Gestion De La Incetidumbre

CHAPTER 1
Limits and Their Properties
Section 1.1

A Preview of Calculus . . . . . . . . . . . . . . . . . . . 305

Section 1.2

Finding Limits Graphically and Numerically . . . . . . . 305

Section 1.3

Evaluating Limits Analytically . . . . . . . . . . . . . . . 309

Section 1.4

Continuity and One-Sided Limits

Section 1.5

Infinite Limits

Review Exercises

. . . . . .. . . . . . . 315

. . . . . . . . . . . . . . . . . . . . . . . 320

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327

CHAPTER 1
Limits and Their Properties
Section 1.1

A Preview of Calculus

Solutions to Even-Numbered Exercises
2. Calculus: velocity is not constant
Distance

4.Precalculus: rate of change

20 ft sec 15 seconds

6. Precalculus: Area

2

slope

0.08

300 feet

2

8. Precalculus: Volume

3 26

54

2

10. (a) Area

5
2

5

5
4
5
2

5
1.5

1
5
2

Area

5
3

10.417
5
2.5

5
3

5
3.5

5
4

5
4.5

9.145

(b) You could improve the approximation by using more rectangles.

Section 1.2
2.

FindingLimits Graphically and Numerically

x

1.9

1.99

1.999

2.001

2.01

2.1

fx

0.2564

0.2506

0.2501

0.2499

0.2494

0.2439

lim

x→2

4.

x
x2

2
4

Actual limit is 1 .
4

0.25

x

3.1

3.01

3.001

2.999

2.99

2.9

fx

0.2485

0.2498

0.2500

0.2500

0.2502

0.2516

lim

x→ 3

6.

1
x

x2
3

0.25

Actuallimit is

1
4.

x

3.9

3.99

3.999

4.001

4.01

4.1

fx

0.0408

0.0401

0.0400

0.0400

0.0399

0.0392

lim

xx

x→4

8.

1
x

45
4

0.04

1

Actual limit is 25 .

x

0.1

0.01

0.001

0.001

0.01

0.1

fx

0.0500

0.0050

0.0005

0.0005

0.0050

0.0500

lim

x→0

cos x
x

1

0.0000

(Actual limit is 0.)(Make sure you use radian mode.)

305

306

Chapter 1

10. lim x2

Limits and Their Properties

2

x→1

3

x→1

1
does not exist since the
x3
function increases and decreases
without bound as x approaches 3.

14. lim

16. lim sec x

x→3

20. C t

0.35

(a)

lim x2

12. lim f x

0.12

t

x→1

1

x→0

2

3

18. lim sin x
x→1

0

1

10

5
0

(b)

t

3

3.4

3.5

3.6

3.7

4

0.59

Ct

3.3
0.71

0.71

0.71

0.71

0.71

0.71

lim C t

0.71

t → 3.5

(c)

3

2.5

2.9

3

3.1

3.5

4

0.47

0.59

0.59

0.59

0.71

0.71

0.71

t
Ct

lim C t does not exist. The values of C jump from 0.59 to 0.71 at t

3.

t → 3.5

22. You need to find such that 0 < x2 < implies
fx
3
x2 1 3
x2 4 < 0.2. That is,
0.2
4 0.2
3.8
3.8
3.8 2
So take

< x2
4
< x2
< x2
<
x
0:

2

Assuming 1 < x < 9, you can choose
0< x

1

<

29
3

fx

x

1

fx
3
2

1<

2
3

9

> 0:

1

1<

x

<

L<

Hence, for any

3
2

Hence, if 0 < x

2

1

Hence, any

32 .

Hence, let

, you have

0<

1<

1<

1<

2
3

9

5

1

x→2

2

6<

fx

> 0:

2
3x

Given

3<

x

L < 0.01

9

2

2.

Hence, let

29 < 0.01

2
3x

x→4

3<

25 < 0.01

fx

34. lim

x

0.01
1
5<
<
0.01
11
x5

4

x

3<

5 < 0.01

x2

Given

<

6<

0.0009.

0.01
, you have
11

5x

2
3x

1

2x

0.01
x5

x2

x→1

5
2x

5<

x

30.lim

1

> 0:

< 0.01

If we assume 4 < x < 6, then

x

5

x→ 3

4

x

Finding Limits Graphically and Numerically

2

0<
3<
.

Hence for 0 < x
x

2 0:

Given
x2

3x
xx

3

x
If we assume

xx
2

x

3<

x
x2

−4

The domain is all x 1, 3. The graphing utility does not
show the hole at 3, 1 .
2

x

3

2, then
<

4.
4

<

0<...