Chapter2 - Problems of Chapter 2

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EARLY TRANSCENDENTALS

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Edited by Vu Thu Giang

CALCULUS

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Section 2.2 and 2.3

The definition of the limit of a function

One – side limits

Infinite limits

The limit laws

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Section 2.2 – Exercise 4

For the function whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why.

f

0 3 3

3

lim lim lim

lim 3

x x x

x

a f x b f x c f x

d f x e f

SolutionEasy to find

d. Since the left and right limits are different, we conclude that does not exist.

0 3 3lim 3 lim 4 lim 2

3 3

x x xa f x b f x c f x

e f

3

limx

f x

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Section 2.2 – Exercise 15

23 3lim 4, lim 2, lim 2

3 3, 2 1

xx xf x f x f x

f f

Sketch the graph of an example of a function f that satisfies all of the given conditions

means

f(x) is close to l when x is close to a from the right. The left – side limit and both – side limit are similar.

limx a

f x l

What will the graph of f(x) be?

limx a

f x l

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Section 2.2 – Exercise 27

Determine the infinite limit

21

2lim

1x

x

x

SolutionIf x is close to 1, then the denominator is a small positive number

and close to 1. So the quotient is a large positive

number. The more the value x is close to 1, the larger the quotient is. Thus, we have

21x

2 x 2

2

1

x

x

21

2lim

1x

x

x

x close to 1, how about (2-x) and 2

1x

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Section 2.2 – Exercise 35

a. Estimate the value of the limit to five decimal places. Does this number look familiar?

b. Illustrate part (a) by graphing the function

1/

0lim 1

x

xx

1/1

xy x

SolutionBy calculating we have

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Section 2.2 – Exercise 35 cont These numbers are more and more close to a constant number. This

number look familiar and we use to call number e .In fact, we have

The graph of the function is given following. From the graph we imply that

1/

0

1lim 1 lim 1

nx

x nx e

n

1/1

xy x

1/

0lim 1 2.7

x

xx

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Section 2.3 – Exercise 7 Evaluate the limits and justify each step by indicating the appropriate Limit Law(s)

4 4

2 2

4

2 2 2

4

lim 3 6 lim 3 6 11

lim lim 3 lim 6 1

2 3. 2 6 3,9

4

u u

u u u

u u u u by law

u u by law

by law

Solution

4

2lim 3 6u

u u

Limit law(s)???11 laws from page 99 to page 101

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Section 2.3 –Exercise 15.19.20 Evaluate the limit, if it exists

2

2 33 2

3

0

9 215.lim 19. lim

2 7 3 8

2 820.lim

t x

h

t x

t t x

h

h

Solution

2

23 3 3

3 39 315. lim lim lim

2 7 3 3 2 1 2 1

6

5t t t

t tt t

t t t t t

3 222 2 2

2 2 119. li

1

1m lim lim

8 2 42 2 4 2x x x

x x

x x xx x x

3 2 3

2

0 0 0

2 8 8 12 6 820.lim lim lim 6 12 21

h h h

h h h hh h

h h

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Section 2.3 –Exercise 35 If for , find

24 9 4 7x f x x x 0x

4limx

f x

SolutionSince we have and the assumption

for then

2

4 4lim 4 9 lim 4 7 7x x

x x x

24 9 4 7x f x x x 0x 4

lim 7x

f x

Can you calculate the limit of and when ?

They are equal!So, What will happen?

4 9x 2 4 7x x 4x

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Section 2.3 –Exercise 39.40

Find the limit, if it exists. If the limit does not exist, explain why.

3 6

2 1239.lim 2 3 40. lim

6x x

xx x

x

Solution

3 3

3 3

3

39. lim 2 3 lim 2 3 6

lim 2 3 lim 2 3 6

lim 2 3 6

x x

x x

x

x x x x

x x x x

Then x x

6

6 6

6 6

2 12The one-side

2 12 2 1240. lim l

limits are no

im 26 6

2

t equal so the limit li

12 2 12lim lim 2

6 6

m does not exist.6

x x

x x

x

x x

x x

x x

x x

x

x

The absolute value may be different when x goes to a from the left and the right!

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Section 2.3 –Exercise 49 a. If the symbol denotes the greatest integer function define in

Example 10, evaluate

b. If n is an integer, evaluate

c. For what values of a does exist?

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2.4

lim 2 lim 1

Thus we see that the one-side limits are

not equal and the limit does not exist.

lim 2

xx

x

i x ii x

iii x

lim limx n x n

i x ii x

limx a

x

Solutiona. From the graph of the greatest

integer function we have

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Section 2.3 –Exercise 49 Cont b. If n is an integer,

c. If and only if a is not an integer.

lim 1 limx n x n

i x n ii x n

limx a

x

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Section 2.3 –Exercise 58

Show by means of an example that may exist,

even through neither nor exists.

limx a

f x g x

limx a

f x

limx a

g x

SolutionFor example, let ,

x x xf x g x

x x

00 0 0 0

lim lim 1, lim lim 1 limxx x x x

x xf x f x f x

x x

00 0 0 0

lim lim 0, lim lim 2 limxx x x x

x x x xg x g x f x

x x

0 0

But lim lim1 1x x

f x g x

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Section 2.3 –Exercise 61

Is there a number a such that

exist? If so, find the value of a and the value of the limit.

2

22

3 3lim

2x

x ax a

x x

Solution approaches to 0 as x approaches to -2Thus, the limit exist implies -2 is a solution of equation

This means

When we have

2 2 1 2x x x x

23 3 0x ax a

2153. 2 2 3 0a a a

15a

2

22 2

3 3 23 15 18lim lim

21

1 2x x

x xx x

x x x x

When will the limit exist?

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Problem 2.3.6Problem 2.3.6

The figure shows a fixed circle C1 with equation (x - 1)2 + y2 = 1 and a

shrinking circle C2 with radius r and center the origin. P is the point

(0,r), Q is the upper point of intersection of the two circles, and R is the point of intersection of the line PQ and the x-axis. What happens to R as C2 shrinks, that is, as r 0+ ?

solution 1

Determine the coordinates of Q

Qx r 21

2 Qy r r 21

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The equation of the line though PQ

ry r x

r

24 20

Set y = 0 in order to find x-intercept, we getR

rx r

r

22

2... 2 4

2 4

Take the limit as r 0+, lim xR = 4. The limiting position of R is (40)

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Problem 2.3.62Problem 2.3.62

T

solution 2

PSQ = PRO (?) OQS = TQR (?) PSQ = OQS (?)

S

TQ = TR

When r 0 then Q O

TR 2 or xT 4

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Section 2.5

Three conditions for the continuity at a number a of a function

A function is continuous from the left and from the right at a number a

Some continuous functions

The Intermediate Value Theorem

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Section 2.5 –Exercise 7

Solutiona. The graph break when x=-2

or x=2 or x=4. Thus, f is discontinuous at these numbers.

b. From the graph we imply that

2 2 4

lim 2 , lim 2 , lim 4x x x

f x f f x f f x f

So f is continuous from the left at -2, and continuous from the right at 2 and 4.

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Section 2.5 –Exercise 7 A parking lost charges $3 for the first hour (or a part of an hour) and $2 for each succeeding hour (or part), up to daily maximum of $10.a. Sketch a graph of the cost of parking at this lot as a function of the

time parked there.b. Discuss this discontinuities of this function and their significance to

someone who parks in the lot.Solution

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Section 2.5 –Exercise 11 Use the definition of the continuity and the properties of the limits to show that the function is continuous at the given number a.

432 , 1f x x x a

Solution• f(x) is continuous at -1 if

• We have

•

• Therefore and f(x) is continuous at -1.

1

lim 1x

f x f

1 81f

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1 1lim lim 2 81x x

f x x x

1

lim 1x

f x f

When will f(x) is

continuous at a?

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Section 2.5 –Exercise 18 Explain why the function is discontinuous at the given number a. Sketch the graph of the function

2

21

111 1

x xif x

f x axif x

SolutionWe have

Since then f(x) is

discontinuous at 1

2

21 1 1

11 1, lim lim lim

1 1 2x x x

x x xf f x

x x

1

lim 1x

f x f

When will f(x) is discontinuous

at a?

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Section 2.5 –Exercise 27 Explain, using Theorem 4,5,7, and 9, why the function is continuous at every number in its domain. State the domain.

4ln 1G t t

SolutionEasy to imply the continuity of the function.By theorem 7 we have polynomial and logarithmic functions are continuous at every number in theirs domain. Therefore are continuous at every number in theirs domain.And By theorem 9 we obtain G(t) is continuous at every number in its domain.The domain of G(t) is

4 1 and lnf t t g u u

G t g f t

; 1 1;

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Section 2.5 –Exercise 32 Using the continuity to evaluate the limit limsin sin

xx x

SolutionThe function sin(x+sinx) is continuous function at every points in its domain. So limsin sin sin sin 0

xx x

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Section 2.5 –Exercise 39 Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, from the left, or neither? Sketch the graph of f

2 0

0 1

2 1

x

x if x

f x e if x

x if x

Solutionf is continuous at every number x<0, 0<x<1, x>1.We examine the continuity at 0 and 1At 0

At 1

0 0 0 0

0 0 is discontinuous at 0

But is continuous from the ri

lim lim 1 0 , lim

gh

lim 2 2

lim lim

t at 0

x

x x x x

x x

f x e f f x x

f x f x f

f

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1 1 1 1

1 1 is discontinuous at 1

But is continuous fro

lim lim 2 1, lim lim 1

lim

m the left at 0

lim

x

x x x x

x x

f x x f x e e f

f x f x f

f

Section 2.5 –Exercise 39 Cont

The graph of function f:

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Section 2.5 –Exercise 41 For what value of the constant c is the function f continuous on ;

2

3

2 2

2

cx x if xf x

x cx if x

Solutionf(x) is continuous at every numbers x<2 and x>2.We find the value of c satisfies the continuity of f at 2The condition is:

We have:

Thus, the condition is equivalent to:

Therefore with , the function f(x) is continuous on

2 2

lim 4 4, lim 8 2 2x x

f x c f x c f

2 2

lim limx x

f x f x f x

24 4 8 2

3c c c

2

3c ;

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Section 2.5 –Exercise 47 Use the intermediate value Theorem to show that there is a root of the given equation in the specified interval

4 3 0, 1,2x x

SolutionConsider function in the interval (1,2). Easy to see f(x) is the sum of three continuous functions, thus, f(x) is continuous.We have:

Therefore, by the intermediate value Theorem, we imply that the function has at least a root in the interval (1,2).

3 3f x x x

1 1; 2 15 1 2 0f f f f

0f x

The Intermediate Value Theorem

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Section 2.5 –Exercise 61 HintIs there a number that is exactly 1 more than its cube?

Consider the equation 3 1 0x

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Section 2.6 –Exercise 7 Hint

Sketch the graph of an example of a function f that satisfies all of the given conditions.

2

0 0

lim , lim , lim 0

lim , limx x x

x x

f x f x f x

f x f x

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Section 2.6 –Exercise 19.25.55 Find the limit

3

23 2

519.lim 25.lim 9 3

2 4x x

x xx x x

x x

Solution

3

3 2

2

2

5 119.lim

2 4 21

25.lim 9 3 lim69 3

x

x x

x x

x xx

x x xx x x

55. Let P and Q be polynomials. Find If the degree of P is a. Less than the degree of Qb. Greater than the degree of Q

limx

P x

Q x

Solutiona. The limit is equal to 0b. The limit is equal to

How about if the degrees of P and Q are equal?

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Section 2.6 –Exercise 41 Find the horizontal and vertical asymptotes of each curve. If you have a graphing device, check your work by graphing the curve and estimating the asymptotes.

2

2

2 1

2

x xy

x x

SolutionWe have

So the horizontal asymptote of this function is In other case,

Therefore, the vertical asymptotes of f(x) are and

2 2

2

2

1 122 1

lim lim 21 22 1

x x

x x x xx x

x x

2y

2

21 1

2

22 2

1 2 12 1lim lim

2 1 2

1 2 12 1lim lim

2 1 2

x x

x x

x xx x

x x x x

x xx x

x x x x

1x 2x

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Section 2.6 –Exercise 41 Cont The graph of and its asymptotes.

2

2

2 1

2

x xy

x x

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Section 2.6 –Exercise 57 HintFind if, for all lim

xf x

1x

10 21 5

2 1

x

x

e xf x

e x

10 21 5lim lim 5

2 1

x

xx x

e x

e x

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Section 2.7 –Exercise 5.8 Find an equation of the tangent line to the curve at the given point

15. , 3,2 7. , 1,1

2

xy y x

x

Solution5. The tangent line to the curve at point (3,2) has the slope m, where

So the equation of the tangent line is:

1

2

xy

x

3 3 3

123 32lim lim lim 1

3 3 3 2x x x

xf x f xxm

x x x x

3 2 5y x y x

6. The tangent line to the curve at point (1,1) has the slope m, where

So the equation of the tangent line is:

y x

1 1 1

1 1 1 1lim lim lim

1 1 21 1x x x

f x f x xm

x x x x

1 1 11 1

2 2 2y x y x

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Section 2.7 –Exercise 9 Hinta. Find the slope of the tangent to the

curve at the point where

b. Find the equations of the tangent lines at the point and

c. Graph the curve and both tangents an a common screen

2 33 4 2y x x x a

1,5 2.3

Similar with ex 5-8

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Section 2.7 –Exercise 13 If a ball is thrown into the air with a velocity of , its height (in feet) after t seconds is given by . Find the velocity when

40 /ft s240 16y t t 2t

SolutionUsing the equation of height, we have 240 16y f t t t

2

2 2

2

2 40 16 162 lim lim

2 28 2 1 2

lim 24 /2

t t

t

f t f t tv

t tt t

ft st

The velocity of the ball at time t=2 is negative, what can you imply about the motion?

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Section 2.7 –Exercise 17 HintFor the function g whose graph is given, arrange the following numbers in increasing order and explain your reasoning

0 ' 2 ' 0 ' 2 ' 4g g g g

g’(a) is also the slope of the tangent line to the curve at point (a, f(a))

Nearly x=-2, x=0, x=2, x=4, When the function g increases, decreases?Thus, which derivatives of x are positive, negative?

The trigonometric function y=tanx is increase in the domain

, so which derivative is larger?

;2 2

H4

H2

' 0 0 ' 4 ' 2 ' 2g g g g

g’(a) is instantaneous rate of change of g(x) with respect to x when x=a

H1

H3

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Section 2.6 –Exercise 18 a. Find an equation of the tangent line to the graph of

b. If the tangent line to y=f(x) at (4,3) passes through the point (0,2), find f(4) and f’(4).

5 5 3 ' 5 4y g x at x if g and g

Solutiona. g(5)=-3 so the tangent line passes through the point (5,-3). From the

assumption g’(5)=4 we imply that the tangent line to y=g(x) at x=5 has slope m=4. Therefore, the equation we need is

b. Since the tangent line to y=f(x) at point (4,3) then it passes through the point (4,3) and f(4)=3. In other, it passes through (0,2). Therefore, the slope of this tangent line is

and by the slope – intercept form, the equation is

4 5 3 4 23y x y x

3 2 1' 4

4 0 4m f

12

4y x

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Section 2.7 –Exercise 19 HintSketch the graph of a function f for which

0 0, ' 0 3, ' 1 0, ' 2 1f f f and f

What can you infer about function f at the point x=a if f’(a)<0, f’(a)>0, f’(a)=0

If f’(a)>0 then the function f(x) may be increase nearly x=a

If f’(a)<0 then the function f(x) may be decrease nearly x=a

If f’(a)=0 then the function f(x) may be constant or the rates of change are

different from left side and right side of x=a

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Section 2.7 –Exercise 23 Find f’(a)

2 1

3

tf t

t

Solution

2

55 52 2

3 33 3' lim lim

5 5lim

3 3 3

t a t a

t a

t a

t at af a

t a t a

t a a

2 1 2 6 5 52

3 3 3

t tf t

t t t

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Section 2.7 –Exercise 35 HintEach limit represents the derivative of some function f at some number a. State such an f and a in each case

0

cos 1limh

h

h

0

' limh

f a h f af a

h

cos 1

cos ,f x x a

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Section 2.7 –Exercise 38 HintA warm can of soda is placed in a cold refrigerator. Sketch the graph of the temperature of the soda as a function of time. Is the initial rate of change of temperature greater or less than the rate of change after an hour?

Solution

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Section 2.7 –Exercise 43 The cost (in dollars) of producing x units of a certain commodity is

a. Find the average rate of change of C with respect to x when the production level is changed

b. Find the instantaneous rate of change of C with respect to x when x=100 (This is called the marginal cost. Its significance will be explained in section 3.7)

25000 10 0.05C x x x

) 100 105

) 100 101

i from x to x

ii from x to x

Solutiona.

b. The instantaneous rate of change of C with respect to x when x=100 is C’(100) = 2000

1

1

105 100 101.2520.25

105 100 5

C CC

x

2

2

101 100 20.0520.05

101 100 1

C CC

x

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Section 2.7 –Exercise 45 Hint

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Section 2.7 –Exercise 47 Hint

Answer

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Section 2.7 –Exercise 51 Determine whether exists. ' 0f

1

sin 0

0 0

x if xf x x

if x

Solution

We have

The limit does not exist, therefore the derivative does not exist.

0 0 0

1sin0 1

' 0 lim lim limsinx x x

xf x f xfx x x