1. Using the zoom feature,
we find that c 0.88. Using a graphing utility, we find
that c 0.8819.
2
s3
h is continuous on 12, 1.
72. hs 5 h
12 5 2
11 and h1 5 2 3.
1
8
By the Intermediate Valve Theorem, hs 0 for at
least one value of c between 12 and 1 and hence in 0, 1.
Using the zoom feature, we find that c 0.74.
Using the graphing utility, we find that c 0.7368.
74. f x x2 6x 8
73. f x x2 x 1
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 < 0 < 8
1 < 11 < 29
The Intermediate Value Theorem applies.
The Intermediate Value Theorem applies.
x2 6x 8 0
x2 x 1 11
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.)
c 2 (x 4 is not in the interval.)
Thus, f 2 0.
Thus, f 3 11.
75. f x x3 x2 x 2
f is continuous on 0, 3.
f 0 2 and f 3 19.
2 < 4 < 19
x3 x2 x 2 4
x3 x2 x 6 0
x 2x2 x 3 0
x2
(x2 x 3 0 has no real solution.)
c2
x2 x
x1
f is continuous on 2 , 4. The nonremovable discontinuity,
x 1, lies outside the interval.
5
f
The Intermediate Value Theorem applies.
Thus, f 2 4.
76. f x 52 356 and f 4 203.
20
35
< 6 <
6
3
The Intermediate Value Theorem applies.
x2 x
6
x1
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.
278
Chapter 3
Limits and Their Properties
(c) The limit exists at x c, but it is not equal to the
value of the function at x c.
77. (a) The limit does not exist at x c.
(b) The function is not defined at x c.
(d) The limit does not exist at x c.
78. 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.
79.
y
5
4
3
2
1
x 2
(a) f x x2
x2 2x
(b) f x x2
1
3 4 5 6 7
−2
−3
if x ≥ 2
if 2 < x < 2
if x 2
if x < 2
1,
0,
(c) f x 1,
0,
x
−2 −1
The function is not continuous at x 3 because
lim f x 1 0 lim f x.
x→3
x→3
y
3
2
1
−3
−2
−1
x
1
−1
2
3
−2
−3
80. If f and g are continuous for all real x, then so is f g (Theorem 3.9, 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.
81. The functions agree for integer values of x:
gx 3 x 3 x 3 x
f x 3 x 3 x
for x an integer
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,
82. 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
t 2 2 t
83. Nt 25 2
Nonremovable discontinuity at each integer greater than 2.
t
0
1
1.8
2
3
3.8
You can also write C as
Nt
50
25
5
50
25
5
C
1.04,
1.04 0.362 t,
0 < t ≤ 2
.
t > 2
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
4
6
8
10 12
Time (in months)
Section 3.4
Continuity and One-Sided Limits
279
84. 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.
f 0 s0 r0 0 k < 0.
When t 0 (8:00 A.M.),
f 10 s10 r10 > 0.
When t 10 (8:10 A.M.),
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.
85. 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.)
87. 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.
86. 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.
88. True
1. f c L is defined.
2. lim f x L exists.
x→c
3. f c lim f x
x→c
All of the conditions for continuity are met.
89. True; if f x gx, x c, then lim f x lim gx (if
x→c
x→c
they exist) and at least one of these limits then does not
equal the corresponding function value at x c.
90. 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.
91. False; f 1 is not defined and lim f x does not exist.
x→1
92. (a)
(b) There appears to be a limiting speed and a possible
cause is air resistance.
S
60
50
40
30
20
10
t
5
10
15 20 25 30
0 ≤ x < b
b < x ≤ 2b
0
93. (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
2b
NOT continuous at x b.
Continuous on 0, 2b.
280
Chapter 3
Limits and Their Properties
94. hx xx
15
h has nonremovable discontinuities at
x ± 1, ± 2, ± 3, . . . .
−3
3
−3
95. 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
1
x c2 c2
1
lim
xx c2 c x→0 x c2 c 2c
Define f 0 12c to make f continuous at x 0.
96. 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
Section 3.5
1.
Infinite Limits
lim 2
x
x2 4
lim 2
x
x2 4
x→2
x→2
2.
1
x2 1
lim
x→2 x 2
lim
x→2
3. 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
4. f x x
f x
x
x2 9
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
1.077
5.082
50.08
500.1
499.9
49.92
4.915
0.9091
lim f x x→3
lim f x x→3
Section 3.5
5. f x x2
x2 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
Infinite Limits
lim f x x→3
lim f x x→3
6. f x x2
3.5
3.1
3.01
3.001
2.999
2.99
2.9
2.5
13.19
48.84
453.76
4503.8
4496.3
446.26
41.34
5.68
x
f x
x3
x3
9 x 3x 3
lim f x x→3
lim f x x→3
7. lim
x→2
lim
x→2
x2 2
x 2x 1 x2 2
x 2x 1
8. f x x3
x3
x 1 x 1x 1
2
Vertical asymptotes: x 1, 1
Therefore, x 2 is a vertical asymptote.
lim x2 2
x 2x 1 lim
x2 2
x 2x 1
x→1
x→1
Therefore, x 1 is a vertical asymptote.
9. lim
x→0
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. f x x2
x2
x2
x 6 x 3x 2
Vertical asymptotes: x 3, 2
12. lim
x→0
2x
2x
lim
x21 x x→0 x21 x
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.
281
282
13.
Chapter 3
lim
x→2
Limits and Their Properties
x2
x2
and lim 2
.
x→2
x 4
x 4
2
Therefore, x 2 is a vertical asymptote.
lim
x→2
x2
14. f x 4x
x2 4
x2 4 0. No vertical asymptotes.
x2
x2
and lim 2
.
x→2 x 4
4
Therefore, x 2 is a vertical asymptote.
t1
t2 1
15. gt 16. hs Vertical asymptotes: s 5, 5
t 2 1 0. No vertical asymptotes.
17. lim 1 t→0
4
4
lim 1 2
t→0
t2
t
2s 3
2s 3
s2 25 s 5s 5
Therefore, t 0 is a vertical asymptote.
18. gx 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.
19.
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
Therefore, x 1 is a vertical asymptote.
20. f x 4x2 x 6
4x 3x 2
4
, x 3, 2
xx3 2x2 9x 18 xx 2x2 9 xx 3
Vertical asymptotes at x 0 and x 3. The graph has holes at x 3 and x 2.
21. g x x3 1 x 1x2 x 1
x1
x1
has no vertical asymptote since
lim f x lim x2 x 1 3.
x→1
x→1
22. hx x3
has no vertical asymptote since
lim hx lim
x→2
23. f x x 5x 3
x3
,x5
x 5x2 1 x2 1
No vertical asymptotes. The graph has a hole at x 5.
x2 4
x 2x 2
2x2 x 2 x 2x2 1
24. ht x→2
x2
4
.
x2 1
5
tt 2
t
,t2
t 2t 2t 2 4 t 2t 2 4
Vertical asymptote at t 2. The graph has a hole at
t 2.
Section 3.5
25. lim
x→1
x2 1
lim x 1 2
x→1
x1
26. lim
x→1
283
x2 6x 7
lim x 7 8
x→1
x1
10
2
−3
−10
3
10
−10
−12
Removable discontinuity at x 1.
Removable discontinuity at x 1.
27.
Infinite Limits
lim x2 1
x1
lim
x2 1
x1
x→1
x→1
28.
lim
x1
x1
lim
x1
x1 x→1 x→1
Vertical asymptote at x 1.
4
−6
6
Vertical asymptote at
x 1.
−4
10
−10
10
−10
29. lim
x3
x2
32. lim
x2
x→2
x→4
34.
lim
x→3
x2 1
x2
9
x→3
1
x
x→3
1
x2 x
x
lim 2
x→1 x 1x 1
x→1 x 1
2
35. lim
2
x→4
x→4
x→4
44. lim
gf xx lim x
2
x→4
1
x
39. lim f x lim
x→4
x2 5x
x 42
5xx 42 0
x2
x 3x 3 x2 2x 3
x1 4
lim x→3 x 2
x2 x 6
5
lim x→0
42. lim f xgx lim
x→4
31. lim
37. lim 1 40. lim gx lim x2 5x 16 20 4
x→4
2x
1x
33.
6x2 x 1
3x 1 5
lim
4x2 4x 3 x→12 2x 3 8
38. lim x2 x→0
x→1
x2
1
16 2
x→12
36. lim
30. lim
x→4
1
x 42 41. lim f x gx lim
x→4
43. lim
x→4
x→4
f x
x 1 4
2
gx lim x 4 x
1
x→4
2
2
x2 5x 5x
284
Chapter 3
45. f x Limits and Their Properties
x2 x 1
x3 1
lim f x lim
x→1
x→1
46. f x 1
x1 x3 1
x2 x 1
1
x2 25
lim f x lim f x lim x 1 0
x→1
x→5
x→1
0.3
4
3
−4
−8
−8
5
8
8
−0.3
−4
−3
48. f x 47. f x 6x
50. The line x a is a vertical
asymptote of the graph of f if f x
tends to or as x tends to a
from the left or right.
49. 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
x 3
6
−1
11
lim f x x→c
says how the limit fails to exist.
−2
lim f x x→3 51. One answer is f x x3
x3
.
x 6x 2 x2 4x 12
52. No. For example, f x 1
has no
x2 1
vertical asymptote.
53.
k
V
54. P y
3
lim
2
V→0
k
V 1
−2
x
−1
1
3
In this case we know that k > 0.
−1
−2
55. C 80,000p
, 0 ≤ p < 100
100 p
56. C 528x
, 0 ≤ x < 100
100 x
(a) C 15 $14,118
(a) C25 $176 million
(b) C 50 $80,000
(b) C50 $528 million
(c) C 90 $720,000
(c) C75 $1584 million
(d)
lim C p→100 (d)
lim
x→100
528
Thus, it is not possible.
100 x
The cost increases without bound.
57. m m0
1 v2c2
lim m lim
v→c
v→c
m0
1 v2c2
58. (a) r (b) r 27
625 49
7
ftsec
12
215
3
ftsec
2
625 225
(c) lim
x→25
2x
625 x2
Section 3.5
59. (a) Average speed Total distance
Total time
50 2d
dx dy
50 2xy
yx
60. (a)
30 30
1
x
y
30y 30x xy
30x xy 30y x 30y
y
50x 2xy 50y
50x 2yx 25
30x
together with
x 30
y 65 to obtain x ≥ 55.7 mph.
(c) lim y x→30
Domain: x > 25
x
30
40
50
60
y
150
66.667
50
42.857
x→25
30x
x 30
(b) If y ≤ 65, then graph y 25x
y
x 25
(c) lim
285
150 150 300
5
x
y
60
50y 50x 2xy
(b)
Infinite Limits
25x
x 25 As x gets close to 25 mph, y becomes larger and larger.
62. False; for instance, let
61. False; for instance, let
f x x2 1
.
x1
f x The graph of f has a hole at 1, 2, not a vertical asymptote.
x2 1
.
x1
The graph of f has a hole at 1, 2, not a vertical asymptote.
64. True
63. True
65. Let f x 1
1
and gx 4, and c 0.
x2
x
1
1
1
1
x2 1
0.
2 and lim 4 , but lim
2 4 lim
x→0 x
x→0 x
x→0 x
x→0
x
x4
lim
66. 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
gx
f x
<
< , we have
3L2
.
3L2
Therefore, lim
x→c
gx
0.
f x
286
Chapter 3
Limits and Their Properties
67. Given lim f x , let gx 1. Then lim
x →c
68. Given lim
x→c
x →c
gx
0 by Theorem 3.13.
fx
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 .
1
0.
f x
69. f x Suppose lim f x exists and equals L. Then,
x→c
lim 1
1
1
x→c
0.
x→c f x
lim f x L
Equivalently, x 3 <
lim
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 This is not possible. Thus, lim f x does not exist.
x→c
1
1
is defined for all x < 4. Let N < 0 be given. We need > 0 such that f x < N whenever 4 < x < 4.
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
70. f x Review Exercises for Chapter 3
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.
2. Precalculus. L 9 12 3 12 8.25
3.
f x
2
0.1
0.01
0.001
0.001
0.01
0.1
1.053
1.005
1.001
0.9995
0.995
0.9524
x
−2
lim f x 1
x→0
4.
0.1
x
f x
−2
0.01
1.432
1.416
0.001
0.001
0.01
0.1
1.414
1.414
1.412
1.397
2
lim f x 1.414
− 1.5
x→0
5. hx x2 2x
x
6. gx 3x
x2
(a) lim gx does not exist.
(b) lim hx 3
(b) lim gx 0
x→0
1.5
0
(a) lim hx 2
x→1
2
x→2
x→0
Review Exercises for Chapter 3
7. lim 3 x 3 1 2
8. lim x 9 3.
x→9
x→1
Let > 0 be given. Choose . Then for
Let > 0 be given. We need
x 3 < ⇒ x 3x 3 < x 3
x 9 < x 3 .
0 < x 1 < , you have
1 x < 3 x 2 < f x L < .
x1 < Assuming 4 < x < 16, you can choose 5.
Hence, for 0 < x 9 < 5, you have
x 9 < 5 < x 3 x 3 < f x L < .
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 < .
10. lim 9 9. Let > 0 be given. can be any positive
x→5
11. lim t 2 4 2 6 2.45
t→4
number. Hence, for 0 < x 5 < , you have
9 9 < f x L < .
12. lim 3 y 1 3 4 1 9
13. lim
t2 9
lim t 3 6
t→3
t3
15. lim
y→4
14. 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
16. lim
x→0
4 x 2
x
lim
x→0
lim
x→0
287
4 x 2
x
1
4 x 2
4 x 2
4 x 2
1
4
17. lim
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
lim
1
x→0 x 1
1
4
288
Chapter 3
18. lim
s→0
Limits and Their Properties
11 s 1 lim 11 s 1 11 s 1
s→0 s
s
11 s 1 lim
s→0
19. lim
x→5
11 s 1
1
1
lim
s 11 s 1 s→0 1 s 11 s 1 2
x3 125
x 5x2 5x 25
lim
x→5
x5
x5
20. lim
x→2
x2 4
x 2x 2
lim
x3 8 x→2 x 2x2 2x 4
lim x2 5x 25
lim
x→5
x→2
75
21. lim f x gx 4 3 2
3
2
1
x→c
x2
x2 2x 4
1
4
12
3
7
22. lim f x 2gx 34 223 12
x→c
sa st
4.942 200 4.9t2 200
lim
t→a
t→4
at
4t
23. lim
lim
t→4
4.9t 4t 4
4t
lim 4.9t 4 39.2 msec
t→4
24. 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
25. lim
x→3
x 3 x3
lim
x→3 x 3
1
x3
27. lim f x 0
x→1
29. lim ht does not exist because lim ht 1 1 2 and
t→1
lim ht t→1
x→4
The graph jumps from 2 to 3 at x 4.
28. lim gx 1 1 2
x→2
t→1
1
2 1
26. lim x 1 does not exist.
30. lim f s 2
s→2
1 1.
31. f x x 3
32. f x lim x 3 k 3 where k is an integer.
3x2 x 2 ,
x1
0,
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.
lim f x lim
x→1
x→1
x1
x1
3x2 x 2
x1
lim 3x 2 5 0
x→1
Removable discontinuity at x 1.
Continuous on , 1 1, .
Review Exercises for Chapter 3
33. f x 3x2 x 2 3x 2x 1
x1
x1
34. f x lim f x lim 3x 2 5
x→1
52xx,3,
x ≤ 2
x > 2
lim 5 x 3
x→2
x→1
Removable discontinuity at x 1.
lim 2x 3 1
x→2
Continuous on , 1 1, .
Nonremovable discontinuity at x 2.
Continuous on , 2 2, .
35. f x lim
x→2
1
x 22
36. f x 1
x 22 lim
x→0
Nonremovable discontinuity at x 2.
x x 1 1 1x
1 1x Domain: , 1, 0, Continuous on , 2 2, .
Nonremovable discontinuity at x 0.
Continuous on , 1 0, .
37. f x 3
x1
38. f x lim f x x→1
lim
x→1
lim f x x→1
Continuous on , 1 1, .
Continuous on , 1 1, .
(a)
2x 1 3
x1
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
x1
1
2x 1 2
Removable discontinuity at x 1.
Nonremovable discontinuity at x 1.
39. f x x1
2x 2
2x 1 3
x1
2x 1 3
x1
0.577
lim
Actual limit is 33.
2x 1 3
x1
x→1
2x 1 3
2x 1 3
lim
2x 1 3
x 1 2x 1 3 lim
2
2x 1 3
x→1
x→1
2
23
1
3
3
3
(b)
2
−1
2
0
289
290
Chapter 3
40. f x (a)
3
1 x
x1
x
f x
lim
x→1
(c) lim
x→1
Limits and Their Properties
1.1
1.01
0.3228
0.3322
3
1 x
0.333
x1
1.001
0.3332
1.0001
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
3
42. lim x 1 2
41. f 2 5
x→1
Find c so that lim cx 6 5.
lim x 1 4
x→2
x→3
c2 6 5
Find b and c so that
2c 1
c
lim x2 bx c 2 and
x→1
1
2
lim x2 bx c 4.
x→3
Consequently we get
1bc2
and 9 3b c 4.
Solving simultaneously,
b 3 and
c 4.
43. 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.
45. f x 44. C 9.80 2.50x 1, x > 0
9.80 2.50x 1
C has a nonremovable discontinuity at each integer.
x2
x2 4
x 2
x2
x2
(a) lim f x 4
x→2
(b) lim f x 4
30
x→2
(c) lim f x does not exist.
x→2
0
5
0
46. f x x 1x
(a) Domain: , 0 1, (b) lim f x 0
x→0
(c) lim f x 0
x→1
Problem Solving for Chapter 3
47. gx 1 2
x
48. hx Vertical asymptote at x 0.
49. f x 51.
lim x→2
8
x 102
Vertical asymptote at x 10.
Vertical asymptotes at x 2
and x 2.
x3
xx2 1
50. f x 4x
4 x2
2x2 x 1
x2
52.
lim
x→ 12
x
2x 1 x2 1 0. Vertical asymptote:
x0
53.
lim
x→1
55. lim
x→1
1
x1
1
lim
x3 1 x→1 x2 x 1 3
59. r x1
1
1
lim
x4 1 x→1 x2 1x 1
4
lim
x2 2x 1
x1
56.
58. lim
1
x3
2L
(b) When L 13, r 225
252 144
50481
2.28 ft/sec
481
213
26
5.2 ft/sec
2
5
13 144
lim r L→12 Problem Solving for Chapter 3
1. (a) Perimeter PAO x2 y 12 x2 y2 1
x2 x2 12 x2 x 4 1
Perimeter PBO x 12 y2 x2 y2 1
x 12 x 4 x2 x 4 1
(b) rx x2 x2 12 x2 x4 1
x 12 x4 x2 x4 1
x
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
(c) lim r x x→0
x→1
x→2
L2 144
(a) When L 25, r (c)
lim
x→1
x2 2x 1
x1
57. lim x x→0
54.
101 2
1
101 2
1
3 2
x 4
291
292
Chapter 3
Limits and Their Properties
2. (a) There are 6 triangles, each with a central angle of
60 3. Hence,
3. (a) Slope 12bh 6 121 sin 3 (b) Slope Area hexagon 6
3
Tangent line: y 4 x 3
4
(c) Let Q x, y x, 25 x2
h = sin θ
h = sin 60°
mx 1
25 x2 4
x3
θ
60°
3
4
3
25
y x
4
4
33
2.598.
2
1
40 4
30 3
(d) lim mx lim
x→3
33
Error: 0.5435.
2
12bh n 121 sin n (c)
x3
25 x2 4
25 x2 4
25 x2 16
x→3 x 3 25 x2 4 lim
(b) There are n triangles, each with central angle of
2n. Hence,
2
An n
25 x2 4
x→3
lim
x→3
n sin 2n
.
2
lim
x→3
n
6
12
24
48
96
An
2.598
3
3.106
3.133
3.139
3 x3 x
x 325 x2 4
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 .
4. (a) Slope 12
5
5.
5
(b) Slope of tangent line is .
12
5
y 12 x 5
12
y
mx x
a bx 3
x
x2
lim
12 169 x2
x→5
x5
(d) lim mx lim
3 bx 3
x
x→0
x→0
x→0
12 169 x2
12 169 x2
144 169 x2
x→5 x 5 12 169 x2 lim
x2 25
x 512 169 x2
lim
x 5
12 169 x2
5
10
12 12 12
This is the same slope as part (b).
lim
lim
lim
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
(c) Q x, y x, 169 a bx 3
Setting
b
3 3
bx
x 3 bx 3
b
3 bx 3
3, you obtain b 6.
Thus, a 3 and b 6.
.
Problem Solving for Chapter 3
6. (a) 3 x13 ≥ 0
(b)
(c)
0.5
lim f x x→27
x13 ≥ 3
x ≥ 27
− 30
Domain: x ≥ 27, x 1
(d) lim f x lim
x→1
lim
x→1
lim
x1
x→1
1
x23
2
1
0.0714
28 14
3 x13 2
3 x13 2
x13 1
x13 13 x13 2
1
x23 x13 13 x13 2
1
1
1 1 12 2 12
7. (a) lim f x 3: g1, g4
(c) lim f x 3: g1, g3, g4
(b) f continuous at 2: g1
x→2
8. (a)
27 1
3 x13 4
x 13 x13 2
x→1 x13
lim
3 2713 2
− 0.1
3 x13 2
x→1
12
x→2
9.
y
y
4
3
2
2
1
1
−4 −3 −2 −1
x
1
2
3
4
−2
−3
x
a
(b) (i)
(ii)
b
lim Pa, bx 1
x→a
−4
(a)
f 12 0 1 1
(iii) lim Pa, bx 0
x→b
f 2.7 3 2 1
(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.
f 1 1 1 1 1 0
f 0 0
lim Pa, bx 0
x→a
(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, . . .
293
294
Chapter 3
Limits and Their Properties
v2 11. (a)
y
10.
192,000
v02 48
r
3
192,000
v2 v02 48
r
2
1
x
−1
−1
1
r
−2
lim r v→0
(a) f 14 4 4
192,000
48 v02
Let v0 48 43 milessec.
f 3 13 0
f 1 1 1
192,000
v2 v02 48
(b)
(b) lim f x 1
x→1
v2 1920
v02 2.17
r
1920
v2 v02 2.17
r
lim f x 0
x→1
lim f x x→0
r
lim f x x→0
(c) f is continuous for all real numbers except
lim r v→0
1
1
x 0, ± 1, ± 2, ± 3, . . ..
1920
v2 v02 2.17
1920
2.17 v02
Let v0 2.17 misec 1.47 misec.
r
(c)
lim r v→0
10,600
v2 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.