Solucionario De Calculo Integral Larson 7Edicion

324

Chapter 1

Limits and Their Properties

Review Exercises for Chapter 1
2. Precalculus. L
4.

9

2

1

3

1

2

8.25

0.5

x

0.1

0.01

0.001

0.001

0.01

0.1

fx

0.358

0.354

0.354

0.354

0.353

0.349

lim f x

0.2

x→0

x

8. lim

(b) lim g x

(a) lim g x does not exist.

9

Assuming 4 < x < 16, you can choose

3.Hence, for 0 < x

> 0 be given. We need

Let

0

x→0

x→2

2

x

x→9

1

− 0.5

3x

6. g x

−1

⇒

9<

x

x

3

3<

x

3

x

9 0 be given.

9. Let

x→5

number. Hence, for 0 < x

L<

t2
t

1

y→4

34

x

2

1

9
3

lim t

3

t→3

6

4

16. lim

x→0

lim

1

s→0

s

1

lim

s→0

20. lim

x→ 2x2
x3

4
8

1

s

1

11 s
s1 1 s

1
1

lim

s→0

s

lim

x→ 2

lim

x→ 2

4
12

x
x

s

2x 2
2 x2 2x 4

x
x2

1

2
2x
1
3

4

1
1

1
1

s
s

1

s

1
1

lim

s→0

1

22.

1
1

lim

x→

s

4

4

x→0

x

lim

1

9

5 < , you have

x→0

18. lim

3

L<

fx

t→3

12. lim 3 y

can beany positive

x

9<

9

14. lim

5 , you have

3<

fx

10. lim 9

5.

1
2

1

4x
tan x

4

4
1

x

2

x

2

4
4

x
1
4

1
4

x
x

2
2

Review Exercises for Chapter 1
cos

24. lim

x

x→0

1

cos

lim

x→0

0

(a)

01
2
3

2

sin x

1

1
sin x
x

lim sin
x→0

0

7
12

3

1
x

28. f x

3
42g x

x→c

sin
x

cos x
x

lim

26. lim f x

cos x

x→0

x

x
1

x

1.1

1.01

1.001

fx

0.3228

0.3322

0.3332

(b)

1.0001
0.3333

3

1
x

lim

x→1

x
1

3

1
x

(c) lim
x→1

0.333

x
1

1
x

lim

x→1

Actual limit is
3

x
1

3

1
1

3

1

lim

x→1

x

11

1

3

3

x
x

3

3

−3

31
3.
3

x
x

2

−3

2

x
2
x

x2

1

lim

x→1

3

x

x

2

1
3
0⇒

30. s t

When t
lim
t→a

4.9t2

200

sa
a

st
t

lim
t→a

lim

32. lim x

4.9 a

6.39 sec

t

4.9 6.39

6.39

62.6 m sec.

1 does not exist. The graph jumps from 2 to 3

x→4

34. lim g x

1

x→1

1

2.

4.

36. lim f s
s→ 2

40.816 ⇒ t6.39, the velocity is approximately

t → 6.39

at x

0 ⇒ t2

2

38. f x

3x2
0,

x
x

1

2,

lim f x

x→1

lim

3x

x→1

lim 3x

x
x

1

x

2

x

1

2
1
5

0

Removable discontinuity at x
Continuous on
,1
1,

1

x→1

2

325

326

Chapter 1

Limits and Their Properties
x≤2
x>2

5 x,
2x 3,

40. f x
lim 5

x

x→2

32x

lim

3

1

lim

1
x

1

,

Domain:

x
2x

2

x
12x

46. f x

1
1

1 , 0,

Nonremovable discontinuity at x
Continuous on
,1
0,

1
2

lim

1
x

1

Nonremovable discontinuity at x
Continuous on
,2
2,

44. f x

1

x

x→0

x→2

x→

x

42. f x

0

tan 2x

Nonremovable discontinuities when

1
2

2n

x

Removablediscontinuity at x
Continuous on
,1

1
4

1
1,

Continuous on
2n

1
4

2n

,

1
4

for all integers n.

48. lim x

1

2

lim x

1

4

x→1

x→3

Find b and c so that lim x2

bx

x→1

Consequently we get

1

b

Solving simultaneously,
50. C

9.80

2.50

9.80

2.50

c

x→3

2

b

x
x

2 and lim x2

c

and 9
3 and

3bbx

c

c

4.

c

4.

4.
52. f x

1, x > 0

x

1x
,0

(a) Domain:

1

C has a nonremovable discontinuity at each integer.

(b) lim f x

0

(c) lim f x

0

x→0

30

x→1

0

1,

5
0

4x

54. h x

56. f x

x2

4

Vertical asymptotes at x

58.

62.

lim

x→ 1 2

lim

x→ 1

66. lim
x→0

x
2x
x2

sec x
x

1
1

Vertical...