Solution manual for calculus of a single variable early transcendental functions 6th edition sample

C H A P T E R2

Limits and Their Properties

Download Full Solutions Manual for Calculus of a Single Variable Early Transcendental Functions 6th Edition

https://getbooksolutions.com/download/solution-manual-for-calculus-of-a-single-variable-early-transcendental-functions-6th-edition

Section 2.1A Preview of Calculus81

Section 2.2Finding Limits Graphically and Numerically82

Section 2.3Evaluating Limits Analytically93

Section 2.4Continuity and One-Sided Limits105

Section 2.5Infinite Limits117

Review Exercises125

Problem Solving133

2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

C H A P T E R 2 Limits and Their Properties

Section 2.1A Preview of Calculus

1. Precalculus: (20 ft/sec)(15 sec) = 300 ft

2. Calculus required: Velocity is not constant.

Distance (20 ft/sec)(15 sec) = 300 ft

3. Calculus required: Slope of the tangent line at x = 2 is the rate of change, and equals about 0.16.

4. Precalculus: rate of change = slope = 0.08

5. (a) Precalculus: Area = 12 bh = 12 (5)(4) = 10 sq. units

(b) Calculus required: Area = bh

2( 2.5)

= 5 sq. units

6.f (x )=x

(a)y

P(4, 2)

2

x

12345

(b)slope = m =x 2

x 4

=x 2

( x + 2)(x 2)

=1, x 4

x + 2

x = 1: m =1= 1

1 + 2

3

x = 3: m =1 0.2679

3 + 2

x = 5: m =1 0.2361

5 + 2

(c)At P(4, 2) the slope is1=1= 0.25.

4 +24

You can improve your approximation of the slope at x = 4 by considering x-values very close to 4.

7. f (x ) = 6x x2

(a)y

10

8P

6

2x

248

(b)slope = m = (6x x 2 ) 8= ( x 2)( 4 x)

x 2x 2

= ( 4 x ), x 2

For x = 3, m = 4 3 = 1

For x = 2.5, m = 4 2.5 =1.5 =3

2

For x = 1.5, m = 4 1.5 = 2.5 = 52

(c) At P(2, 8), the slope is 2. You can improve your approximation by considering values of x close to 2.

8. Answers will vary. Sample answer:

The instantaneous rate of change of an automobiles position is the velocity of the automobile, and can be determined by the speedometer.

2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.81

82Chapter 2Limits and Their Properties

9. (a) Area 5 + 52 + 53 + 54 10.417

Area 12 ( 5 + 1.55 + 52 + 2.55 + 53 + 3.55 + 54 + 4.55 ) 9.145

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

10. (a) D1 =( 5 1) 2 + (1 5)2 =16 + 16 5.662( 2 )2( 23 )2( 34 )2( 4)2

(b) D =1 +5+1 +55+1 +55+1 +5 1

2.693 + 1.302 + 1.083 + 1.031 6.11

(c) Increase the number of line segments.

Section 2.2Finding Limits Graphically and Numerically

1.

x3.93.993.9994.0014.014.1

f (x)0.20410.20040.20000.20000.19960.1961

x 4 0.20001

limActual limit is.

2 3x 45

x 4 x

2.

x0.10.010.00100.0010.010.1

f (x)0.51320.50130.5001?0.49990.49880.4881

x + 1 1 0.50001

limActual limit is.

x2

x 0

3.

x0.10.010.0010.0010.010.1

f (x)0.99830.999981.00001.00000.999980.9983

limsin x 1.0000( Actual limit is 1.) ( Make sure you use radian mode.)

x 0x

4.

x0.10.010.0010.0010.010.1

f (x)0.05000.00500.00050.00050.00500.0500

limcos x 1 0.0000( Actual limit is 0.) ( Make sure you use radian mode.)

x 0x

5.

x0.10.010.0010.0010.010.1

f (x)0.95160.99500.99951.00051.00501.0517

limex 1 1.0000( Actual limit is 1.)

x 0x

6.

x0.10.010.0010.0010.010.1

f (x)1.05361.00501.00050.99950.99500.9531

limln( x + 1) 1.0000( Actual limit is 1.)

x 0x


Section 2.2 Finding Limits Graphically and Numerically83

7.

x0.90.990.9991.0011.011.1

f (x)0.25640.25060.25010.24990.24940.2439

x 2 0.25001

limActual limit is.

x2+ x 64

x 1

8.

x4.14.014.00143.9993.993.9

f (x)1.11111.01011.0010?0.99900.99010.9091

limx + 4 1.0000( Actual limit is 1.)

x2 + 9x +20

x 4

9.

x0.90.990.9991.0011.011.1

f (x)0.73400.67330.66730.66600.66000.6015

x4 12

lim0.6666Actual limit is.

x6 1

x 13

10.

x3.13.013.00132.9992.992.9

f (x)27.9127.090127.0090?26.991026.910126.11

limx3+ 27 27.0000( Actual limit is 27.)

x + 3

x 3

11.

x6.16.016.00165.9995.995.9

f (x)0.12480.12500.1250?0.12500.12500.1252

10 x 41

lim 0.1250Actual limit is .

x + 68

x 6

12.

x1.91.991.99922.0012.012.1

f (x)0.11490.1150.1111?0.11110.11070.1075

x (x + 1) 2 31


x 29

x 2

13.

x0.10.010.0010.0010.010.1

f (x)1.98671.99992.00002.00001.99991.9867

limsin 2x2.0000( Actual limit is 2.) ( Make sure you use radian mode.)

x

x 0

14.

x0.10.010.0010.0010.010.1

f (x)0.49500.50000.50000.50000.50000.4950

tan x1


tan 2x2

x 0



15.

x1.91.991.9992.0012.012.1

f (x)0.51290.50130.50010.49990.49880.4879

ln x ln 21


x 22

x 2

16.

x0.10.010.0010.0010.010.1

f (x)3.9998244000.00018

lim4does not exist.

1 +e1 x

x 0

17.lim(4 x)= 125. (a)f (1) exists. The black dot at (1, 2) indicates that

x 3f (1)

18.lim sec x= 1= 2.

(x) does not exist. As x approaches 1 from the

x 0(b)limf

f (x)(4 x)x 1

19.lim= lim= 2left,f (x) approaches 3.5, whereas as x approaches 1

x 2x 2from the right,f (x) approaches 1.

20. lim f ( x ) = lim(x2+ 3= 4(c)f (4) does not exist. The hollow circle at

x 1x 1)(4, 2) indicates that f is not defined at 4.

21.limx 2does not exist.(d)lim f(x) exists. As x approaches 4, f (x) approaches

x 2

x 2x 4

x 22: lim f (x)= 2.

For values of x to the left of 2,= 1, whereasx 4

( x 2)26. (a)f (2) does not exist. The vertical dotted line

x 2

for values of x to the right of 2,= 1.indicates that fis not defined at 2.

( x 2)

(b)limf (x) does not exist. As x approaches 2, the

4x 2()

22.limdoes not exist. The function approachesvalues of fxdo not approach a specific number.

2 +e1 x

x 0

2 from the left side of 0 by it approaches 0 from the left(c)f (0)exists. The black dot at ( 0, 4) indicates that

side of 0.f (0)= 4.

23. lim cos(1 x) does not exist because the function(d)lim f(x) does not exist. As x approaches 0 from the

x 0

oscillates between 1 and 1 as x approaches 0.x 0

left,f (x) approaches1, whereas as x approaches 0

24.lim tan x does not exist because the function increases2

from the right, f (x) approaches 4.

x 2

without bound as x approaches(e)f (2) does not exist. The hollow circle at

2 from the left and(2,1) indicates that f ( 2) is not defined.

decreases without bound as x approachesfrom2

2(f )lim f(x) exists. As x approaches 2, f ( x ) approaches

the right.

x 2

1: lim f (x) =1.

22

x 2

(g)f (4) exists. The black dot at ( 4, 2) indicates that

f (4) = 2.

(h) lim f(x) does not exist. As x approaches 4, the

x 4

values of f (x) do not approach a specific number.


Section 2.2 Finding Limits Graphically and Numerically85

27.y32. You needf ( x ) 1=1 1= 2 x < 0.01.

6x 1x 1

5Let =1. If 0 0.

By the Intermediate Value Theorem, g(c) = 0 for at

least one value of c between 0 and 1. Using a graphing utility to zoom in on the graph of g(x), you find that

x0.49. Using the root feature, you find that

x0.4918.

99.f (x) = x2 + x 1

f is continuous on [0, 5].

f (0) = 1 and f (5) = 29 1 < 11 < 29

The Intermediate Value Theorem applies.

x2 + x 1= 11

x2 + x 12= 0

( x + 4)( x 3)= 0

x = 4 or x= 3

c = 3( x= 4 is not in the interval.)

So, f ( 3)= 11.

100.f (x) = x2 6x + 8

f is continuous on [0, 3]. f (0) = 8 and f (3) = 11 < 0 < 8


x2 6x + 8 = 0

( x 2)( x 4) = 0

x = 2 or x = 4

c = 2 ( x = 4 is not in the interval.)

So, f ( 2) = 0.

Section 2.4Continuity and One-Sided Limits113

101.f (x) = x3 x2 + x 2

f is continuous on [0, 3].

f (0) = 2 and f (3) = 19 2 < 4 < 19The Intermediate Value Theorem applies.

x3 x2 + x 2 = 4 x3 x2 + x 6 = 0 ( x 2)( x2 + x + 3) = 0

x = 2

( x2 + x + 3 has no real solution.)

c = 2

So, f (2) = 4.

102. f ( x ) = x2 + xx 1

f is continuous on5, 4 . The nonremovable

2

discontinuity, x = 1, lies outside the interval.

535and f ( 4)20

f==

63

2

356 < 6 < 203


x2 + x=6

x 1

x2 + x=6x 6

x2 5x + 6=0

( x 2)( x 3) = 0

x = 2 or x= 3

c = 3 ( x= 2 is not in the interval.)

So, f (3)= 6.

103. (a) The limit does not exist at x = c.

(b) The function is not defined at x = c.

(c) The limit exists at x = c, but it is not equal to the value of the function at x = c.

(d) The limit does not exist at x = c.

104. Answers will vary. Sample answer:

y

5

4

3

2

1

x

2 113 4 5 6 7

2

3

The function is not continuous at x = 3 becauselim f (x) = 1 0 = lim f ( x ).x 3+x 3


114 Chapter 2 Limits and Their Properties

105. If f and g are continuous for all real x, then so is f + g (Theorem 2.11, part 2). However, f g might not be

continuous if g (x) = 0. For example, let f (x ) = x and

g(x)= x2 1. Then f and g are continuous for all real

x, butf g is not continuous at x = 1.

106. A discontinuity at c is removable if the function f can be made continuous at c by appropriately defining (or redefining) f (c). Otherwise, the discontinuity is nonremovable.(a) f ( x ) = x 4x 4

(b) f ( x ) = sin( x + 4)x + 4

1,x 4

4 < x < 4

(c) f ( x ) =0,

x= 4

1,

x< 4

0,

x= 4 is nonremovable,x= 4 is removable

y

4

3

2

1

6 4 2 1x

246

2

107. 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.

108. True. If f ( x ) = g ( x ), x c, then

lim f ( x ) = lim g( x ) (if they exist) and at least one of

x cx c

these limits then does not equal the corresponding function value at x = c.

109. False. A rational function can be written asP ( x )Q ( x ) where P and Q are polynomials of degree m

and n, respectively. It can have, at most, n discontinuities.

110. False. f (1) is not defined and lim f ( x ) does not exist.

x 1

111. The functions agree for integer values of x:

g(x) = 3 x = 3 ( x) = 3 + x

f ( x ) = 3 + x = 3 + xfor x an integer

However, for non-integer values of x, the functions differ by 1.

f (x ) = 3 +x= g (x ) 1 = 2 x .

For example,

f (12 ) = 3 + 0 = 3, g(12 ) = 3 ( 1) = 4.

112. lim f (t)28

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 is now about 56 oz.

0.40,0 < t 10

113. C ( t ) =+ 0.05 t 9 ,t> 10, t not an integer

0.40

0.40+ 0.05( t 10),t> 10, t an integer

C

0.7

0.6

0.50.4

0.3

0.2

0.1

t246810 12 14

There is a nonremovable discontinuity at each integer greater than or equal to 10.

Note: You could also express C as

0.40,0 < t 10

C ( t )=

0.05 10 t ,t > 10

0.40

114. N ( t )=t + 2 t

25 2

2

t011.8233.8

N (t)5025550255

Discontinuous at every positive even integer. The company replenishes its inventory every two months.

N

50

of units40

30

Number20

10

t

24681012

Time (in months)


115. Let s( t ) be the position function for the run up to the campsite. s ( 0) = 0 (t = 0 corresponds to 8:00 A.M., s( 20) = k (distance to campsite)). Let r ( t ) be the

position function for the run back down the mountain: r ( 0) = k , r(10) = 0. Let f (t ) = s (t ) r (t).

When t= 0 (8:00 A.M.),

f (0) = s (0) r (0) = 0 k < 0.

When t= 10 (8:00 A.M.), f (10) = s (10) r(10) > 0.

Becausef (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 s ( t ) r ( t ) = 0, which gives us

s(t )= r (t). Therefore, at some time t, where

0 t 10, the position functions for the run up and the run down are equal.

Section 2.4Continuity and One-Sided Limits 115

1,ifx< 0

120. sgn( x )=ifx= 0

0,

ifx> 0

1,

(a) lim sgn(x) = 1x 0

(b) lim sgn(x) = 1x 0+

(c) limsgn(x) does not exist.x 0

y

4

3

2

1

x 4 3 2 11 2 3 4

2

3

4

116. Let V =4 r3 be the volume of a sphere with radius r.121. (a)S

3= 500 523.6 and60

V is continuous on [5, 8]. V ( 5)50

3

40

V (8) =204830

2144.7. Because20

3

10

523.6 < 1500 < 2144.7, the Intermediate Valuet

Theorem guarantees that there is at least one value r51015202530

(b)There appears to be a limiting speed and a possible

between 5 and 8 such that V ( r )= 1500. (In fact,

r 7.1012.)cause is air resistance.

117. Suppose there exists x1 in [a , b] such that122. (a) f ( x ) =0,0x< b

b 0 and there exists x2 in [a , b] such thatb,

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). So, 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].

118. Let c be any real number. Then lim f (x) does not exist

x c

because there are both rational and irrational numbers arbitrarily close to c. Therefore, f is not continuous at c.

119. If x = 0, then f ( 0) = 0 and lim f (x) = 0. So, f is

x 0

y

2b

bx

b2b

NOT continuous at x = b.

x,0 x b

2

(b)g( x ) =

x

b , b < x 2b

2

continuous at x= 0.

If x 0, thenlim f ( t )=0 for x rational, whereas

t x

lim f ( t ) = limkt = kx0 for x irrational. So, f is not

t xt x

continuous for all x 0.

y

2b

b

b2b

x

Continuous on [0, 2b].



x2,x c

123. f ( x ) =1

x,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 obtainc = 1 1 + 4=1 5.

22

125. f ( x ) =x + c 2 c, c> 0

x)

x + c2 0 x c2 and

Domain:x 0, c 2, 0

124. Let y be a real number. If y = 0, then x = 0. If

y> 0, then let 0 < x0< 2 such that

M= tan x0 > y (thisis possible since the tangent

function increases without bound on [0, 2)). By the Intermediate Value Theorem, f ( x ) = tan x is continuous on [0, x0] and 0 < y < M , which implies that there exists x between 0 and x0 such that

tan x = y. The argument is similar if y < 0.

(0, )

limx + c 2 c=limx + c 2 cx + c 2+ c

xxx + c 2+ c

x 0x 0

Define f ( 0)= 1 ( 2c) to make f continuous at x = 0.

126. 1.f ( c ) is defined.

2.lim f ( x )=limf ( c+ x ) =f ( c ) exists.

x cx 0

[Let x = c + x. As x c, x 0]

3.lim f ( x )=f ( c ).

x c

Therefore, fis continuous at x= c.

=lim(x + c 2) c2= lim

c 2 + c

x 0 xx +x0

127. h (x )= x x

15

3

3

1=1

x + c 2 + c2c

3

h has nonremovable discontinuities at x = 1, 2, 3, .

128.(a)Definef ( x ) =f 2 ( x ) f1( x ). Becausef1 andf2 are continuous on [a , b], so is f.

f (a ) =f 2 (a ) f1(a) > 0 and f (b ) =f 2 (b ) f1(b) < 0

By the Intermediate Value Theorem, there exists c in [a , b]such that f (c) = 0.

f (c ) = f 2 (c ) f1 (c ) = 0 f1 (c ) = f 2 (c)

(b)Let f1( x ) = x and f2 (x ) = cos x, continuous on [0, 2],f1 (0) f2 ( 2).

So by part (a), there exists c in [0, 2] such that c = cos(c).

Using a graphing utility, c 0.739.

129.The statement is true.

Ify 0 and y 1, then y(y 1) 0 x2 , as desired. So assumey > 1. There are now two cases.

Case l: If x y 12 , then 2x + 1 2 y and y(y 1) = y(y + 1) 2 y ( x + 1)2 2 y

x2 + 2x + 1 2 y

x2 + 2 y 2 y

x2

In both cases, y( y 1) x2.

Case 2: If x y 12

x2 ( y 12 )2

= y2 y + 14 > y2 y= y( y 1)



130. P (1) =(0 2)= P( 0 )2 + 1 = 1

P+ 1

P ( 2 ) =()= P(1) 2 + 1 = 2

P 12+ 1

P ( 5) =(2 2)= P( 2 )2 + 1 = 5

P+ 1

Continuing this pattern, you see that P (x ) = x for infinitely many values of x.

So, the finite degree polynomial must be constant: P (x ) = x for all x.

Section 2.5Infinite Limits

1.lim2x=

x2 4

x 2+

lim2x=

x2 4

x 2

2.lim1=

x 2+ x + 2

lim1=

x 2 x + 2

3.limtan x=

4

x 2+

limtan x=

x 24

4.limsec x=

x 2+4

limsec x=

x 24

5.f ( x )=1

x 4

As x approaches 4 from the left, x 4 is a small negative number. So,lim f (x) = x 4

As x approaches 4 from the right, x 4 is a small positive number. So,lim f ( x ) = x 4+

9.f ( x ) =1

x2 9

6.f ( x ) =1

x 4

As x approaches 4 from the left, x 4 is a small negative number. So,lim f (x) = .x 4

As x approaches 4 from the right, x 4 is a small positive number. So,lim f (x) = .x 4+

7. f ( x ) =1

( x 4)2

As x approaches 4 from the left or right, ( x 4)2 is a small positive number. So,limf (x) = lim f (x) = .

x 4 +x 4

8. f ( x )=1

( x 4)2

As x approaches 4 from the left or right, ( x 4)2 is a small positive number. So,lim f (x) =lim f (x) = .

x 4 x 4+

x3.53.13.013.0012.9992.992.92.5

f ( x )0.3081.63916.64166.6166.716.691.6950.364

lim f (x)= 2

x 3=

lim+ f ( x ) 66

x 3

2


118Chapter 2 Limits and Their Properties

10. f ( x )=x

x2 9

x3.53.13.013.0012.9992.992.92.5

f (x)1.0775.08250.08500.1499.949.924.9150.9091

limf (x)= 2

x 3

lim+ f ( x )= 66

x 3

2

11. f ( x )=x2

x2 9

x3.53.13.013.0012.9992.992.92.5

f ( x )3.76915.75150.815011499149.314.252.273

limf (x)= 4

x 3

lim+ f ( x )= 66

x 3

4

12.f ( x )= cot x

3

x3.53.13.013.0012.9992.992.92.5

f (x)1.73219.51495.49954.9954.995.499.5141.7321

limf (x)= 4

x 3

limf ( x )= 66

x 3+

4

13. f ( x) = 1

x 2

lim1= = lim1

x 0 + x2x 0 x2

Therefore,x= 0 is a vertical asymptote.

14. f (x)=2

(x 3)3

lim2=

( x3)3

x 3

lim2=

( x3)3

x 3+


15.f (x)=x2=x2

x2 4(x + 2)(x 2)

limx2= andlimx2=

44

x 2 x2x 2+ x2


limx2= andlimx2=

44

x 2 x2x 2+ x2


16.f ( x)=3x

x2+ 9

No vertical asymptotes because the denominator is never zero.


17. g(t) =t 1

t2+ 1

No vertical asymptotes because the denominator

is never zero.

18.h(s) =3s + 4=3s + 4

s2 16(s 4)(s + 4)

lim3s + 4= and lim3s + 4=

s 4 s2 16s 4+ s2 16

Therefore,s= 4 is a vertical asymptote.

lim3s+ 4= and lim3s+ 4=

s 4 s2 16s 4+ s2 16

Therefore,s= 4 is a vertical asymptote.

19.f (x) =3=3

x2+ x 2(x + 2)(x 1)

lim3= andlim3=

+ x 2x 2

x 2 x2x 2+ x2 +


lim3= andlim3=

+ x 22 +x 2

x 1 x2x 1+ x


20.g ( x)=x 3 8=( x 2)( x 2+ 2 x + 4)

x 2x 2

= x 2 + 2 x + 4, x 2lim g ( x) = 4 + 4 + 4 = 12

x 2

There are no vertical asymptotes. The graph has a hole at x = 2.

21. f ( x ) =x2 2x 15

x3 5x2 + x 5

=( x 5)( x + 3)

( x 5) x2+ 1

()

=x + 3, x 5

x2+ 1

lim f (x)=5 + 3=15

52 + 1

x 526

There are no vertical asymptotes. The graph has a hole at x = 5.

Section 2.5Infinite Limits 119

22. h ( x )=x2 9

x 3 + 3 x 2 x 3

=( x 3)( x + 3)

( x 1)( x + 1)( x + 3)

=x 3, x 3

(x + 1)(x 1)

lim h (x )= andlim+ h (x) =

x 1x 1


lim h (x )= and lim h (x) =

x 1x 1+


lim h(x)= 3 3= 3

( 3 + 1)( 3 1)4

x 3

Therefore, the graph has a hole at x= 3.

23. f ( x )=e 2x

x 1

limf (x)= andlim=

x 1x 1+


24. g(x) = xe 2x

The function is continuous for all x. Therefore, there are no vertical asymptotes.

ln t2+ 1

25. h( t )=()

t+ 2

limh(t)= andlim=

t 2 t 2 +

Therefore, t= 2 is a vertical asymptote.

26. f (z )= ln(z2 4)= ln ( z+ 2)(z 2)

= ln( z + 2) + ln( z 2)

The function is undefined for 2 < z < 2.

Therefore, the graph has holes atz = 2.

27. f ( x )=1

ex 1

limf (x)= andlimf (x)=

x 0 x 0+


28. f (x )= ln(x + 3)

limf (x)=

x 3




29.f ( x ) = csc x =1

sin x

Let n be any integer.

lim f (x) = or x n

Therefore, the graph has vertical asymptotes at x = n.

30. f ( x )= tan x =sin x

cos x

cos x = 0 for x = 2n + 1, where n is an integer.

2

limf (x) = or

x 2 n +1

2

Therefore, the graph has vertical asymptotes at

x =2 n + 1.

2

34.limx2 2x 8=

x +1

x 1

limx2 2x 8=

x +1

x 1+

Vertical asymptote atx= 1

35.limx2+ 1=

x+ 1

x 1+

limx2+ 1=

x+ 1

x 1

Vertical asymptote atx= 1

ln(x2+ 1

36.lim)=

x + 1

x 1+

ln(x2+1

lim)=

x + 1

x 1

4

108

8

8

33

8

3

53

31. s( t )=t

sin t

sin t= 0 for t = n , where n is an integer.

lim s(t) = or (for n 0)

t n

Therefore, the graph has vertical asymptotes at t = n , for n 0.

lim s(t)= 1

t 0

Therefore, the graph has a hole at t = 0.

32. g( )=tan =sin

cos

cos = 0 for = + n , where n is an integer.

2

limg( ) = or 2 + n

Therefore, the graph has vertical asymptotes at

= 2 + n .

lim g( ) = 1

0

Therefore, the graph has a hole at = 0.

33. limx2 1=lim ( x 1)= 2

x 1x + 1x 1

Removable discontinuity atx = 1

2

33

5

Vertical asymptote at x= 1 5

37.lim1=

1

x 1+ x +

38.lim1=

( x 1)2

x 1

39.limx=

2

x 2+ x

40.limx2=4=1

4 + 42

x 2 x2 + 4

41.limx + 3=limx + 3

+ x 6)+ 3)( x 2)

x 3 ( x 2x 3 ( x

=lim1= 1

25

x 3 x

42.lim6x2 + x 1=lim( 3x 1)(2 x + 1)

x (1 2)+ 4x2 4x 3x (1 2)+ ( 2x 3)( 2x + 1)

=lim3x 1=5

8

x (1 2)+ 2x 3

43.+1=

lim1

x

x 0

44.1=

lim6

x3

x 0+

2

45.limx2+=

x +

x 44

46.limx+ cot x=

2

x 3+3


47.lim2=

x 0+ sin x

48.lim2=

x ( 2)+ cos x

49.limex=

( x8)3

x 8

50.lim+ ln(x2 16) =

x 4


57. f ( x )=10.3

x2 25

limf ( x )= 88

x 5

= sec x 0.3

58. f ( x )6

f ( x )8

lim= 99

x 4+

6

51.cos x

limln= lncos= ln 0 =

x ( 2)2

52.lime 0.5x sin x= 1(0) = 0

x 0+

59. 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 symbollim f (x) =

x c

53.limxsec x=limx=

cos x

x (1 2 )x (1 2)

54.limx2tan x=

x (1 2)+

55.f ( x )=x2 + x + 1=x2 + x + 1

()

x3 12++

( x1)xx1

lim f ( x )=lim1=

1

x 1+x 1+ x

says how the limit fails to exist.

60. The line x = c is a vertical asymptote if the graph of f approaches as x approaches c.

61. One answer is

f ( x ) =x 3=x 3

.

( x 6)( x + 2)x 2 4x 12

62. No. For example, f ( x )=1has no vertical

x2 + 1

3asymptote.

4563.y

3

32

( x 1)x+ x + 11

56.f ( x )=x3 1=(2) 2 1x

13

x 2+x + 1x 2 +x + 1 1

lim f ( x ) =lim( x 1)= 0 2

x 1x 1

4m0

64. m =

881 ( v 2c2 )

m0

limm =lim=

1 ( v)

4v c v c2c2

65. (a)

x10.50.20.10.010.0010.0001

f (x)0.15850.04110.00670.0017 0 0 0

0.5lim x sin x= 0

x 0+x

1.51.5

0.25


Total time 2d

Total distance122 Chapter 2Limits and Their Properties

(b)

x10.50.20.10.010.0010.0001

f (x)0.15850.08230.03330.01670.0017 0 0

0.25lim x sin x= 0

x 0+x2

1.51.5

0.25

(c)

x10.50.20.10.010.0010.0001

f (x)0.15850.16460.16630.16660.16670.16670.1667

0.25lim x sin x= 0.1667 (1 6)

x 0+x3

1.51.5

0.25

(d)

x10.50.20.10.010.0010.0001

f (x)0.15850.32920.83171.665816.67166.71667.0

1.5lim x sin x= or n > 3, limx sin x = .

x 0+x4x 0+xn

1.51.5

1.5

66.lim P = V 0+

As the volume of the gas decreases, the pressure increases.

67. (a)r=2(7)=7ft sec

625 4912

(b)r=2(15)= 3ft sec

625 2252

(c)lim2x=

625

x 25x2

68. (a)Average speed =

50 = ( d x) + ( d y)

50=2xy

y + x

50 y + 50x= 2xy

50x= 2xy 50 y

50x= 2 y( x 25)

25x= y

x 25

Domain: x> 25

(b)

x30405060

y15066.6675042.857

(c) lim25x=

x 25

x 25+

As x gets close to 25 mi/h, y becomes larger and larger.



69. (a)A = 12bh 12r2 = 12(10)(10 tan ) 12(10)2 = 50 tan 50

Domain:0,

2

(b)

0.30.60.91.21.5

f ( )0.474.2118.068.6630.1

100

01.50

(c) lim A = 2

70. (a) Because the circumference of the motor is half that of the saw arbor, the saw makes 17002 = 850 revolutions per minute.

(b) The direction of rotation is reversed.

(c)2(20 cot )+ 2(10 cot ): straight sections. The angle subtended in each circle is 2= + 2.

2

2

So, the length of the belt around the pulleys is 20( + 2 ) + 10( + 2 ) = 30( + 2).

Total length= 60 cot + 30( + 2)

Domain: 0,

2

(d)

0.30.60.91.21.5

L306.2217.9195.9189.6188.5

(e)

450

0p

2

0

(f ) limL = 60 188.5

( 2)

(All the belts are around pulleys.)

(g) lim L = 0+

71. False. For instance, let

f ( x )=x2 1or

x 1

g ( x )=x

.

x2+ 1

72. True

73. False. The graphs of y = tan x , y = cot x , y= sec x

and y= csc x have vertical asymptotes.

74. False. Let

1,x 0

f ( x )

= x

x= 0.

3,

The graph offhas a vertical asymptote at x= 0, but

f (0)= 3.


124 Chapter 2 Limits and Their Properties

75. Let f ( x ) = x12 and g( x ) = x14 , and c = 0.

1111x2 1

lim= and lim= , butlim= lim= 0.

242x4x4

x 0 xx 0 xx 0 xx 0

76. Givenlim f(x)= and lim g(x)= L:

x cx c

(1)Difference:

Let h (x ) = g (x). Then lim h(x)= L, and lim f(x ) g (x ) = lim f (x ) + h (x) = , by the Sum Property.

x cx cx c

(2)Product:

If L > 0, then for = L 2 > 0 there exists 1> 0 such that g(x) L< L 2 whenever 0 0, there exists 2 > 0 such that

f ( x ) > M ( 2 L) wheneverx c

x c< 2. Let be the smaller of 1 and 2. Then for 0 M (2 L )(L 2)= M . Therefore lim f (x)g(x)= . The proof is similar forL < 0.

x c

(3)Quotient: Let > 0 be given.

There exists 1> 0 such thatf ( x )> 3L 2 whenever 0 0 such that

g( x ) L< L 2 whenever 0 0 such that

x cf ( x )f ( x ) =1> M whenever 3 < x< 3 + .

1( x ) exists and equals L.x 3

78. Given lim= 0. Suppose lim f1

f( x )

x cx c

lim 1Equivalently,x 3 < M whenever

11x 3< , x> 3.

Then, lim=x c== 0.

f( x )lim f( x )L

x c1

x cSo take =. Then for x> 3 and

This is not possible. So,lim f (x) does not exist.M

x 3< ,1> 1 = M and sof ( x ) > M .

x c

1x 381

80. f ( x ) =is defined for all x < 5. LetN< 0 be given. You need >0 such thatf ( x ) =< N whenever

x 5x 5

5 < x< 5. Equivalently,x 5>1wheneverx 5< , x < 5. Equivalently,1< 1whenever

x 5< , x1NN < 0. Forx 5x 5N

< 5. So take = . Note that > 0 because< and

N

x < 5,1>1= N, and1= 1< N.

x 5x 5x 5


Review Exercises for Chapter 2125

Review Exercises for Chapter 2

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.

11

99

1

2.Precalculus. L =( 9 1) 2 + ( 3 1)2 8.25

3. f ( x ) =x 3

x 2 7 x + 12

x2.92.992.99933.0013.013.1

f (x)0.90910.99010.9990?1.00101.01011.1111

lim f ( x ) 1.0000 (Actual limit is 1.)

x 3

6

612

6

4.f ( x ) =x + 4 2

x

x0.10.010.00100.0010.010.1

f (x)0.25160.25020.2500?0.25000.24980.2485

x 0 0.2500(Actual limit is4 )

lim f (x)1.

0.5

55

0

5. h( x )=4x x2=x(4 x)= 4 x, x 06. f ( t ) =ln(t + 2)

x

xt

(a)lim h( x )= 4 0 = 4(a)limf (t)does not exist because lim f (t) =

x 0t 0f (t) = .t 0

(b)lim h( x )= 4 ( 1) = 5andlim

x 1t 0 +

(b)limf ( t )=ln 1=0

t 11



7.lim( x + 4) = 1 + 4 = 5

x 1

Let > 0 be given. Choose = . Then for 0 < x 1 < = , you have x 1 <

x + 4) 5 < f ( x ) L < .

8. limx =9 = 3

x 9

Let > 0 be given. You need

x 3< x + 3x 3< x + 3

Assuming 4 < x < 16, you can choose = 5.

So, for 0 0 such that if 0

Education

Solution manual for calculus of a single variable early transcendental functions 6th edition sample