Thomas Calculus Early Transcendentals 14th Edition Hass Solutions Manual by a394695247

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CHAPTER 2 2.1

LIMITS AND CONTINUITY

RATES OF CHANGE AND TANGENTS TO CURVES

1. (a)

f x



f (3)  f (2) 3 2

2. (a)

g x



g (3)  g (1) 3 1

3. (a)

h t



g t



4. (a)

 2819  19

(b)

f x



f (1)  f ( 1) 1( 1)

3  ( 1) 2

2

(b)

g x



g (4)  g ( 2) 4  ( 2)

 34 h 4   11   4

(b)

h t



g t



3 4







g ( )  g (0)  0



(2 1) (2 1)  0

  2

R 



R (2)  R (0) 20

 812 1  321  1

P 



P (2)  P (1) 21



7. (a) (b) 8. (a) (b)

(816 10)  (1 4  5) 1

y x



((2 h )2 5) (22 5) h

y x



(7(2 h )2 ) (7 22 ) h

(b)

 22 0  1  8 6 8  0

 2 h 6   0   2



g ( )  g (  )  (  )



 3 3

(2 1) (2 1) 2

0

 22  0

2 2  4 4h hh 51  4h h h  4  h. As h  0, 4  h  4  at P (2, 1) the slope is 4. y  ( 1)  4( x  2)  y  1  4 x  8  y  4 x  9

2 2  744hh h 3  4 hhh  4  h. As h  0,  4  h  4  at P (2, 3) the slope

is 4. y  3  ( 4)( x  2)  y  3  4 x  8  y  4 x  11 y

((2  h )2  2(2  h ) 3)  (22  2(2) 3)

y

((1 h ) 2  4(1 h )) (12  4(1))

y

(2  h )3  23

y

2 (1 h )3 (2 13 )

4  4 h  h 2  4  2 h 3( 3)

 9. (a) x  h h P(2,  3) the slope is 2. (b) y  (3)  2( x  2)  y  3  2 x  4  y  2 x  7. 1 2 h  h 2  4  4 h  ( 3)

2  2h h h  2  h. As h  0, 2  h  2  at

10. (a) x    h h 2h  h  2. As h  0, h  2  2  at P (1,  3) the h h slope is 2. (b) y  ( 3)  ( 2)( x  1)  y  3  2 x  2  y  2 x  1. 2

11. (a) x   812h  4hh  h 8  12 h  4hh  h  12  4h  h 2 . As h  0, 12  4h  h 2  12,  at P (2, 8) h the slope is 12. (b) y  8  12( x  2)  y  8  12 x  24  y  12 x  16. 2

12. (a) x   213h h3h  h 1  3h 3hh  h  3  3h  h 2 . As h  0, 3 3h  h 2  3,  at h P(1, 1) the slope is 3. (b) y  1  ( 3)( x  1)  y  1  3 x  3  y  3x  4.

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Chapter 2 Limits and Continuity 3

2 3 y (1 h ) 12(1 h ) (1 12(1)) 13h 3h  h 1212 h ( 11) 13. (a) x    9h 3hh  h  9  3h  h 2 . h h As h  0, 9  3h  h 2  9  at P(1,  11) the slope is 9. (b) y  (11)  (9)( x  1)  y  11  9 x  9  y  9 x  2.

(2  h )3 3(2  h ) 2  4 (23 3(2)2  4)

y

12 h 3h 14. (a) x   812h  6h  h 12 h h 2 As h  0, 3h  h  0  at P (2, 0) the slope is 0. (b) y  0  0( x  2)  y  0.

15. (a)

y x



1  1 2  h 2

As h  0, (b)

16. (a)

 

17. (a)

1 2( 2  h )

y x



(4 h )  4 2 (4 h ) 2 4

h 1 2 h

18. (a)

y x



 41 ,  at P 2, 21 the slope is 41 .





4  h  2( 2  h ) 1  42hh  12  1h   h  21h  21 h . 2  h 1, 2



 at P (4,  2) the slope is

1. 2

(4 h )  4 1  4 hh  4  4hh  2  4 h  2   . 4  h  2 h ( 4  h  2) 4 h  2

1 4 h  2



1 4 2

 14 ,  at P (4, 2) the slope is

1. 4

y  2  14 ( x  4)  y  2  14 x  1  y  14 x  1 y x



7 ( 2  h )  7 ( 2) h

As h  0, (b)

2 3  3h h h  3h  h 2 .

y  (2)  12 ( x  4)  y  2  12 x  2  y  12 x  4

As h  0, (b)



 40

y  21  41 ( x  (2))  y  12  41 x  12  y  41 x  1

As h  0, (b)

2  ( 2  h )  2( 2 h )  1h  2( 21 h) .

1 9 h 3



9  h 3 h



9  h 3 9  h  3  h 9 h 3



(9  h ) 9 h ( 9  h 3)



1 . 9 h 3

 1  61 ,  at P (2, 3) the slope is 61 . 9 3

y  3  61 ( x  (2))  y  3  61 x  13  y  61 x  83

19. (a)

Q Q1 (10, 225)

Q2 (14,375) Q3 (16.5, 475) Q4 (18,550)

Slope of PQ  650  225 2010 650 375 20 14 650  475 20 16.5 650 550 2018

p t

 42.5 m/sec  45.83 m/sec  50.00 m/sec  50.00 m/sec

(b) At t  20, the sportscar was traveling approximately 50 m/sec or 180 km/h.

20. (a)

Q Q1 (5, 20) Q2 (7, 39)

Q3 (8.5,58) Q4 (9.5, 72)

p

Slope of PQ  t 80  20  12 m/sec 10 5

80 39  13.7 m/sec 10 7 80 58  14.7 m/sec 10 8.5 8072  16 m/sec 10 9.5

(b) Approximately 16 m/sec

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Section 2.1 Rates of Change and Tangents to Curves 21. (a)

Profit (1000s)

200 160 120 80 40 0

(b)

p t

2010 2011 2012 2013 2014 Ye ar



174 62 2014 2012

 112  56 thousand dollars per year 2 p

62 27 20122011

 35 thousand dollars per year.

p

11162  49 thousand dollars per year. The average rate of change from 2012 to 2013 is t  2013 2012 So, the rate at which profits were changing in 2012 is approximately 12 (35  49)  42 thousand dollars per year.

22. (a) F ( x)  ( x  2)/( x  2) x 1.2 1.1 F ( x) 4.0 3.4 F x F x F x



 

1.01 3.04

1.001 3.004

4.0  ( 3)  5.0; 1.2 1 3.04 ( 3)  4.04; 1.011 3.0004 ( 3)  4.0004; 1.00011

F x F x



g x



1.0001 3.0004

1 3

3.4 ( 3)  4.4; 1.11 3.004 ( 3)  4.004; 1.0011

(b) The rate of change of F ( x) at x  1 is 4. g

23. (a) x  g x



g (2)  g (1)  2211  0.414213 2 1 g (1 h )  g (1)  1hh 1 (1 h ) 1

g (1.5)  g (1) 1.51



1.5 1 0.5

 0.449489

(b) g ( x)  x 1 h



1 h



1  h  1 /h

1.1

1.01

1.001

1.0001

1.00001

1.000001

1.04880

1.004987

1.0004998

1.0000499

1.000005

1.0000005

0.4880

0.4987

0.4998

0.499

0.5

h 0

24. (a) i) ii)

1 h 1 h

 12 .

11 1 f (3)  f (2)  3 1 2  16   16 3 2 1 1 2 T f (T )  f (2)  TT  22  2TT  22T  2T2(TT 2) T 2

T  2T2(2   21T , T  2 T )

(b) T f (T ) ( f (T )  f (2))/(T  2)

2.1 2.01 2.001 0.476190 0.497512 0.499750 0.2381 0.2488 0.2500 (c) The table indicates the rate of change is 0.25 at t  2. (d) lim 21T   14 T 2

2.0001 0.4999750 0.2500

2.00001 0.499997 0.2500

 

NOTE: Answers will vary in Exercises 25 and 26. 0  15 mph; [1, 2.5]: s  25. (a) [0, 1]: st  15 t 10

20 15 2.51

30 20  10 mph  10 mph; [2.5, 3.5]: st  3.5  2.5 3

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2.000001 0.499999 0.2500

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Chapter 2 Limits and Continuity (b) At P

 12 , 7.5 : Since the portion of the graph from t  0 to t  1 is nearly linear, the instantaneous rate of

change will be almost the same as the average rate of change, thus the instantaneous speed at t  12 is 157.5  15 mi/hr. At P (2, 20): Since the portion of the graph from t  2 to t  2.5 is nearly linear, the 10.5  20  0 mi/hr. instantaneous rate of change will be nearly the same as the average rate of change, thus v  20 2.5 2 For values of t less than 2, we have Q Q1 (1, 15)

Q2 (1.5, 19) Q3 (1.9, 19.9)

Slope of PQ 

s t

15 20  5 mi/hr 1 2 19  20  2 mi/hr 1.5 2 19.9  20  1 mi/hr 1.9  2

Thus, it appears that the instantaneous speed at t  2 is 0 mi/hr. At P(3, 22): Q Q1 (4, 35) Q2 (3.5, 30) Q3 (3.1, 23)

Slope of PQ  35 22 4 3 30  22 3.53 23 22 3.13

s t

 13 mi/hr

Q1 (2, 20)

 16 mi/hr

Q2 (2.5, 20)

 10 mi/hr

Q3 (2.9, 21.6)

Slope of PQ 

s t

20 22  2 mi/hr 2 3 20 22  4 mi/hr 2.53 21.6 22  4 mi/hr 2.9 3

Thus, it appears that the instantaneous speed at t  3 is about 7 mi/hr. (c) It appears that the curve is increasing the fastest at t  3.5. Thus for P(3.5, 30) Slope of PQ  st Slope of PQ  Q Q Q1 (4, 35) Q2 (3.75, 34)

Q3 (3.6, 32)

3530  10 mi/hr 43.5 34 30  16 mi/hr 3.753.5 32 30  20 mi/hr 3.6 3.5

Q1 (3, 22) Q2 (3.25, 25) Q3 (3.4, 28)

s t

2230  16 mi/hr 33.5 2530  20 mi/hr 3.253.5 2830  20 mi/hr 3.4 3.5

Thus, it appears that the instantaneous speed at t  3.5 is about 20 mi/hr. gal

 1.67 day ; [0, 5]: At  26. (a) [0, 3]: At  10315 0

(b) At P(1, 14): Q Q1 (2, 12.2) Q2 (1.5, 13.2) Q3 (1.1, 13.85)

Slope of PQ 

A t

12.2 14  1.8 gal/day 21 13.2 14  1.6 gal/day 1.51 13.8514  1.5 gal/day 1.11

3.915 5 0

gal

 2.2 day ; [7, 10]: At  0101.4  0.5 day 7

Q Q1 (0, 15) Q2 (0.5, 14.6)

Q3 (0.9, 14.86)

Slope of PQ 

A t

1514  1 gal/day 0 1 14.6 14  1.2 gal/day 0.51 14.86 14  1.4 gal/day 0.9 1

Thus, it appears that the instantaneous rate of consumption at t  1 is about 1.45 gal/day. At P(4, 6): Slope of PQ  At Q Slope of PQ  A Q t

Q1 (5, 3.9)

Q2 (4.5, 4.8) Q3 (4.1, 5.7)

3.9 6 5 4 4.86 4.5 4 5.7 6 4.1 4

 2.1 gal/day

Q1 (3, 10)

 2.4 gal/day

Q2 (3.5, 7.8)

 3 gal/day

Q3 (3.9, 6.3)

10 6  4 gal/day 3 4 7.86  3.6 gal/day 3.5 4 6.36  3 gal/day 3.9  4

Thus, it appears that the instantaneous rate of consumption at t  1 is 3 gal/day. (solution continues on next page)

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Section 2.2 Limit of a Function and Limit Laws At P(8, 1): Q Q1 (9, 0.5)

Q2 (8.5, 0.7) Q3 (8.1, 0.95)

Slope of PQ 

A t

Q1 (7, 1.4)

0.51  0.5 gal/day 9 8 0.7 1  0.6 gal/day 8.58 0.951  0.5 gal/day 8.18

Q2 (7.5, 1.3) Q3 (7.9, 1.04)

Slope of PQ 

A t

1.4 1  0.6 gal/day 7 8 1.31  0.6 gal/day 7.58 1.04 1  0.6 gal/day 7.98

Thus, it appears that the instantaneous rate of consumption at t  1 is 0.55 gal/day. (c) It appears that the curve (the consumption) is decreasing the fastest at t  3.5. Thus for P(3.5, 7.8) Slope of PQ  st Q Slope of PQ  At Q 11.2 7.8   3.4 gal/day Q1 (2.5, 11.2) 4.8 7.8  3 gal/day Q1 (4.5, 4.8) 2.53.5 Q2 (4, 6)

Q3 (3.6, 7.4)

4.53.5 6 7.8  3.6 gal/day 4 3.5 7.4  7.8  4 gal/day 3.6 3.5

Q2 (3, 10)

Q3 (3.4, 8.2)

10 7.8  4.4 gal/day 33.5 8.2 7.8  4 gal/day 3.4 3.5

Thus, it appears that the rate of consumption at t  3.5 is about 4 gal/day. 2.2

LIMIT OF A FUNCTION AND LIMIT LAWS

1. (a) Does not exist. As x approaches 1 from the right, g ( x) approaches 0. As x approaches 1 from the left, g ( x) approaches 1. There is no single number L that all the values g ( x) get arbitrarily close to as x  1. (b) 1 (c) 0 (d) 0.5 2. (a) 0 (b) 1 (c) Does not exist. As t approaches 0 from the left, f (t ) approaches 1. As t approaches 0 from the right, f (t ) approaches 1. There is no single number L that f (t ) gets arbitrarily close to as t  0. (d) 1 3. (a) (d) (g) (j)

True False True True

(b) (e) (h) (k)

4. (a) False (d) True (g) False x x0 | x |

(b) False (e) True (h) True

 1 if x  0 and | xx |  xx   1 if x  0. As x approaches 0 from the left, | xx | approaches 1. As x approaches 0 from the right, | xx | approaches 1. There is no single number L that all the

5. lim

does not exist because | xx | 

True False False False

x x

function values get arbitrarily close to as x  0.

1 become increasingly large and negative. As x approaches 1 6. As x approaches 1 from the left, the values of x 1 from the right, the values become increasingly large and positive. There is no number L that all the function values get arbitrarily close to as x  1, so lim x11 does not exist. x 1

7. Nothing can be said about f ( x) because the existence of a limit as x  x0 does not depend on how the function is defined at x0 . In order for a limit to exist, f ( x) must be arbitrarily close to a single real number L when x is close enough to x0 . That is, the existence of a limit depends on the values of f ( x) for x near x0 , not on the definition of f ( x) at x0 itself. Copyright  2018 Pearson Education, Inc.

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Chapter 2 Limits and Continuity

8. Nothing can be said. In order for lim f ( x) to exist, f ( x) must close to a single value for x near 0 regardless of x 0

the value f (0) itself.

9. No, the definition does not require that f be defined at x  1 in order for a limiting value to exist there. If f (1) is defined, it can be any real number, so we can conclude nothing about f (1) from lim f ( x)  5. x 1

10. No, because the existence of a limit depends on the values of f ( x) when x is near 1, not on f (1) itself. If lim f ( x) exists, its value may be some number other than f (1)  5. We can conclude nothing about lim f ( x), x 1

x 1

whether it exists or what its value is if it does exist, from knowing the value of f (1) alone. 11.

lim ( x 2  13)  ( 3)2  13  9  13  4

x3

12. lim ( x 2  5 x  2)  (2)2  5(2)  2  4  10  2  4 x 2

13. lim 8(t  5)(t  7)  8(6  5)(6  7)  8 t 6

14.

15.

16.

17.

lim ( x3  2 x 2  4 x  8)  (2)3  2(2)2  4( 2)  8  8  8  8  8  16

x 2

lim 2 x 53 x2 11 x



2(2) 5 11(2)3



lim (8  3s )(2 s  1)  8  5

t 2/3

    

y2



2 y 2 y 5 y  6

20.

2 2 (2) 2 5(2)  6



 ( 2)  23  4



y 3

lim

z 4

22. lim

5h  4  2 h

h 0

x 5 2 x 5 x  25

23. lim

 

2  ( 2) 25   25 2



 24  16

z 2  10  42  10  16  10  6 3 3h 1 1

h 0



4 4 1  410  20 6 5

lim (5  y ) 4/3  [5  (3)]4/3  (8)4/3  (8)1/3

21. lim

24.

 23    2  23   1  (8  2)   43   1  (6)  13   2

lim 4 x (3x  4)2  4  12 3  12  4 x 1/2

18. lim

19.

 93  3



3 3(0) 1 1

 lim

h 0

 3  32 1 1

5 h  4  2 5h  4  2  h 5h  4  2

 lim

h 0 h

(5h  4)  4



5h  4  2



 lim

h 0 h



5h 5h  4  2



 lim

h0

5 5h 4  2



x 5  lim ( x 5)(  lim 1  1  1 x 5) x 5 x 5 55 10 x 5

lim 2 x 3 x 3 x  4 x 3

x 3  lim ( x 3)(  lim 1  1   12 x 1) x 3 x 1 31 x 3

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5 4 2

 54

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Section 2.2 Limit of a Function and Limit Laws 25.

2 ( x 5)( x  2) lim x x3x510  lim  lim ( x  2)  5  2  7 x 5 x 5 x 5

x 5

x 2 7 x 10 x2 x 2

26. lim

t 2 t  2 2 t 1 t 1

27. lim

( x 5)( x  2) x2

 lim

x2

 lim ( x  5)  2  5  3 x 2

(t  2)(t 1)

 lim (t 1)(t 1)  lim tt12  1112  32 t 1 t 1

28.

2 (t  2)(t 1) lim t 23t  2  lim (t  2)(t 1)  lim tt  22  11 22   13 t t 2   t 1 t 1 t 1

29.

lim 32 x  42 x 2 x  2 x

30. lim

2( x  2)

 lim

 lim 22  42   12 x 2 x

2 x 2 x ( x  2)

5 y 3 8 y 2

y 2 (5 y 8)

 lim

4 2 y 0 3 y 16 y

 816   12

2 y 0 3 y 16



1 x

5 y 8

 lim

2 2 y 0 y (3 y 16)



31.

1 lim xx 11  lim x x 1  lim 1xx  x11  lim  1x  1 x1 x 1 x1 x1

32.

lim

x 0

1  1 x 1 x 1

u 4 1 3 u 1 u 1

33. lim

 lim

4 x  x2 x 4 2 x

 lim

x 1

lim

x 1

x 1 x 3  2

 lim

x 2 8 3 x 1

x 1

( x 1)



x 3  2

 lim

x 1

 lim

39. lim

x 2

x 2 12  4 x 2



( x 1)

x 1





x 2 12  4

 lim

x2

x2

x 2 12  4

x 1

x 2 8  3



  lim



 lim

x 8  3

x 2 8  3

( x  2)



x 1

x 2



x 3  2

x 2 8 3

1 9 3



x (2 x )(2 x ) 2 x

x 3  2



1 x 3

x 9

x 4

(11)(11) 111

v 2  2v  4 2 v 2 (v  2)( v  4)

 lim

2 x 0 ( x 1)( x 1)

 lim

v 2 (v  2)(v  2)( v  4) x 3 x 9 ( x 3)( x 3)

  lim 

u 2 u 1

u 1

x (4  x ) x 4 2 x

(u 2 1)(u 1)

 lim

(v  2)( v 2  2v  4)

lim x 3 x 9 x 9

37. lim

38.

2 u 1 (u u 1)(u 1)

 lim



2x 1 x 0 ( x 1)( x 1) x

 lim

(u 2 1)(u 1)(u 1)

v3 8 4 v  2 v 16

36. lim

x 0

34. lim

35.

( x 1)  ( x 1) ( x 1)( x 1)

 lim

 2

2 3 3



2 1

 2

4 3

 12 3 32 8

 16





x 3  2

  lim

 lim x 2  x  4(2  2)  16 x 4

( x 1)



( x 3)  4

lim

x 1

( x 2 8) 9

x 1 ( x 1)



x 2 8  3





x32  4 24

 lim

( x 1)( x 1)

x 1 ( x 1)



x 2 8  3



 13



x 2 12  4

x 12  4





4 4  4 (4)(8)





4 16  4

  lim

( x 2 12) 16

x 2 ( x  2)



1 2



x 12  4



 lim

( x  2)( x  2)

x 2 ( x  2)



x 2 12  4



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68 40.

Chapter 2 Limits and Continuity x2

lim

x 2

x  5 3

( x  2)

 lim

x 2



x 2

2 lim 2 x x3 5 x 3

 lim

x 2 5 3

x 5 3



9 3 4

x  5 3

 2

x 3





(3 x )(3 x )





4 x 2

x  4 5 x  9

 lim





(4  x ) 5 x 2 9







(4  x ) 5 x 2 9 (4  x )(4 x )

x 4

x 2



1 x 0 cos x

47. lim

x 0

1 x sin x 3cos x



1 cos 0

  lim

( x  2)



x 2 5 3

( x  2)( x  2)

x 2

3 x x 3 2  x 2 5





x 3 ( x  3) 2  x 5

 lim

x4

  lim 5 x 4





6 2 4

 32

(4  x ) 5 x 2 9 2

25( x 9)

x2 9 4 x



5 25 8



9 x 2

 lim



x 3 ( x 3) 2  x 2 5

  lim (4 x)5 x4





16  x

x 2 9 2



5 4 2

44.

x 0

x 2 5 3

4 ( x 2 5)

43. lim (2sin x  1)  2 sin 0  1  0  1  1 45. lim sec x  lim



( x 2 5) 9

 lim

x  4 5 x 2  9 5 x 2  9

 lim

  lim

x 2 5 2  x 2 5

x 3 ( x 3) 2 x 2 5

42. lim



( x  2)

  32

( x 3) 2  x 5

 lim





x 2 5 3 x 2

 lim

41.



 11  1

46.

lim sin 2 x   lim sin x   (sin 0) 2  02  0 x 0  x 0  sin x x 0 cos x

lim tan x  lim

x 0



sin 0 cos 0



0 1

0

0 sin 0  1 0  0  1  13cos 0 3 3

48. lim ( x 2  1)(2  cos x)  (02  1)(2  cos 0)  (1)(2  1)  (1)(1)  1 x 0

49.

lim

x 

x  4 cos( x   )  lim

50. lim 7  sec2 x  x 0

x  4  lim cos( x   )    4  cos 0  4    1  4  

x 

lim (7  sec2 x)  7  lim sec2 x  7  sec2 0  7  (1) 2  2 2

x 0

51. (a) quotient rule (c) sum and constant multiple rules

(b) difference and power rules

52. (a) quotient rule (c) difference and constant multiple rules

(b)

power and product rules

53. (a) lim f ( x) g ( x)   lim f ( x)   lim g ( x)   (5)(2)  10 x c  x c   x c  (b) lim 2 f ( x) g ( x)  2  lim f ( x )   lim g ( x)   2(5)( 2)  20 x c  x c   x c  (c) lim [ f ( x )  3 g ( x)]  lim f ( x)  3 lim g ( x)  5  3( 2)  1 x c

(d)

f ( x) lim x c f ( x )  g ( x )



x c lim f ( x )

xc

lim f ( x )  lim g ( x )

x c



5 5 5( 2) 7

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Section 2.2 Limit of a Function and Limit Laws

54. (a) lim [ g ( x)  3]  lim g ( x)  lim 3  3  3  0 x 4

x4

x 4

(b) lim xf ( x )  lim x  lim f ( x)  (4)(0)  0 x 4

x4

(d) lim



lim g ( x )



x 4

lim f ( x )  lim 1

x4

x 4

3 0 1

3

55. (a) lim [ f ( x)  g ( x )]  lim f ( x)  lim g ( x )  7  (3)  4 x b

x b

(b) lim f ( x)  g ( x)   lim f ( x )   lim g ( x)   (7)(3)  21 x b  x b   xb  (c) lim 4 g ( x)   lim 4   lim g ( x)   (4)(3)  12 x b  x b   x b  (d) lim f ( x)/g ( x)  lim f ( x)/ lim g ( x)  73   73 x b

56. (a)

x b

lim [ p ( x)  r ( x)  s ( x)]  lim p ( x)  lim r ( x)  lim s ( x )  4  0  ( 3)  1

x 2

(b) lim p ( x)  r ( x)  s ( x )   lim p ( x )   lim r ( x)   lim s ( x)   (4)(0)(3)  0 x 2  x 2   x2   x 2    (c) lim [4 p ( x)  5r ( x)]/s ( x)   4 lim p ( x)  5 lim r ( x )  lim s ( x)  [4(4)  5(0)]/  3  16 3 x 2 x 2  x2  x2 (1 h ) 2 12 h h 0

57. lim 58. lim

h 0

1 2 h  h 2 1 h h 0

 lim

( 2  h ) ( 2) h

 lim

h 0

[3(2  h )  4][3(2)  4] h h 0

59. lim

60. lim

h 0

61. lim

h 0

 21 h  12   lim 7h  7 h

 lim

3h h 0 h

 lim

2 1 2  h

h 0



63. lim

x 0

h 0

2 ( 2  h )



7 h  7



h 0

h ( h  4) h

 lim

h0 2 h ( 2  h )

7h  7

 lim (2  h)  2  lim (h  4)  4 h 0

3

 lim

3(0  h ) 1  3(0) 1  lim h h 0 h 0

62. lim

h 0

4 4h  h2  4 h

h 0 2 h

h (2  h ) h

 lim





  lim

h 0 h



3h 1 1 h

h h 0 h (4  2 h )



(7  h ) 7 7h  7

  lim

3h 1 1



3h 1 1

5  2 x 2  5  2(0)2  5 and lim

x0

  14

 lim

h 0 h



 lim

h 0 h

(3h 1) 1





3h 1 1



h 7h  7

 lim

h 0 h





 lim

h0

3h 3h 1 1



1 7h  7

 lim

h0

3 3h11



x 0

65. (a) lim 1  x6  1  06  1 and lim 1  1; by the sandwich theorem, lim x0

x0



x 0



1 2 7

3 2

5  x 2  5  (0) 2  5; by the sandwich theorem, lim f ( x)  5

64. lim (2  x 2 )  2  0  2 and lim 2 cos x  2(1)  2; by the sandwich theorem, lim g ( x)  2 x 0



x sin x x 0 2 2 cos x

1

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Chapter 2 Limits and Continuity (b) For x  0, y  ( x sin x)/(2  2 cos x) lies between the other two graphs in the figure, and the graphs converge as x  0.



1 x 0 2

66. (a) lim



x2 24

  lim

1 x 0 2

x2 24 x 0

 lim



1 2

0

1 2

1 x0 2

and lim

(b) For all x  0, the graph of f ( x)  (1  cos x)/x 2 lies between the line y  12 and the parabola

 12 ; by the sandwich theorem, lim

x 0

1cos x x2

y  12  x 2 /24, and the graphs converge as

x  0.

67. (a)

f ( x)  ( x 2  9)/( x  3)

3.1

3.01

3.001

3.0001

3.00001

3.000001

f ( x)

6.1

6.01

6.001

6.0001

6.00001

6.000001

2.9

2.99

2.999

2.9999

2.99999

2.999999

f ( x)

5.9

5.99

5.999

5.9999

5.99999

5.999999

The estimate is lim f ( x)  6. x 3

(b)

x 2 9 x 3



( x 3)( x 3) x 3



68. (a) g ( x)  ( x 2  2)/ x  2

x g ( x)

1.4 2.81421

 x  3 if x  3, and lim ( x  3)  3  3  6. x 3



1.41 2.82421

1.414 2.82821

1.4142 2.828413

1.41421 2.828423

1.414213 2.828426

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 12 .

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Section 2.2 Limit of a Function and Limit Laws (b)

x2 2 x 2



 x  2  x 2   x   x 2 

69. (a) G ( x)  ( x  6)/( x 2  4 x  12) x 5.9 5.99 G ( x ) .126582 .1251564

x G ( x)

6.1 .123456

6.01 .124843

2 if x  2, and lim

x 2

x  2 

2  2  2 2.

5.999 .1250156

5.9999 .1250015

5.99999 5.999999 .1250001 .1250000

6.001 .124984

6.0001 .124998

6.00001 .124999

6.000001 .124999

(b)

x 6 ( x 2  4 x 12)



x6 ( x  6)( x  2)



1 x2

1 x 6 x  2

if x  6, and lim



1 6  2

  18  0.125.

70. (a) h( x)  ( x 2  2 x  3)/( x 2  4 x  3)

x h( x )

2.9 2.052631

2.99 2.005025

2.999 2.000500

2.9999 2.000050

2.99999 2.000005

2.999999 2.0000005

x 3.1 h( x) 1.952380

3.01 1.995024

3.001 1.999500

3.0001 1.999950

3.00001 1.999995

3.000001 1.999999

(b)

x 2  2 x 3 x 2  4 x 3



( x 3)( x 1) ( x 3)( x 1)



x 1 if x 1

x 1 x 3 x 1

x  3, and lim



31 31



4 2

 2.

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Chapter 2 Limits and Continuity

71. (a) f ( x )  ( x 2  1)/(| x |  1)

x f ( x)

1.1 2.1

1.01 2.01

1.001 2.001

1.0001 2.0001

1.00001 2.00001

1.000001 2.000001

x f ( x)

.9 1.9

.99 1.99

.999 1.999

.9999 1.9999

.99999 1.99999

.999999 1.999999

(b)

x 2 1 x 1

 ( x 1)( x 1)  x  1, x  0 and x  1  x 1  , and lim (1  x )  1  ( 1)  2. ( x 1)( x 1) x 1  ( x 1)  1  x, x  0 and x  1 

72. (a) F ( x)  ( x 2  3 x  2)/(2 | x |)

x F ( x)

2.1 1.1

2.01 1.01

2.001 1.001

2.0001 1.0001

2.00001 1.00001

2.000001 1.000001

x F ( x)

1.9 .9

1.99 .99

1.999 .999

1.9999 .9999

1.99999 .99999

1.999999 .999999

(b)

x2 3 x  2 2 x

 ( x  2)( x 1) , x0  2 x  , and lim ( x  1)  2  1  1. ( x  2)( x 1) x 2       x 1, x 0 and x 2   2 x

73. (a) g ( )  (sin  )/

 g ( )



.1 .998334

.1 g ( ) .998334 lim g( )  1

.01 .999983

.001 .999999

.01 .001 .999983 .999999

.0001 .999999

.00001 .999999

.000001 .999999

.0001 .999999

.00001 .999999

.000001 .999999

 0

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Section 2.2 Limit of a Function and Limit Laws (b)

74. (a) G (t )  (1  cos t )/t 2

t G (t )

.1 .499583

.01 .499995

.001 .499999

.0001 .5

t G (t )

.1 .499583

.01 .499995

.001 .499999

.0001 .00001 .000001 .5 .5 .5

.00001 .5

.000001 .5

lim G (t )  0.5

t 0

(b)

75. (a)

f ( x)  x1/(1 x ) x .9 .99 f(x) .348678 .366032

.999 .367695

.9999 .367861

.99999 .367877

.999999 .367879

x f(x)

1.001 .368063

1.0001 .367897

1.00001 .367881

1.000001 .367878

1.1 .385543

1.01 .369711

lim f ( x)  0.36788

x 1

(b)

Graph is NOT TO SCALE. Also, the intersection of the axes is not the origin: the axes intersect at the point (1, 2.71820).

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Chapter 2 Limits and Continuity

76. (a)

f ( x)  (3x  1)/x x .1 .01 f(x) 1.161231 1.104669

.1 1.040415

x f(x)

.01 1.092599

.001 1.099215

.0001 1.098672

.00001 1.098618

.000001 1.098612

.001 1.098009

.0001 1.098551

.00001 1.098606

.000001 1.098611

lim f ( x)  1.0986

x 1

(b)

77. lim f ( x) exists at those points c where lim x 4  lim x 2 . Thus, c 4  c 2  c 2 (1  c 2 )  0  c  0, 1, or 1. x c

x c

Moreover, lim f ( x)  lim x 2  0 and lim f ( x )  lim f ( x)  1. x 0

x 0

x 1

x 1

78. Nothing can be concluded about the values of f , g , and h at x  2. Yes, f (2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f ( x )  5  0. x 2

f ( x ) 5 x 4 x  2

79. 1  lim

80. (a) 1  lim



lim f ( x )  lim 5

x4

lim x  lim 2

x 4

f ( x)

2 x 2 x

(b) 1  lim

x4

f ( x)

2 x 2 x



lim f ( x )

x 2

lim x 2

x 2



4 2

lim f ( x )

x 2

f ( x)  x 

f ( x ) 5 81. (a) 0  3  0   lim x  2   x 2   lim f ( x)  5  lim

(b) 0  4  0   lim  x 2

x 4

x4

  lim  x 2

x2

lim f ( x ) 5



 lim f ( x)  5  2(1)  lim f ( x)  2  5  7. x 4

 lim f ( x)  4. x 2

 lim 1    lim  x 2 x   x 2

 lim ( x  2)   lim   x2  x 2  f ( x)  5.

x 2 f ( x ) 5   lim ( x  2)  x  2   x 2 

x 0

f ( x)  x2



f ( x)  x 

f ( x ) 5 x2

f ( x)  2.  12   xlim 2 x

 ( x  2)  lim [ f ( x)  5] x2

 lim f ( x)  5 as in part (a).

f ( x) 82. (a) 0  1  0   lim 2   lim x    lim  x0  x 0 x   x 0  That is, lim f ( x)  0.

(b) 0  1  0   lim  x 0

x 4

x 2

f ( x)  x2

f ( x)  x 2   lim  2  x 2   lim f ( x).   xlim  x 0 0  x0  x

 lim x   lim  f ( x )  x   lim  x0   x 0  x0  x 2

f ( x) . That x

is, lim

x0

f ( x) x

 0.

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Section 2.3 The Precise Definition of a Limit

83. (a) lim x sin 1x  0 x 0

(b) 1  sin 1x  1 for x  0: x  0   x  x sin 1x  x  lim x sin 1x  0 by the sandwich theorem; x 0

x  0   x  x sin 1x  x  lim x sin 1x  0 by the sandwich theorem. x 0

84. (a) lim x 2 cos x 0

(b) 1  cos

 0 1 x3

   1 for x  0   x 1 x3

 x 2 cos

lim x 2  0.

  x 1 x3

 lim x 2 cos x 0

   0 by the sandwich theorem since 1 x3

x 0

Example CAS commands: 85–90. Maple: f : x - (x^4  16)/(x  2); x0 : 2; plot( f (x), x  x0-1..x0 1, color  black, title  "Section 2.2, #85(a)" ); limit( f (x), x  x 0 );

In Exercise 87, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be overcome in Maple by entering the function as f : x - (surd(x 1, 3)  1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f , x] f[x _]: (x 3  x 2  5x  3)/(x  1)2 x0  1; h  0.1; Plot[f[x],{x, x0  h, x0  h}] Limit[f[x], x  x0]

2.3

THE PRECISE DEFINITION OF A LIMIT

1. Step 1: x  5      x  5      5  x    5 Step 2:   5  7    2,or   5  1    4. The value of δ which assures x  5    1  x  7 is the smaller value,   2. Copyright  2018 Pearson Education, Inc.

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Chapter 2 Limits and Continuity

Step 1: Step 2:

x  2      x  2      2  x    2   2  1    1, or   2  7    5. The value of  which assures x  2    1  x  7 is the smaller value,   1.

Step 1: Step 2:

x  (3)      x  3      3  x    3   3   72    12 , or   3   12    52 . The value of  which assures x  (3)     72  x   12 is the smaller value,   12 .

Step 1:

x   32      x  32      32  x    32

Step 2:

 

  32   72    2, or   32   12    1.

 

The value of  which assures x   32     72  x   12 is the smaller value,   1.

5. Step 1: Step 2:

x  12      x  12      12  x    12

  12 

4 9

1 , or   1     18 2

4 7

1 .    14

The value of  which assures x  12    94  x 

4 7

1. is the smaller value,   18

Step 1: Step 2:

7. Step 1: Step 2:

x  3      x  3      3  x    3   3  2.7591    0.2409, or   3  3.2391    0.2391. The value of  which assures x  3    2.7591  x  3.2391 is the smaller value,   0.2391. x  5      x  5      5  x    5 From the graph,   5  4.9    0.1, or   5  5.1    0.1; thus   0.1 in either case.

8. Step 1: Step 2:

x  (3)      x  3      3  x    3 From the graph,   3  3.1    0.1, or   3  2.9    0.1; thus   0.1.

9. Step 1: Step 2:

x  1      x  1      1  x    1 9    7 , or   1  25    9 ; thus   7 . From the graph,   1  16 16 16 16 16

10. Step 1: Step 2:

x  3      x  3      3  x    3 From the graph,   3  2.61    0.39, or   3  3.41    0.41; thus   0.39.

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Section 2.3 The Precise Definition of a Limit 11. Step 1: Step 2:

x  2      x  2      2  x    2 From the graph,  2  3    2  3  0.2679, or   2  5    5  2  0.2361; thus   5  2.

12. Step 1:

x  (1)      x  1      1  x    1

Step 2:

From the graph,   1   thus  

5 2 . 2

5 2

 

5 2 2

 0.118 or   1  

13. Step 1: Step 2:

x  (1)      x  1      1  x    1 From the graph,   1   16    79  0.77, or   1   16  9 25

14. Step 1:

x  12      x  12      12  x    12

Step 2:

From the graph,   12  thus   0.00248.

1  2.01

3 2

 

9 25

2 3 2

 0.36; thus  

( x  1)  5  0.01  x  4  0.01  0.01  x  4  0.01  3.99  x  4.01 x  4      x  4      4  x    4    0.01.

16. Step 1:

(2 x  2)  (6)  0.02  2 x  4  0.02  0.02  2 x  4  0.02  4.02  2 x  3.98  2.01  x  1.99 x  (2)      x  2      2  x    2    0.01.

17. Step 1:

x  1  1  0.1  0.1  x  1  1  0.1  0.9  x  1  1.1  0.81  x  1  1.21

Step 2:

 0.19  x  0.21 x  0      x   . Then,   0.19    0.19 or   0.21; thus,   0.19. x  12  0.1  0.1  x  12  0.1  0.4  x  0.6  0.16  x  0.36

18. Step 1: Step 2:

x  14      x  14    Then   14  0.16    0.09

 4  x  19  16  15  x  3 or 3  x  15 x  10      x  10      10  x    10. Then   10  3    7, or   10  15    5; thus   5.

x  7  4  1  1  x  7  4  1  3  x  7  5  9  x  7  25  16  x  32

20. Step 1: Step 2:

21. Step 1: Step 2:

or   14  0.36    0.11; thus   0.09.

19  x  3  1  1  19  x  3  1  2  19  x  4  4  19  x  16

19. Step 1: Step 2:

9 25

x  23      x  23      23  x    23. Then   23  16    7, or   23  32    9; thus   7. 1 x

 0.36.

1  0.00248, or   1  1    1  1  0.00251;  12  2.01 2 1.99 1.99 2

15. Step 1: Step 2:

Step 2:

 0.1340;

 14  0.05  0.05  1x  14  0.05  0.2 

1 x

 0.3  10  x  10 or 10  x  5. 2 3 3

x  4      x  4      4  x    4. Then   4  10 or   23 , or   4  5 or   1; thus   23 . 3

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Chapter 2 Limits and Continuity

22. Step 1: Step 2:

x 2  3  0.1  0.1  x 2  3  0.1  2.9  x 2  3.1  2.9  x  3.1 x  3      x  3      3  x    3.

Then   3  2.9    3  2.9  0.0291, or   3  3.1    3.1  3  0.0286; thus   0.0286 23. Step 1: Step 2:

24. Step 1: Step 2:

25. Step 1: Step 2:

26. Step 1: Step 2:

x 2  4  0.5  0.5  x 2  4  0.5  3.5  x 2  4.5  3.5  x  4.5   4.5  x   3.5,

for x near 2. x  (2)      x  2      2  x    2. Then   2   4.5    4.5  2  0.1213, or   2   3.5    2  3.5  0.1292; thus   4.5  2  0.12. 1 x

11   (1)  0.1  0.1  1x  1  0.1   10

1 x

9   10  x   10 or  10  x   10 .   10 11 9 9 11

x  (1)      x  1      1  x    1. 1 ; thus   1 . Then   1   10    19 , or   1   10    11 9 11 11

( x 2  5)  11  1  x 2  16  1  1  x 2  16  1  15  x 2  17  15  x  17.

x  4      x  4      4  x    4. Then   4  15    4  15  0.1270, or   4  17    17  4  0.1231; thus   17  4  0.12. 120 x

 5  1  1  120  5  1  4  120 6 x x

1 4

x   120

1 6

 30  x  20 or 20  x  30.

x  24      x  24      24  x    24. Then   24  20    4, or   24  30    6; thus    4.

27. Step 1: Step 2:

mx  2m  0.03  0.03  mx  2m  0.03  0.03  2m  mx  0.03  2m  2  0.03  x  2 m x  2      x  2      2  x    2. Then   2  2  0.03    0.03 , or   2  2  0.03    0.03 . In either case,   0.03 . m m m m m

28. Step 1: Step 2:

mx  3m  c  c  mx  3m  c  c  3m  mx  c  3m  3  mc  x  3  mc x  3      x  3      3  x    3. Then   3  3  mc    mc , or   3  3  mc    mc . In either case,   mc .

29. Step 1:

(mx  b) 

Step 2:

 m2  b   c  c  mx  m2  c  c  m2  mx  c  m2  12  mc  x  12  mc .

x  12      x  12      12  x    12 .

Then   12  12  mc    30. Step 1: Step 2:

c, m

or   12 

1 2

 mc   

c . In m

either case,  

c. m

(mx  b)  (m  b)  0.05  0.05  mx  m  0.05  0.05  m  mx  0.05  m  1  0.05  x  1  0.05 . m m

x  1      x  1      1  x    1. Then   1  1  0.05    0.05 , or   1  1  0.05   m m m

0.05 . In m

either case,  

0.05 . m

31. lim (3  2 x)  3  2(3)  3 x 3

Step 1:

(3  2 x )  (3)  0.02  0.02  6  2 x  0.02  6.02  2 x  5.98  3.01  x  2.99 or 2.99  x  3.01.

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0.03 . m

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Section 2.3 The Precise Definition of a Limit 0  x  3      x  3      3  x    3. Then   3  2.99    0.01, or   3  3.01    0.01; thus   0.01.

Step 2:

32.

lim (3 x  2)  (3)(1)  2  1

x  1

Step 1: Step 2:

(3 x  2)  1  0.03   0.03  3 x  3  0.03  0.01  x  1  0.01  1.01  x  0.99. x  (1)      x  1      1  x    1. Then   1  1.01    0.01, or   1  0.99    0.01; thus   0.01.

x2 4 x 2 x  2

 lim

33. lim

x 2 x2 4 x 2

lim

x 2 6x 5 x 5



Step 1:

4  0.05  3.95  x  2  4.05, x  2

 lim

x5

x  6 x 5 x 5

(x 5)(x 1) (x 5)

 lim ( x  1)  4, x  5. x 5

  (4)  0.05  0.05 

( x 5)( x 1) ( x 5)

 4  0.05  4.05  x  1  3.95, x  5

1  5 x  4  0.5  0.5  1  5 x 4  0.5  3.5  1  5 x  4.5  12.25  1  5 x  20.25

Step 1: Step 2:

4 x2 x

( x  2)( x  2) ( x 2)

1  5 x  1  5(3)  16  4

lim

x  3

36. lim

x 2

 5.05  x   4.95, x  5. x  (5)      x  5      5  x    5. Then   5  5.05    0.05, or   5  4.95    0.05; thus   0.05.

Step 2:

35.

 lim ( x  2)  2  2  4, x  2

 1.95  x  2.05, x  2. x  2      x  2      2  x    2. Then   2  1.95    0.05, or   2  2.05    0.05; thus   0.05.

Step 2:

x 5

( x  2)( x  2) ( x  2)

   4  0.05  0.05 

Step 1:

34.



Step 1: Step 2:

 11.25  5 x  19.25  3.85  x  2.25. x  (3)      x  3      3  x    3. Then   3  3.85    0.85, or   3  2.25  0.75; thus   0.75. 4 2

2 4 x

 2  0.4  0.4  4x  2  0.4  1.6 

4 x

 2.4  10  16

x 4

 10  10  x  10 or 53  x  52 . 24 4 6

x  2      x  2      2  x    2. Then   2  53    13 , or   2  52    12 ; thus   13 .

37. Step 1: Step 2:

(9  x)  5      4  x      4   x    4    4  x  4    4    x  4  . x  4      x  4      4  x    4. Then   4    4    , or   4    4    . Thus choose   .

38. Step 1: Step 2:

(3x  7)  2      3x  9    9    3 x  9    3  3  x  3  3 . x  3      x  3      3  x    3. Then   3  3  3    3 , or   3  3  3    3 . Thus choose   3 .

39. Step 1: Step 2:

x  5  2      x  5 2    2    x  5  2    (2   ) 2  x  5  (2   )2  (2   ) 2  5  x  (2   ) 2  5. x  9      x  9      9  x    9.

Then   9   2  4  9    4   2 , or   9   2  4  9    4   2 . Thus choose the smaller distance,   4   2 . Copyright  2018 Pearson Education, Inc.

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Chapter 2 Limits and Continuity 4  x  2      4  x  2    2    4  x  2    (2   ) 2  4  x  (2   ) 2

40. Step 1: Step 2:

41. Step 1: Step 2:

  (2   ) 2  x  4  (2   )2  (2   )2  4  x  (2   )2  4. x  0      x   . Then   (2   ) 2  4   2  4    4   2 , or    (2   ) 2  4  4   2 . Thus choose the smaller distance,   4   2 .

For x  1, x 2  1      x 2  1    1    x 2  1    1    x  1    1    x  1   near x  1. x  1      x  1      1  x    1. Then   1  1      1  1   , or   1  1      1    1. Choose





  min 1  1   , 1    1 , that is, the smaller of the two distances. 42. Step 1: Step 2:

For x  2, x 2  4      x 2  4    4    x 2  4    4    x  4     4    x   4   near x  2. x  (2)      x  2      2  x    2. Then   2   4      4    2, or   2   4      2  4   . Choose   min 4    2, 2  4   .



43. Step 1: Step 2:

44. Step 1:

1 x



 1      1x  1    1    1x  1    11  x  11 .

x  1      x  1    1    x  1   . Then 1    11    1  11  1  , or 1    11    11  1  1  . Choose   1  , the smaller of the two distances. 1 x2

 13      12  13    13    x

 133  x 2  133 

Step 2:

Choose   min

Step 2:

46. Step 1: Step 2:

47. Step 1: Step 2:

 x 

 13    133  12  133 x

3 , 13

3 13

x

3 13

for x near 3.

x  3      x  3    3    x  3   .

Then 3   

45. Step 1:

3 13

1 x2

3 13



   3  133 , or 3    133    133  3. 3  133 , 133  3 .



   (6)      ( x  3)  6  , x  3    x  3      3  x    3. x 2 9 x 3

x  (3)      x  3      3  x    3. Then   3    3    , or   3    3    . Choose   .

   2      (x  1)  2  , x  1  1    x  1  . x 2 1 x 1

x  1      x  1    1    x  1   . Then 1    1      , or 1    1      . Choose   . x  1: (4  2 x)  2    0  2  2 x   since x  1. Thus, 1  2  x  0; x  1: (6 x  4)  2    0  6 x 6   since x  1. Thus, 1  x  1  6 . x  1      x  1    1    x  1   . Then 1    1  2    2 , or 1    1  6    6 . Choose   6 .

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Section 2.3 The Precise Definition of a Limit 48. Step 1: Step 2:

x  0: 2 x  0      2 x  0   2  x  0;

x  0: 2x  0    0  x  2. x  0      x   . Then    2    2 , or   2   2. Choose   2 .

49. By the figure,  x  x sin 1x  x for all x  0 and  x  x sin 1x  x for x  0. Since lim ( x)  lim x  0, then by x 0

the sandwich theorem, in either case, lim x sin 1x  0.

x 0

50. By the figure,  x 2  x 2 sin 1x  x 2 for all x except possibly at x  0. Since lim ( x 2 )  lim x 2  0, then by the x 0

sandwich theorem, lim x 2 sin 1x  0.

x 0

51. As x approaches the value 0, the values of g ( x) approach k. Thus for every number   0, there exists a   0 such that 0  x  0    g ( x)  k  . 52. Write x  h  c. Then 0  x  c      x  c   , x  c    ( h  c) c   , h  c  c    h   , h  0  0  h  0   . Thus, lim f ( x)  L  for any   0, there exists   0 such that f ( x)  L   whenever 0  x  c    x c

f (h  c)  L   whenever 0  h  0    lim f (h  c)  L. h 0

53. Let f ( x)  x 2 . The function values do get closer to 1 as x approaches 0, but lim f ( x)  0, not 1. The x 0

function f ( x)  x 2 never gets arbitrarily close to 1 for x near 0.

54. Let f ( x)  sin x, L  12 , and x0  0. There exists a value of x (namely x  6 ) for which sin x  12   for any

given   0. However, lim sin x  0, not 12 . The wrong statement does not require x to be arbitrarily close to x0 . x 0

As another example, let g ( x)  sin 1x , L  12 , and x0  0. We can choose infinitely many values of x near 0 such that sin 1x  12 as you can see from the accompanying figure. However, lim sin 1x fails to exist. The wrong x 0

statement does not require all values of x arbitrarily close to x0  0 to lie within   0 of L  12 . Again you can see from the figure that there are also infinitely many values of x near 0 such that sin 1x  0. If we choose   14

we cannot satisfy the inequality sin 1x  12   for all values of x sufficiently near x0  0.

55.

 

2 π x2 A  9  0.01  0.01   2x  9  0.01  8.99  4  9.01  π4 (8.99)  x 2  π4 (9.01)

 2 8.99  x  2 9.01 or 3.384  x  3.387. To be safe, the left endpoint was rounded up and  

the right endpoint was rounded down.

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Chapter 2 Limits and Continuity

56. V  RI  VR  I  

(120)(10) 51

R

V 5 R (120)(10) 49

 0.1  0.1  120  5  0.1  4.9  120  5.1  10  R  10 R R 49 120 51  23.53  R  24.48.

To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 57. (a)   x  1  0  1    x  1  f ( x)  x. Then f ( x)  2  x  2  2  x  2  1  1. That is, f ( x)  2  1 

1 2

no matter how small  is taken when 1    x  1  lim f ( x)  2. x 1

(b) 0  x  1    1  x  1    f ( x)  x  1. Then f ( x)  1  ( x  1)  1  x  x  1. That is, f ( x)  1  1 no matter how small  is taken when 1  x  1    lim f ( x)  1. x 1

(c)   x  1  0  1    x  1  f ( x)  x. Then f ( x )  1.5  x  1.5  1.5  x  1.5  1  0.5. Also, 0  x  1    1  x  1    f ( x)  x  1. Then f ( x)  1.5  ( x  1)  1.5  x  0.5  x  0.5  1  0.5  0.5. Thus, no matter how small  is taken, there exists a value of x such that   x  1   but f ( x )  1.5  12  lim f ( x)  1.5. x 1

58. (a) For 2  x  2    h( x)  2  h( x)  4  2. Thus for   2, h( x)  4   whenever 2  x  2   no matter how small we choose   0  lim h( x)  4. x2

(b) For 2  x  2    h( x )  2  h( x)  3  1. Thus for   1, h( x)  3   whenever 2  x  2   no matter how small we choose   0  lim h( x)  3. x2

(c) For 2    x  2  h( x)  x 2 so h( x)  2  x 2  2 . No matter how small   0 is chosen, x 2 is close to 4 when x is near 2 and to the left on the real line  x 2  2 will be close to 2. Thus if   1, h( x)  2   whenever 2    x  2 no matter how small we choose   0  lim h( x)  2. x2

59. (a) For 3    x  3  f ( x)  4.8  f ( x)  4  0.8. Thus for   0.8, f ( x)  4   whenever 3    x  3 no matter how small we choose   0  lim f ( x)  4. x 3

(b) For 3  x  3    f ( x)  3  f ( x)  4.8  1.8. Thus for   1.8, f ( x)  4.8   whenever 3  x  3   no matter how small we choose   0  lim f ( x)  4.8. x 3

(c) For 3    x  3  f ( x)  4.8  f ( x)  3  1.8. Again, for   1.8, f ( x)  3   whenever 3    x  3 no matter how small we choose   0  lim f ( x)  3. x 3

60. (a) No matter how small we choose   0, for x near 1 satisfying 1    x  1   , the values of g ( x) are near 1  g ( x)  2 is near 1. Then, for   12 we have g ( x)  2  12 for some x satisfying 1    x  1   , or 0  x  1    lim g ( x)  2. x 1

(b) Yes, lim g ( x )  1 because from the graph we can find a   0 such that g ( x)  1   if 0  x  (1)   . x 1

61–66. Example CAS commands (values of del may vary for a specified eps): Maple: f : x - (x^4-81)/(x-3); x0 : 3;

plot( f (x), x  x0-1..x0 1, color  black, title "Section 2.3, #61(a)" );

# (a)

L : limit( f (x), x  x0 );

# (b)

epsilon : 0.2; # (c) plot( [f (x), L-epsilon,L  epsilon], x  x0-0.01..x0  0.01, color  black, linestyle [1,3,3], title "Section 2.3, #61(c)" );

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Section 2.4 One-Sided Limits q : fsolve( abs( f (x)-L )  epsilon, x  x0-1..x0 1 ); # (d) delta : abs(x0-q); plot( [f (x), L-epsilon, L  epsilon], x  x0-delta..x0  delta, color  black, title "Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e) q : fsolve( abs( f (x)-L )  eps, x  x0-1..x0 1 ); delta : abs(x0-q); head : sprintf ("Section 2.3, #61(e)\n epsilon  %5f , delta  %5f \n", eps, delta ); print(plot( [f (x), L-eps, L  eps], x  x0-delta..x0  delta, color  black, linestyle [1,3,3], title  head ));

end do: Mathematica (assigned function and values for x0, eps and del may vary): Clear[f , x] y1:  L  eps; y2:  L  eps; x0  1; f[x _]:  (3x 2  (7x  1)Sqrt[x]  5)/(x  1) Plot[f [x], {x, x0  0.2, x0  0.2}] L:  Limit[f [x], x  x0] eps  0.1; del  0.2; Plot[{f [x], y1, y2}, {x, x0  del, x0  del}, PlotRange  {L  2eps, L  2eps}]

2.4

ONE-SIDED LIMITS

1. (a) True (e) True (i) False

(b) True (f) True (j) False

(d) True (h) False (l) False

2. (a) True (e) True (i) True

(b) False (f) True (j) False

(d) True (h) True

3. (a)

lim f ( x)  22  1  2, lim f ( x)  3  2  1 x  2

x 2

(b) No, lim f ( x) does not exist because lim f ( x)  lim f ( x) x  2

x 2

(c)

x 2

lim f ( x )  42  1  3, lim f ( x)  42  1  3

x 4

x 4

(d) Yes, lim f ( x)  3 because 3  lim f ( x)  lim f ( x) x  4

x 4

4. (a)

x  4

lim f ( x)  22  1, lim f ( x)  3  2  1, f (2)  2 x 2

x  2

(b) Yes, lim f ( x)  1 because 1  lim f ( x)  lim f ( x ) x 2

x 2

(c)

lim

x 1

f ( x)  3  (1)  4, lim

x 1

f ( x)  3  (1)  4

(d) Yes, lim f ( x )  4 because 4  lim

x 1

x 1

5. (a) No, lim f ( x) does not exist since sin (b)

x 0

x  2



f ( x)  lim

x 1

f ( x)

 1x  does not approach any single value as x approaches 0

lim f ( x)  lim 0  0

x 0



x 0

x 0 

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Chapter 2 Limits and Continuity

6. (a) Yes, lim g ( x)  0 by the sandwich theorem since  x  g ( x)  x when x  0 x 0 

(b) No, lim g ( x) does not exist since x is not defined for x  0 x 0 

x 0

7. (a)

(b)

lim f ( x)  1  lim f ( x )

x 1

x 1

limits exist and equal 1

8. (a)

(b)

lim f ( x)  0  lim f ( x )

x 1

x 1

limits exist and equal 0

9. (a) domain: 0  x  2 range: 0  y  1 and y  2 (b) lim f ( x) exists for c belonging to (0, 1)  (1, 2) x c

10. (a) domain:   x   range: 1  y  1 (b) lim f ( x) exists for c belonging to x c

( ,  1)  ( 1, 1)  (1, ) (c) none (d) none

11.

13.

lim

x 0.5

lim

x 2



x2 x 1



0.5 2 0.51



3/2 1/2

 3

12.

lim

x 1

x 1 x2

 1112  0  0

 xx1   2x x5x    221   (2(2)2)(52)   (2)  12   1 2

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Section 2.4 One-Sided Limits 14.

15.

 x11   xx 6   37 x    111   116   371    12   17   72   1

lim

x 1

lim

h 0 

h2  4h 5  5 h

 2   2  ( h 2  4 h 5) 5  lim  h  4 hh5  5   h 2  4h 5  5   lim h 0    h  4 h  5  5  h 0 h h 2  4 h  5  5

 lim

h 0  h

16.

lim

h 0



6  5h 2 11h  6 h

h ( h  4)



h2  4h 5  5





0 4 5 5





2 5



2 2      lim  6  5hh 11h  6   6  5h2 11h  6   h 0    6  5h 11h  6 

 lim



h 0 h



6 (5h 2 11h  6)

6  5h 2 11h  6

x2

 lim



h 0 h



 h (5h 11) 6  5h 2 11h  6





(0 11) 6 6

  11

2 6

( x  2)

lim ( x  3) x  2  lim ( x  3) ( x  2) (|x  2|  ( x  2) for x  2) x 2 x 2  lim ( x  3)  ((2)  3)  1

17. (a)

lim ( x  3)

(b)

x 2



x2 x2

x 2

( x  2)  lim ( x  3)  ( x  2)  (|x  2|  ( x  2) for x  2)    x 2  lim ( x  3)(1)  (2  3)  1 x 2

18. (a)

2 x ( x 1) x 1

lim



x 1

(b)

 lim

2 x ( x 1) ( x 1)



x 1

 lim 2 x ( x 1) x 1

lim



x 1

x 1

 lim



x 1

(|x  1|  x  1 for x  1)

2x  2 2 x ( x 1) ( x 1)

(|x  1|  ( x  1) for x  1)

 lim  2 x   2 x 1

19. (a) If 0  x  2 , then sin x  0, so that lim x 0



sin x sin x

 lim

x 0

sin x sin x

 lim 1  1 x 0 

sin x sin x  lim 1  1 (b) If 2  x  0, then sin x  0, so that lim sin x  lim sin x  x 0 x 0  x 0  cos x  lim 1cos x  lim 1  1 20. (a) If 0  x  2 , then cos x  1, so that lim 1cos x  lim 1(cos 1cos x x 1) x 0 cos x 1 x 0 x 0  x 0  cos x 1  lim 1  1 (b) If 2  x  0, then cos x  1, so that lim cos x 1  lim (cos x 1) x 0 cos x 1

 

lim    33  1  3 

21. (a)

lim (t  t  )  4  4  0

22. (a)

23.

lim sin 2  lim sinx x  1

t 0

(b)

t  4

 0

24. lim

(b)

2

sin kt t

x 0

x 0 

x 0

 

lim    23  3 

lim (t  t  )  4  3  1

t  4

(where x  2 )

kt  lim k sin   k lim sin   k  1  k  lim k sin  0   0  t 0 kt

(where   kt )

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Chapter 2 Limits and Continuity sin 3 y y 0 4 y

26.

lim

h 0 

h sin 3h

tan 2 x x 0 x

27. lim

2t t 0 tan t

28. lim

1 lim 3sin 3 y 4 y 0 3 y



25. lim

h 0 

x csc 2 x x 0 cos 5 x

29. lim

sin 2 x x 0 x cos 2 x

x 0

  sin t cos t

t cos t t 0 sin t

 2 lim



x  1 x 0 sin 2 x cos 5 x

 lim

 sin3h3h 

x  x cos x x 0 sin x cos x



x 2  x sin x 2x x 0



x x 0 2

 lim

 12  12

(where   3y )    1 1  1 3  3 

(where   3h)

 lim 2sin 2 x   1  2  2    x 0 2 x 

  1sin t  lim t  t 0

   2 1 1  2  





 lim 3cos x  sinx x  sin2 x2 x  3 1 1  3 x 0

x  sinx cos  lim x cos x

   lim  sin1 x   lim x 0  x  x 0

32. lim

3 4

2 x   lim 1   1 1 (1)  1    12 xlim    2  2 0 sin 2 x   x 0 cos 5 x 



x x 0 sin x cos x

 lim

1  cos 2 x 

 2  lim cos t   t 0 

6 x 2 cos x x sin 2 x sin x 0

31. lim



 1  13   lim sin    0

  lim  x 0

30. lim 6 x 2 (cot x)(csc 2 x )  lim x 0

3 lim sin  4  0 





sin 2 x  cos 2x   lim

 lim

t 0

3 lim sin 3 y 4 y 0 3 y

 13  sin3h3h   13 hlim 0

 lim

 2 lim





x  1 x 0 sin x cos x

  cos1 x   xlim  0 

sin x x

x   xlim 0 sin x

   (1)(1)  1  2 

 sinx x   0  12  12 (1)  0

1cos   lim (1cos  )(1 cos  )  lim 1cos 2  sin 2   lim  0 sin 2  0 (2sin  cos  )(1 cos  )  0 (2sin  cos  )(1 cos  )  0 (2sin  cos  )(1 cos  )

33. lim

sin   0 0  0 (2 cos  )(1 cos  ) (2)(2)

 lim

x  x cos x 2 x 0 sin 3 x

34. lim

 lim

x 0

sin(1cos t ) t 0 1cos t

35. lim

36. lim

h 0

sin(sin h ) sin h

x (1cos x ) 2

sin 3 x

 lim

x 0

x (1 cos x ) 9 x2 sin 2 3 x

 lim

9 x2

x 0

1 cos x 9x

 sin3 x3 x 





1 lim 1 cos x 9 x0 x 2 sin 3 x lim 3 x x 0







1 (0) 9 2

0

 lim sin   1 since   1  cos t  0 as t  0  0

 lim sin   1 since   sin h  0 as h  0  0









sin   lim sin   2  1 lim sin   2  12 1 1  12 2  0  sin 2  0 sin 2  0 sin 2 2

37. lim

sin 5 x x 0 sin 4 x

38. lim



sin 5 x  4 x  5 x 0 sin 4 x 5 x 4

 lim

  54 xlim  sin5 x5x  sin4 x4 x   54 11  54 0

39. lim  cos   0 1  0  0

40. lim sin  cot 2  lim sin   0

 0

cos 2 sin 2

 lim sin   0

cos 2 2sin  cos 

cos 2  1 2  0 2 cos 

 lim

Full file at https://testbankuniv.eu/Thomas-Calculus-Early-Transcendentals-14th-Edition-Hass-Solutions-Manual

Section 2.4 One-Sided Limits tan 3 x x 0 sin 8 x

41. lim



  xlim  sin 3x  1  8 x  3  0 cos 3 x sin 8 x 3 x 8  83 lim  cos13 x   sin3 x3 x   sin8 x8 x   83 1 1 1  83 x0 sin 3 x  1 x 0 cos 3 x sin 8 x

 lim

sin 3 y cot 5 y y 0 y cot 4 y

 lim

y 0

tan  2  0  cot 3

43. lim

            1 1 1 1 

sin 3 y sin 4 y cos 5 y y 0 y cos 4 y sin 5 y

 lim

42. lim

 lim

 0



sin 3 y 3y

sin  cos  3  2 cos sin 3



sin 4 y 4y

sin 3 y y y 0

sin 4 y cos 4 y

 lim

5y sin 5 y

cos 5 y cos 4 y

34 5



345 y 345 y

12 5



 12 5

  sin33   cos 3cos 3   (1)(1)  113   3

sin  sin 3  lim sin  2  0  cos  cos 3  0 

 lim

cos 5 y sin 5 y

4 2 2 2  cos  cot 4  lim  cos 4 sin 2 2  lim  cos 4 (2sin  cos  )  lim  cos 4 (4sin  cos  ) sin 4  lim 2 2 2 2 2 2 2 2 2  0 sin  cos 2 sin 4  0 sin  cot 2  0 sin 2 cos 2  0 sin  cos 2 sin 4  0 sin  cos 2 sin 4 2

44. lim

sin 2

   111   1

 1   cos 4 cos2  1  cos 4 cos2  4  lim  lim    1  2 2 sin 4   2  0 cos 2 sin 4  0 sin 4  cos 2   0  4   cos 2 

 lim

45.

lim

x 0

1cos3 x 2x

 lim

x 0

 lim

1cos3 x  1 cos3 x 2x 1 cos3 x 3  sin 



 0 2

46.

cos 2 x cos x x2 x 0

lim

 lim



4 cos 4 cos 2

x 0



1cos 2 3 x x 0 2 x (1 cos3 x )

 lim

  3 (1)  1sin cos 2

cos x (cos x 1) x2



 101   0

sin 2 3 x x 0 2 x (1 cos3 x )

 lim

cos x (cos x 1) cos x 1  cos x 1 x2 x 0

 lim



3  sin 3 x  sin 3 x x 0 2 3 x 1 cos3 x

 lim

(where   3x )

cos x  (1)(1)  1   lim  sinx x  sinx x  cos x 1 11 x 0

 lim

x 0

cos x (cos 2 x 1) x 2 (cos x 1)

 lim

x 0

cos x(  sin 2 x ) x 2 (cos x 1)

1 2

47. Yes. If lim f ( x)  L  lim f ( x), then lim f ( x)  L. If lim f ( x )  lim f ( x), then lim f ( x) does not exist.

x a 

xa 

x a 

x a

x a 

x a

48. Since lim f ( x)  L if and only if lim f ( x)  L and lim f ( x)  L, then lim f ( x) can be found by x c

calculating lim f ( x).

x c 

x c 

x c

x c 

49. If f is an odd function of x, then f ( x)   f ( x). Given lim f ( x)  3, then lim f ( x)  3. x 0 

x 0 

50. If f is an even function of x, then f ( x)  f ( x). Given lim f ( x)  7 then lim can be said about lim

x 2

x 2

f ( x ) because we don’t know lim f ( x).

x 2

f ( x)  7. However, nothing

x  2

51. I  (5, 5   )  5  x  5   . Also, x  5    x  5   2  x  5   2 . Choose    2 lim

x  5  0.

52. I  (4   , 4)  4    x  4. Also, 4  x    4  x   2  x  4   2 . Choose    2 lim

4  x  0.

x 5

x  4

Full file at https://testbankuniv.eu/Thomas-Calculus-Early-Transcendentals-14th-Edition-Hass-Solutions-Manual

Chapter 2 Limits and Continuity

53. As x  0 the number x is always negative. Thus,

x x

 (1)   

x x

 1    0   which is always true x x 0  x

independent of the value of x. Hence we can choose any   0 with   x  0  lim 54. Since x  2 we have x  2 and x  2  x  2. Then,

x2 x2

1 

x2 x2

 1    0   which is always true so

long as x  2. Hence we can choose any   0, and thus 2  x  2    55. (a)

lim

x 400

 1.

x2 x2

x2 x 2 x  2

 1  . Thus, lim

 1.

 x   400. Just observe that if 400  x  401, then  x   400. Thus if we choose   1, we have for

any number   0 that 400  x  400     x   400  400  400  0  . (b) lim  x   399. Just observe that if 399  x  400 then  x   399. Thus if we choose   1, we have for x 400

any number   0 that 400    x  400   x   399  399  399  0  . (c) Since lim  x   lim  x  we conclude that lim  x  does not exist. x 400

56. (a)

x 400

lim f ( x)  lim

x 0 

x 400

x  0  0;

x  0      x    0  x   2 for x positive. Choose    2

 lim f ( x)  0. x 0 

(b)

 1x   0 by the sandwich theorem since  x2  x2 sin  1x   x2 for all x  0. Since x 2  0   x 2  0  x 2   whenever x   , we choose    and obtain x 2 sin  1x   0   lim f ( x)  lim x 2 sin

x 0 

x 0

if   x  0. (c) The function f has limit 0 at x0  0 since both the right-hand and left-hand limits exist and equal 0. 2.5

CONTINUITY

1. No, discontinuous at x  2, not defined at x  2 2. No, discontinuous at x  3, 1  lim g ( x)  g (3)  1.5 x 3

3. Continuous on [1, 3] 4. No, discontinuous at x  1, 1.5  lim k ( x)  lim k ( x)  0 x 1

x 1

5. (a) Yes

(b) Yes, lim

(d) Yes

6. (a) Yes, f (1)  1

x 1

f ( x)  0

(b) Yes, lim f ( x)  2

(d) No

7. (a) No

(b) No

x 1

8. [1, 0)  (0, 1)  (1, 2)  (2, 3) 9. f (2)  0, since lim f ( x)  2(2)  4  0  lim f ( x) x 2

x  2

Full file at https://testbankuniv.eu/Thomas-Calculus-Early-Transcendentals-14th-Edition-Hass-Solutions-Manual

Section 2.5 Continuity 10. f (1) should be changed to 2  lim f ( x ) x 1

11. Nonremovable discontinuity at x  1 because lim f ( x) fails to exist ( lim f ( x)  1 and lim f ( x)  0). x 1

x 1

x 1

Removable discontinuity at x  0 by assigning the number lim f ( x)  0 to be the value of f (0) rather x 0

than f (0)  1.

12. Nonremovable discontinuity at x  1 because lim f ( x) fails to exist ( lim f ( x)  2 and lim f ( x )  1). x 1

x 1

x 1

Removable discontinuity at x  2 by assigning the number lim f ( x)  1 to be the value of f (2) rather than x 2

f (2)  2.

14. Discontinuous only when ( x  2)2  0  x  2

13. Discontinuous only when x  2  0  x  2

15. Discontinuous only when x 2  4 x  3  0  ( x  3)( x  1)  0  x  3 or x  1 16. Discontinuous only when x 2  3 x  10  0  ( x  5)( x  2)  0  x  5 or x  2 17. Continuous everywhere. (|x  1|  sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. (|x|  1  0 for all x; limits exist and are equal to function values.) 19. Discontinuous only at x  0 20. Discontinuous at odd integer multiples of 2 , i.e., x  (2n  1) 2 , n an integer, but continuous at all other x. 21. Discontinuous when 2x is an integer multiple of  , i.e., 2 x  n , n an integer  x  n2 , n an integer, but continuous at all other x. 22. Discontinuous when 2x is an odd integer multiple of 2 , i.e., 2x  (2n  1) 2 , n an integer  x  2n  1, n an integer (i.e., x is an odd integer). Continuous everywhere else. 23. Discontinuous at odd integer multiples of 2 , i.e., x  (2n  1) 2 , n an integer, but continuous at all other x. 24. Continuous everywhere since 1  sin x  1  0  sin 2 x  1  1  sin 2 x  1; limits exist and are equal to the function values.



25. Discontinuous when 2 x  3  0 or x   32  continuous on the interval   32 ,  .



26. Discontinuous when 3 x  1  0 or x  13  continuous on the interval  13 ,  . 27. Continuous everywhere: (2 x  1)1/3 is defined for all x; limits exist and are equal to function values. 28. Continuous everywhere: (2  x)1/5 is defined for all x; limits exist and are equal to function values. x 2  x 6 x 3 x 3

29. Continuous everywhere since lim

( x 3)( x  2) x 3 x 3

 lim

 lim ( x  2)  5  g (3) x 3

Full file at https://testbankuniv.eu/Thomas-Calculus-Early-Transcendentals-14th-Edition-Hass-Solutions-Manual

Chapter 2 Limits and Continuity

30. Discontinuous at x  2 since lim f ( x) does not exist while f (2)  4. x 2

31. Discontinuous at x  1; lim ( x 2  2)  3, but lim e x  e, so that lim f ( x) does not exist while f (1)  e; x 1

x 1

x 1

and lim (1  x)  1  lim e , so that lim f ( x)  1  f (0) x 0 

x 0 

x 0

32. Discontinuous at x  ln 2, since 2  e x  0  e x  2  ln e x  ln 2  x  ln 2 33. lim sin( x  sin x)  sin(  sin  )  sin(  0)  sin   0, and function continuous at x   . x 





 

34. lim sin( 2 cos(tan t ))  sin( 2 cos(tan(0)))  sin 2 cos(0)  sin 2  1, and function continuous at t  0. t 0

35. lim sec ( y sec2 y  tan 2 y  1)  lim sec ( y sec2 y  sec2 y )  lim sec (( y  1) sec2 y )  sec ((1  1)sec2 1)  y 1

y 1

sec 0  1, and function continuous at y  1.





 

36. lim tan  4 cos(sin x1/3 )   tan  4 cos(sin(0))   tan 4 cos(0)  tan 4  1, and function continuous at x  0.   x 0  37. lim cos  t 0 





19 3 sec 2t 

  cos  





193 sec 0 

 cos   cos 4  16

 

2 2

, and function continuous at t  0.

 

csc2 x  5 3 tan x  csc2 6  5 3 tan 6  4  5 3

38. lim

x  6

at x  6 . 39.

40.



lim sin 2 e

x 0 

  sin  e   sin   0 2







  1 3

9  3, and function continuous



  1, and the function is continuous at x = 0. 2



lim cos 1 ln x  cos 1 ln 1  cos 1 (0)  2 , and the function is continuous at x = 1.

x 1

41. g ( x)  42. h(t ) 

x 2 9 x 3

( x 3)( x 3) ( x 3)



t 2 3t 10 t 2

(t 5)(t  2) t 2

( s 2  s 1)( s 1) ( s 1)( s 1)

43. f ( s ) 

s 3 1 s 3 1

44. g ( x) 

x 2 16 x 2 3 x  4



 x  3, x  3  g (3)  lim ( x  3)  6



x 3

 t  5, t  2  h(2)  lim (t  5)  7 t 2



( x  4)( x  4) ( x  4)( x 1)

s 2  s 1 , s 1



x 4 , x 1





2 s  1  f (1)  lim s ss11  s 1

x  4  g (4)  lim

x 4

3 2

 xx14   85

45. As defined, lim f ( x)  (3)2  1  8 and lim (2a )(3)  6a. For f ( x) to be continuous we must have x 3

6a  8  a  43 .

x 3

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Section 2.5 Continuity

46. As defined, lim g ( x)  2 and lim g ( x)  b(2)2  4b. For g ( x) to be continuous we must have x 2

x 2

4b  2  b   12 .

47. As defined, lim f ( x)  12 and lim f ( x)  a 2 (2)  2a  2a 2  2a. For f ( x ) to be continuous we must have 2

x  2

x 2

12  2a  2a  a  3 or a  2.

48. As defined, lim g ( x)  b b 1

x 0 

0 b b 1



b b 1

and lim g ( x)  (0) 2  b  b. For g ( x) to be continuous we must have

 b  b  0 or b  2.

49. As defined, lim

x 1

f ( x)  2 and lim

x 0 

x 1

f ( x)  a (1)  b  a  b, and lim f ( x)  a (1)  b  a  b and x 1

lim f ( x)  3. For f ( x) to be continuous we must have 2  a  b and a  b  3  a 

x 1

5 2

and b  12 .

50. As defined, lim g ( x)  a (0)  2b  2b and lim g ( x)  (0)2  3a  b  3a  b, and lim g ( x)  (2)2  3a  b  x 0 

x 0 

x  2

4  3a  b and lim g ( x)  3(2)  5  1. For g ( x) to be continuous we must have 2b  3a  b and 4  3a  b  1  x 0 

a   32 and b   32 .

51. The function can be extended: f (0)  2.3.

52. The function cannot be extended to be continuous at x  0. If f (0)  2.3, it will be continuous from the right. Or if f (0)  2.3, it will be continuous from the left.

53. The function cannot be extended to be continuous at x  0. If f (0)  1, it will be continuous from the right. Or if f (0)  1, it will be continuous from the left.

54. The function can be extended: f (0)  7.39.

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Chapter 2 Limits and Continuity

55. f ( x) is continuous on [0, 1] and f (0)  0, f (1)  0  by the Intermediate Value Theorem f ( x) takes on every value between f (0) and f (1)  the equation f ( x)  0 has at least one solution between x  0 and x  1.

   

 

56. cos x  x  (cos x)  x  0. If x   2 , cos  2   2  0. If x  2 , cos 2  2  0. Thus cos x  x  0 for some x between   and  according to the Intermediate Value Theorem, since the function cos x  x is

continuous.

57. Let f ( x)  x3  15 x  1, which is continuous on [4, 4]. Then f (4)  3, f (1)  15, f (1)  13, and f (4)  5. By the Intermediate Value Theorem, f ( x )  0 for some x in each of the intervals 4  x  1, 1  x  1, and 1  x  4. That is, x3  15 x  1  0 has three solutions in [4, 4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. 58. Without loss of generality, assume that a  b. Then F ( x)  ( x  a )2 ( x  b)2  x is continuous for all values of x, so it is continuous on the interval [a, b]. Moreover F (a )  a and F (b)  b. By the Intermediate Value Theorem, since a  a 2 b  b, there is a number c between a and b such that F ( x)  a 2 b . 59. Answers may vary. Note that f is continuous for every value of x. (a) f (0)  10, f (1)  13  8(1)  10  3. Since 3    10, by the Intermediate Value Theorem, there exists a c so that 0  c  1 and f (c)   . (b) f (0)  10, f (4)  (4)3  8(4)  10  22. Since 22   3  10, by the Intermediate Value Theorem, there exists a c so that 4  c  0 and f (c)   3. (c) f (0)  10, f (1000)  (1000)3  8(1000)  10  999,992, 010. Since 10  5, 000, 000  999,992, 010, by the Intermediate Value Theorem, there exists a c so that 0  c  1000 and f (c)  5, 000, 000. 60. All five statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f ( x )  x3  3 x  1 is a point c where f (c)  0. (b) The point where y  x3 crosses y  3x  1 have the same y-coordinate, or y  x3  3 x  1  f ( x)  x3  3 x  1  0. (c) x3  3x  1  x3  3 x  1  0. The solutions to the equation are the roots of f ( x)  x3  3 x  1. (d) The points where y  x3  3 x crosses y  1 have common y-coordinates, or y  x3  3 x  1  f ( x)  x3  3 x  1  0. (e) The solutions of x3  3x  1  0 are those points where f ( x)  x3  3 x  1 has value 0. sin( x  2)

61. Answers may vary. For example, f ( x)  x  2 is discontinuous at x  2 because it is not defined there. However, the discontinuity can be removed because f has a limit (namely 1) as x  2. 62. Answers may vary. For example, g ( x) 

1 x 1

has a discontinuity at x  1 because lim g ( x) does not exist.

   lim  g ( x)   and lim  g ( x)  .  x 1  x 1 

x 1

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Section 2.5 Continuity

63. (a) Suppose x0 is rational  f ( x0 )  1. Choose   12 . For any   0 there is an irrational number x (actually infinitely many) in the interval ( x0   , x0   )  f ( x)  0. Then 0  |x  x0 |   but | f ( x)  f ( x0 )|  1  12  , so lim f ( x) fails to exist  f is discontinuous at x0 rational. x  x0

On the other hand, x0 irrational  f ( x0 )  0 and there is a rational number x in ( x0   , x0   )  f ( x)  1. Again lim f ( x) fails to exist  f is discontinuous at x0 irrational. That is, f is discontinuous at every point. x  x0

(b) f is neither right-continuous nor left-continuous at any point x0 because in every interval ( x0   , x0 ) or ( x0 , x0   ) there exist both rational and irrational real numbers. Thus neither limits lim f ( x) and x  x0

lim f ( x) exist by the same arguments used in part (a).

x  x0

64. Yes. Both f ( x)  x and g ( x)  x  12 are continuous on [0, 1]. However g

 12   0 

f ( x) g ( x)

is undefined at x 

1 2

since

is discontinuous at x  12 .

65. No. For instance, if f ( x)  0, g ( x)   x  , then h( x)  0   x    0 is continuous at x  0 and g ( x) is not. 66. Let f ( x)  x11 and g ( x)  x  1. Both functions are continuous at x  0. The composition f  g  f ( g ( x))  1  1 is discontinuous at x  0, since it is not defined there. Theorem 10 requires that f ( x ) be continuous ( x 1) 1 x at g (0), which is not the case here since g (0)  1 and f is undefined at 1.

67. Yes, because of the Intermediate Value Theorem. If f (a ) and f (b) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [a, b]. 68. Let f ( x ) be the new position of point x and let d ( x)  f ( x)  x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d ( x)  0 for some point in between. That is, f ( x)  x for some point x, which is then in its original position. 69. If f (0)  0 or f (1)  1, we are done (i.e., c  0 or c  1 in those cases). Then let f (0)  a  0 and f (1)  b  1 because 0  f ( x)  1. Define g ( x)  f ( x)  x  g is continuous on [0, 1]. Moreover, g (0)  f (0)  0  a  0 and g (1)  f (1)  1  b  1  0  by the Intermediate Value Theorem there is a number c in (0, 1) such that g (c)  0  f (c)  c  0 or f (c)  c. f (c )

70. Let   2  0. Since f is continuous at x  c there is a   0 such that x  c    f ( x)  f (c)    f (c)    f ( x)  f (c)  . If f (c)  0, then   12 f (c)  12 f (c)  f ( x)  23 f (c)  f ( x)  0 on the interval (c   , c   ). If f (c)  0, then    12 f (c) 

3 2

f (c )  f ( x ) 

1 2

f (c)  f ( x)  0 on the interval (c   , c   ).

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Chapter 2 Limits and Continuity

71. By Exercise 52 in Section 2.3, we have lim f ( x )  L  lim f (c  h )  L. x c

h 0

Thus, f ( x) is continuous at x  c  lim f ( x)  f (c)  lim f (c  h)  f (c). x c

h 0

72. By Exercise 71, it suffices to show that lim sin(c  h)  sin c and lim cos(c  h)  cos c. h 0

h 0

Now lim sin(c  h)  lim  (sin c)(cos h)  (cos c)(sin h)  (sin c)  lim cos h   (cos c)  lim sin h . h 0 h 0  h 0   h 0  By Example 11 Section 2.2, lim cos h  1 and lim sin h  0. So lim sin(c  h)  sin c and thus f ( x)  sin x is h 0

h 0

continuous at x  c. Similarly,

h 0

lim cos(c  h)  lim  (cos c)(cos h)  (sin c)(sin h)   (cos c)  lim cos h   (sin c)  lim sin h   cos c. Thus, h 0  h 0   h 0  g ( x)  cos x is continuous at x  c.

h 0

73. x  1.8794,  1.5321,  0.3473

74. x  1.4516,  0.8547, 0.4030

75. x  1.7549

76. x  1.5596

77. x  3.5156

78. x  3.9058, 3.8392, 0.0667

79. x  0.7391

80. x  1.8955, 0, 1.8955

2.6

LIMITS INVOLVING INFINITY; ASYMPTOTES OF GRAPHS

1. (a) lim f ( x)  0 (c) (e)

x 2

lim

x 3

(b)

f ( x)  2

lim f ( x)  1

(f)

x 0 

(h)

x 0

lim f ( x)  0

x 

2. (a) lim f ( x)  2 (c) (e) (g) (i)

(b)

x 4

lim f ( x)  1

x  2

lim

x 3

x 3

lim f ( x)  

x 0 

lim f ( x)  1

x 

lim f ( x )  3

x  2

(d) lim f ( x)  does not exist

f ( x)  

(f)

lim f ( x)  

(h)

lim f ( x)  

(j)

x 3

x 0 

(k) lim f ( x)  0

(l)

x 

1 m/ n x  x

Note: In these exercises we use the result lim

Theorem 8 and the power rule in Theorem 1: lim

x 2

lim

x  3

  1

lim f ( x)   

lim f ( x)  does not exist

x 0

lim f ( x)  1

x  m n

(b) 3

4. (a) 

(b)  (b)

 0. This result follows immediately from

 

1 x   x

 lim

3. (a) 3

1 2

f ( x)   

x  0

 0 whenever

m/n x   x

5. (a)

f ( x)  2

(d) lim f ( x)  does not exist

(g) lim f ( x)  does not exist (i)

lim

x 3

m/ n

    lim 1x    x  

m/ n

 0m / n  0.

1 2

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Section 2.6 Limits Involving Infinity; Asymptotes of Graphs

6. (a)

1 8

(b)

7. (a)  53 8. (a)

(b)  53

3 4

9.  1x 

(b)

sin 2 x x

10.  31 

cos  3



1 x

 lim

x 

sin 2 x x

 31  lim  

2t  sin t t  t  cos t

11. lim

 lim

t 

r sin r r  2 r  7 5sin r

12. lim

14. (a)

2 x3  7 3 2 x  x  x  x  7

 0 by the Sandwich Theorem

cos  3

 0 by the Sandwich Theorem

   010  1   10

r 

2 x 3 x  5 x  7

lim

   lim 10  1   r  200 2

sin r r 7 2  r 5 sinr r

1

2  3x

 lim

7 x  5 x

2 5



(b)

2  73  x  7 1 1 x  1 x  2  3

 lim

lim

(b) 2 (same process as part (a))

16. (a)

17. (a)

x 1 2 x  x 3

 lim

3x 7 2 x  x  2

 lim

lim

1 1 x x2

x  1

9 x4  x 4 2 x  2 x 5 x  x  6

19. (a)

10 x5  x 4 31 x6 x 

20.

x 3 7 x 2  2 2 x  x  x 1

(a) lim

0

(b) 0 (same process as part (a))

 lim

x 

 lim

x 

7 9

x3 2  52  13  64 x x x

10  1  31 x x 2 x6

0

 lim

21. (a)

3 x 7 5 x 2 1 3 x 6 x 7 x 3

 lim



9 2

(b)

9 2

(same process as part (a))

(b) 0 (same process as part (a))

 , since x  n  0 and x  7  .

x 7 2 x 1 1 2 x 1 x  x

x 3 7 x 2  2 2 x x  x 1

(b) 7 (same process as part (a))

x 7  2 x 1 1 2 1 x   x  x

(b)

lim

 lim

lim

2

(b) 0 (same process as part (a))

7 3 x  1 x 

lim

(same process as part (a))

0

 lim

18. (a)

lim

2 5

3 7 x x2

x  1

7 x3 2 x  x 3 x  6 x

lim

3 4

2 1 sin t t t 1 cost t

 lim

13. (a)

15. (a)

1 8

 , since x  n  0 and x  7  .

3 x 4  5 x  1  x 3 2 3 x 67 x 3 x

 , since x  n  0 and 3x 4  .

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Chapter 2 Limits and Continuity (b)

22. (a) (b)

23.

7 2 4 1 3 lim 3 x 35 x 1  lim 3 x 5x2  x3  , since x  n  0 and 3x 4  .

x 6 x 7 x 3

8 3 3 2 5 lim 5 x 2 x 5 9  lim 5 x 25 x 49 x  , since x  n  0, 5x 3  , and the denominator  4.

x 3 x 4 x

x 3 x 4 x

8 x 2 3 2 x2  x

lim

x 





25.

1 x3 2 x  x 7 x

26. lim

x 

 lim

x 1  x 4 2 3 x  x  x

 lim



x2 1 1x

x 

 lim

32.

x 5 x  3 2/3 x  2 x  x  4

33.

lim

x 

x 

x 2 1 x 1

lim

x 3

x 

1 1x  12  x   lim  x  8 32 x 

1/3

   



 18000 

1/3



 18 

1/3



1 2



 10000  0  0

28.

1 21   x /15  x  1 1   x 2 /15 

1  7 x19/15 x8/5 3 1 1 3/5  11/10 x x

x  2

1 4 x1/3 x

x 2/3

x 2 1/ x 2 2

lim 2 x x  2  x



 2  1  1/ 2  lim  x  x   2  1  x1/ 2 

 1

1

x 

x 2 1/ x 2

( x 2 1)/ x 2 ( x 1)/ x

11/ x 2 (1 x  1/ x )

 lim



1 0 (1 0)

1

2 ( x 1)/ x  lim ( x 1)/(  x )  lim ( 111/1/xx )  ( 1100)  1 x  x 

( x 3)/ x 2 2



  52

 lim

x  ( x 1)/ x 2

x 

1  2 x 2 x3

 lim

5 3x

 lim

x  1



0

 lim

1 5 x x2

lim

2 x1/15 

x  ( x 1)/ x

4 x  25

 42



x 

 lim

lim

1/3

   

8 0 2 0



x

2 x5/3  x1/3  7 8/5 x  x 3 x  x

x 2 1 x 1

1  2 x 2 x3

1 x (1/5)  (1/3) (1/5)  (1/3) x  1 x

31. lim

lim

x2 2 1x



 2   1   1/ 2   2  lim  x  7 x  3 x x 

x 5 x 3 5 x  x  x

35.

1

x 

29.

30. lim

x 

1 5 x x2

 lim

27.

8

lim

1 x  12  x    5 2  lim  x 7    lim x 7   01  0  x  1 x  x   1 x     

x 2 5 x x3  x  2



 1 1x  12 x  lim  3 x   8 2 x 

1 lim 2 3xx7x x 

lim

x2 2  1x





8

x 

1/3

3 x  x 4

x

 lim

24.

lim

3 x  x 4

x

3 2 5 lim 5 x 2 x 5 9  lim 5 x 25 x 49 x  , since x  n  0, 5x 3  , and the denominator  4. 8

2 lim x 2 x 1 x  8 x 3

34.

x 67 x 3 x

4 x  25 / x

 lim

x 

( x 3)/ x 2

(4 x  25)/ x

 lim

x 

(13/ x ) 4  25/ x



(10) 4 0



1 2

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Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 36.

37.

39.

41.

4 3 x 3

lim

 lim

x 

x 9

40.



negative positive



42.



positive positive



44. lim

lim x2x8   x 8 4 2 x 7 ( x 7)

4 2/5 x 0 x

 

4 1/5 2 x 0 ( x )

 lim

 2 

3 positive negative

x 3

lim x13  



positive positive

lim 2 x3x10   x 5



negative negative

1 2 x 0 x ( x1)



negative positivepositive

lim 21/3 x 0  3 x

(b)

2 lim 1/5 x 0  x

lim

 2 

x



 

  1 1/3 2 x 0 ( x )

 lim



sec x  

lim (1  csc  )  

 0 

lim (2  cot  )   and lim (2  cot  )  , so the limit does not exist

 0 

 0 

53. (a)

lim 21 x  2 x  4

 lim ( x  2)(1 x  2)   x 2

(b)

lim 21 x  2 x  4

 lim ( x  2)(1 x  2)   x 2

(c)

lim 21 x 2 x  4

 lim ( x  2)(1 x  2)   x 2

(d)

lim 21 x 2 x  4

 lim ( x  2)(1 x  2)   x 2

54. (a) (b) (c) (d)

55. (a)

lim 2x x 1 x 1 lim 2x x 1 x 1

 lim ( x 1)(x x 1) x 1

x 2 x 1 x 1 lim 2x x 1 x 1

lim

x 0





x2 2

        

 lim ( x 1)(x x 1)   x 1  lim ( x 1)(x x 1) x 1  lim ( x 1)(x x 1) x 1



 1x  0  lim 1x   x 0 

   

1 positivepositive

   

1 positivenegative 1 positivenegative

1 negativenegative

positive positivepositive

   

positive positivenegative negative positivenegative

negative negativenegative



1 negative





(b)

50.

(0 3) 1 0

lim 25x  

48. lim





x 0 

1 2/3 x 0 x



lim tan x   x

19/ x





47. lim

x 

positive negative

lim x 3 2  

2 lim 1/5 x 0  x

( x 9)/ x

( 4/ x3 3)

38.

x  2

46. (a)

 lim



positive positive



(43 x3 )/(  x3 )

x 



lim 21/3 x 0  3 x

52.

x 9 / x

 lim

lim 31x  

45. (a)

51.

x 0 

43. lim

49.

(4 3 x3 )/ x 6



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 



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Chapter 2 Limits and Continuity (b) (c) (d)

56. (a) (c) (d)

57. (a) (b) (c)

 lim  lim  lim

x 0 

x2 2

x 2

x2 2 2

x 2

x 1

    

22/3 2

1 x

1 2

lim 2xx 14  lim x 1 2 x  1 lim 2 x  4  41

x 1

1 positive



1  21/3  21/3  0  1/3 2

 

 11  23



lim 2xx 14   x 2

( x 1)( x 1) 2 x4

positive positive



(b)

 2204  0



2 ( x  2)( x1) lim x 33 x 22  lim   2

x 0



lim

x 2



x 2 x

x 0

x 2 3 x  2 x3  2 x 2

positive negative

x ( x  2)



 lim

( x  2)( x 1)

x 1 2 x 2 x

 lim

x 2 ( x  2)

x 2

negativenegative positivenegative



 14 , x  2

2 ( x  2)( x 1) lim x 33 x 22  lim  lim x 21  14 , x  2 2

x 2 x  2 x 2

x 3 x  2 3 2 x 2 x  2 x 2

x 3 x  2 3 2 x 0 x  2 x

(e) lim

x  2 x ( x  2) ( x  2)( x 1)

 lim  lim

 lim x 21  14 , x  2

x 2 ( x  2)

x 2

x 2 x

( x  2)( x 1)

x 0

x 2 ( x  2)

x  2 x



 

negativenegative positivenegative



2 ( x  2)( x 1) ( x 1) 1 1 lim x 3 3 x  2  lim x ( x  2)( x  2)  lim x ( x  2)  2(4) 8 x  2 x  4 x x  2 x  2

(b)

2 ( x  2)( x 1) ( x 1) lim x 3 3 x  2  lim x ( x  2)( x  2)  lim x ( x  2)    x 4 x x 2 x 2 x 2

(c)

2 ( x  2)( x 1) ( x 1) lim x 3 3 x  2  lim x ( x  2)( x  2)  lim x ( x  2)      x  4 x x 0 x 0 x 0

(d)

x 2 3 x  2 3 x 1 x  4 x

(e)



lim 2xx 14   x 2

x 0 

(d) lim

58. (a)



 1x  0  lim 1x   x 0 

lim

 lim

x 1

( x  2)( x 1) x ( x  2)( x  2)



lim x (xx12)  

x 0 

and lim x 0



x 1 x ( x  2)





 lim

x 1

( x 1) x ( x  2)

negative positivepositive

negative negativepositive





0 (1)(3)

 

negative negativepositive negative negativepositive

 

0



so the function has no limit as x  0. 59. (a) 60. (a)

3    lim  2  1/3  t 

(b)

1  7   lim  3/5  t 

(b)

t 0 

 1  lim  2/3  2 2/3    ( x 1)  x  1  (c) lim  2/3  2 2/3    ( x 1)  x 1  x

61. (a)

62. (a)

x 0 

 1  lim  1/3  1 4/3    ( x 1)  x

x 0 

3  lim  2  1/3  t 

t 0 

1  7    lim  3/5  t 

t 0 

 1  lim  2/3  2 2/3    ( x 1)  x  1  (d) lim  2/3  2 2/3    ( x 1)  x 1  x

(b)

x 0 

 1  lim  1/3  1 4/3    ( x 1)  x

x 0 

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

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Section 2.6 Limits Involving Infinity; Asymptotes of Graphs (c)

 1  lim  1/3  1 4/3    ( x 1)  x

(d)

x 1

 1  lim  1/3  1 4/3    ( x 1)  x

x 1

63. y 

1 x 1

64. y 

1 x 1

65. y 

1 2 x4

66. y 

3 x 3

67. y 

x 3 x2

68. y 

2x x 1

 1  x 1 2

69. domain  (, ); y in range and y  4 

3x2 , x 2 1

 2  x21

2 2 0  32x  3 and lim 32x  3  range  [4, 7) ; horizontal

x 1

x  x 1

asymptote is y  7 .

70. domain  (, 1)  (1, 1)  (1, ); y in range and y  lim 22 x x 1 x 1

 , and lim 22 x  ; x 1 x 1

lim 22 x x 1 x 1

2x ; x 2 1

if x  0, then y  0,

2x 2 x 1 x 1

  and lim

lim 22 x x  x 1

 0,

   range  (, ); horizontal

asymptote is y  0; vertical asymptotes are x  1, x  1

71. domain  (, ); y in range and y  x

lim 8e x x  2  e

8 e x 2 e x

x x ; if x  ln 8, then y  0, 1  8e x  4, lim 8e x  1, and

2 e

x  2  e

 4  range  (1, 4); horizontal asymptotes are y  1, y  4

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100

Chapter 2 Limits and Continuity

72. domain  (, ); y in range and y  x

lim 4ex  e2 x x  e  e

4e x  e 2 x e x  e2 x

x 2x x 2x , 1  4ex  e2 x  4, lim 4ex  e2 x  1, and

e e

 4  range  (1, 4) ; horizontal asymptotes are y  1, y  4

x2  4 , x

73. domain  (,0 )  (0,  ) ; y in range and y  lim

x 

x2  4 x

x  e  e

 1; if x  0, then

x2  4 x

 1,

lim

x 0



if x  0, then 1 

x2  4 x

x2  4 , x

 , and lim

x 

x2 4 x

lim

x 0 

x2  4 x

 , and

 1 

range  (, 1)  (1, ); horizontal asymptotes are y  1, y  1; vertical asymptote is x  0

74. domain  (, 2 )  (2,  ); y in range and y  x3 x3 8

x3 , lim x3 x 8 x  2  x 3 8 3

x3 x  2  x 8

 , lim

 ,

x3 x  x 8

lim

 1  range  (,1)  (1, ) ; horizontal asymptote is y  1; vertical asymptote is x  2

75. Here is one possibility.

76. Here is one possibility.

77. Here is one possibility.

78. Here is one possibility.

79. Here is one possibility.

80. Here is one possibility.

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 1, and

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Section 2.6 Limits Involving Infinity; Asymptotes of Graphs

81. Here is one possibility.

f ( x) x g ( x )

83. Yes. If lim

101

82. Here is one possibility.

f ( x)

 2 then the ratio the polynomials’ leading coefficients is 2, so lim g ( x )  2 as well. x

84. Yes, it can have a horizontal or oblique asymptote. f ( x) x  g ( x )

85. At most 1 horizontal asymptote: If lim f ( x) x  g ( x )

so lim

86.

 x  lim

 L, then the ratio of the polynomials’ leading coefficients is L,

 L as well.



( x 9) ( x  4) x  9  x  4  lim  x  9  x  4    x 9  x  4   lim   x   x  9 4   x  x  x 9  x  4

 lim

x 

5 x 9  x  4

5 x

 lim

 87. lim  x 2  25  x 2  1   lim  x 2  25  x 2  1       x     x   26

 lim

x 2  25  x 2 1

x 

88.

 lim

x 

 lim

x  1

3 x2



x x2

x  25  x 26 x

1 252  1 x

1 x2

  lim 1  x 

( x 2  25)  ( x 2 1)

x 2  25  x 2 1

 101  0



 3x

lim

x  1

3 x2

1

 101  0

2   (4 x 2 ) (4 x 2 3 x  2) lim  2 x  4 x 2  3 x  2   lim  2 x  4 x 2  3x  2    2 x  4 x 2 3 x  2   lim     2 x  4 x 3 x  2  x  2 x  4 x 2 3 x  2  x   x  

3 x  2

 lim

x  2 x  4 x 2 3 x  2

 lim

x  2 

90.

x 2  25  x 2 1 

 2  ( x 2 3) ( x 2 ) 3 lim  x 2  3  x   lim  x 2  3  x    x 2 3  x   lim  lim   x 3  x  x  x 2 3  x  x   x   x  x 2 3  x 3

89.

 101  0

1 9x  1 4x

x 

3 2x 4  3x  22 x

 lim

x 

3 x  2 x2

2x x2

 4  3x 

 lim

x 

3 x  2 x 2x  4  3x  22 x x

 3202   34

 2  (9 x 2  x )  (9 x 2 ) lim  9 x 2  x  3 x   lim  9 x 2  x  3 x    9 x 2  x 3 x   lim  lim    x    9 x  x 3 x  x  9 x 2  x 3 x x   x   lim

x 

 xx

9 x2  x x2 x2

 3xx

 lim

x 

1 9  1x 3

x 9 x 2  x 3 x

 313   16

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102

Chapter 2 Limits and Continuity

 91. lim  x 2  3 x  x 2  2 x   lim  x 2  3x  x 2  2 x     x     x   5x  lim  lim 2 2 x 

x 3 x  x  2 x

x2 3 x  5 x  1 3x  1 2x

 x 2  x  x 2  x  lim  x 2  x  x 2  x      x  

92. lim

x 

2 x  1 1x  1 1x

 lim

 121  1

x2 3 x  x2 2 x 

x2  x  x2  x  x2  x 

( x 2 3 x ) ( x 2  2 x )

  lim x 2  2 x  x   151  52

  lim x 2  x  x 

x 2 3 x  x 2  2 x

( x 2  x ) ( x 2  x )

 lim

x2  x  x2  x

x 

93. For any   0, take N  1. Then for all x  N we have that f ( x)  k  k  k  0  . 94. For any   0, take N  1. Then for all y   N we have that f ( x)  k  k  k  0  . 95. For every real number  B  0, we must find a   0 such that for all x, 0  x  0    Now,  

1 x2

1 x2

 B  0 

1 x2

B0 x 

1 B

 x 

  B so that lim  12  . x0

1 . B

Choose  

1 B

  B.

, then 0  x    x  1

96. For every real number B  0, we must find a   0 such that for all x, 0  x  0    1 x

1 x2

1 x

 B. Now,

 B  0  x  B1 . Choose   B1 . Then 0  x  0    x  B1  1  B so that lim 1  . x x0 x

97. For every real number  B  0, we must find a   0 such that for all x, 0  x  3    2 ( x 3) 2

( x 3) 2  B  0  2  B1  ( x  3)2 ( x 3) 2 2   B  0 so that lim 2  . 2 ( x 3) 2 x 3 ( x 3)

 B  0 

0  x 3  

 B2  0 

3 

2 ( x 3) 2

Now,

1 ( x  5)2

 x5 

 B  0  ( x  5)2  1 B



1 ( x 5)2

1 B

 x5  1 2 x 5 ( x 5)

 B so that lim

1 . Choose  B



2, B

1 ( x 5) 2

 B.

2 . Choose  B

98. For every real number B  0, we must find a   0 such that for all x, 0  x  (5)   

  B. Now, then

 1 . Then 0  x  (5)   B

 .

99. (a) We say that f ( x) approaches infinity as x approaches x0 from the left, and write lim f ( x)  , x  x0

if for every positive number B, there exists a corresponding number   0 such that for all x, x0    x  x0  f ( x)  B. (b) We say that f ( x) approaches minus infinity as x approaches x0 from the right, and write lim f ( x)  , x  x0

if for every positive number B (or negative number  B) there exists a corresponding number   0 such that for all x, x0  x  x0    f ( x )   B. (c) We say that f ( x) approaches minus infinity as x approaches x0 from the left, and write lim f ( x)  , if x  x0

for every positive number B (or negative number  B) there exists a corresponding number   0 such that for all x, x0    x  x0  f ( x)   B. 100. For B  0,

1 x

 B  0  x  B1 . Choose   B1 . Then 0  x    0  x  B1  1x  B so that lim x 0 

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1 x

 .

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Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 101. For B  0, 1x   B  0   1x  B  0   x   1x   B so that lim 1x  .

1 B

103

  B1  x. Choose   B1 . Then   x  0   B1  x

x 0 

102. For B  0, x 1 2   B   x 1 2  B  ( x  2)  B1  x  2   B1  x  2  B1 . Choose   B1 . Then 2    x  2    x  2  0   B1  x  2  0  x 1 2   B  0 so that lim x 1 2  . x  2

 B  0  x  2  B1 . Choose   B1 . Then 2  x  2    0  x  2    0  x  2  B1  x 1 2  B  0 so that lim x 1 2  .

103. For B  0,

1 x 2

x  2  1, 1 2  1 x

B  1  x 2  B1  (1  x)(1  x)  B1 . Now 12x  1 since x  1. Choose   21B . Then 1    x  1    x  1  0  1  x    21B  (1  x )(1  x )  B1 12x  B1  1 2  B for 0  x  1 and

104. For B  0 and 0  x x near 1 

105. y 

x2 x 1

107. y 

x2 4 x 1

109. y 

x 2 1 x

lim 1 2 x 1 1 x

 

 .

106. y 

x 2 1 x 1

 x  1  x21

 x  1  x31

108. y 

x 2 1 2 x4

 12 x  1  2 x3 4

 x  1x

110. y 

x3 1 x2

 x

 x  1  x11

1 x

1 x2

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104

111. y 

Chapter 2 Limits and Continuity x

112. y 

4 x 2

113. y  x 2/3 

1 x1/3

1 4 x 2

114. y  sin

  

x 2 1

115. (a) y   (see accompanying graph) (b) y   (see accompanying graph) (c) cusps at x   1 (see accompanying graph)

116. (a) y  0 and a cusp at x  0 (see the accompanying graph) (b) y  32 (see accompanying graph) (c) a vertical asymptote at x  1 and contains the



point 1,

3 23 4

 (see accompanying graph)

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Chapter 2 Practice Exercises

CHAPTER 2

PRACTICE EXERCISES

1. At x  1:

lim f ( x)  lim f ( x)  1

x  1

x  1

 lim f ( x)  1  f (1) x 1

 f is continuous at x  1.

At x  0 :

lim f ( x)  lim f ( x)  0

x 0 

x 0

 lim f ( x)  0. x 0

But f (0)  1  lim f ( x) x 0

 f is discontinuous at x  0. If we define f (0)  0, then the discontinuity at x  0 is removable.

At x  1:

lim f ( x)  1 and lim f ( x)  1

x 1

x 1

 lim f ( x ) does not exist x 1

 f is discontinuous at x  1.

2. At x  1:

lim f ( x)  0 and lim f ( x)  1

x  1

 lim f ( x) does not exist

x 1

x  1

 f is discontinuous at x  1.

At x  0 :

lim f ( x )   and lim f ( x)  

x 0 

x 0 

 lim f ( x) does not exist x 0

 f is discontinuous at x  0.

At x  1:

lim f ( x)  lim f ( x)  1  lim f ( x)  1.

x 1

x 1

x 1

But f (1)  0  lim f ( x) x 1

 f is discontinuous at x  1. If we define f (1)  1, then the discontinuity at x  1 is removable.

3. (a) lim (3 f (t ))  3 lim f (t )  3( 7)  21 t t0

t t0

  (b) lim ( f (t ))   lim f (t )   (7) 2  49 t t0  t t0  (c) lim ( f (t )  g (t ))  lim f (t )  lim g (t )  ( 7)(0)  0 2

t t0

f (t ) t t0 g (t ) 7

(d) lim



t t0 lim f (t )

t  t0

lim ( g (t ) 7)



t  t0

t t0 lim f (t ) t  t0

lim g (t )  lim 7

t  t0



7 07

1

  (e) lim cos ( g (t ))  cos  lim g (t )   cos 0  1 t t0 t  t  0 

(f)

lim | f (t )|  lim f (t )  | 7|  7

t t0

(g) lim ( f (t )  g (t ))  lim f (t )  lim g (t )  7  0  7 t t0

(h) lim

t t0

  1 f (t )

t t0

lim f (t )

t  t0



t t0

1 7



 17

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105

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106

Chapter 2 Limits and Continuity

4. (a) lim  g ( x)   lim g ( x)   2 x 0

x 0

(b) lim ( g ( x)  f ( x))  lim g ( x)  lim f ( x)  x 0

x 0

 2   12  

2 2

x 0

(d) lim f 1( x )  lim 1f ( x )  11  2 x 0 2 x 0

(e) lim ( x  f ( x))  lim x  lim f ( x)  0  12  12 x 0 x 0 x 0

(f)

f ( x )cos x x 1

lim

x 0



lim f ( x ) lim cos x

x 0

lim x lim 1

x0

x 0



 12  (1)   1 0 1

5. Since lim x  0 we must have that lim (4  g ( x))  0. Otherwise, if lim (4  g ( x)) is a finite positive number, x 0

x 0

4 g ( x ) 4 g ( x ) we would have lim  x    and lim  x    so the limit could not equal 1 as x  0. Similar       x 0 x 0 reasoning holds if lim (4  g ( x)) is a finite negative number. We conclude that lim g ( x)  4. x 0

x 0

6. 2  lim  x lim g ( x)   lim x  lim  lim g ( x)    4 lim  lim g ( x)    4 lim g ( x) (since lim g ( x) is a x0 x  4  x 0 x  4  x 0 x 0  x  4 x  4  x 0   constant)  lim g ( x )  24   12 . x 0

7. (a) lim f ( x)  lim x1/3  c1/3  f (c) for every real number c  f is continuous on ( , ). x c

x c

1 c 2/3

x c

1 c1/ 6

(b) lim g ( x)  lim x3/4  c3/4  g (c) for every nonnegative real number c  g is continuous on [0, ). (c) lim h( x)  lim x 2/3  (,  ). (d) lim k ( x)  lim x 1/6 

8. (a)

 h(c) for every nonzero real number c  h is continuous on (, 0) and  k (c ) for every positive real number c  k is continuous on (0, )

   n  12   ,  n  12   , where I  the set of all integers.

n I

(b)



(n , (n  1) ), where I  the set of all integers.

n I

(c) (,  )  ( , ) (d) (, 0)  (0, ) x2  4 x  4 ( x  2)( x  2)  lim x ( x  7)( x  2)  lim x (xx27) , x  2; the limit does not exist because 3 2 x  5 x  14 x x 0 x 0 x 0 lim x (xx27)   and lim x (xx27)   x 0  x 0  x2  4 x  4 ( x  2)( x  2) 0 0 lim 3 2  lim x ( x  7)( x  2)  lim x (xx27) , x  2, and lim x (xx27)  2(9) x  5 x  14 x x 2 x2 x 2 x 2

9. (a) lim

(b)

x ( x 1) x2  x  lim 3 2  lim 2 x 1  lim 2 1 , x  0 and x  1. 4 3 x 0 x  2 x  x x 0 x ( x  2 x 1) x 0 x ( x 1)( x 1) x 0 x ( x 1) 2 Now lim 2 1   and lim 2 1    lim 5 x 4x 3  .  x ( x 1)  x ( x 1) 2 x  x x x 0 x 0 x 0 2 x ( x 1) x  x 1 lim 5 4 3  lim 3 2  lim 2 , x  0 and x  1. The limit does not x 1 x  2 x  x x 1 x ( x  2 x 1) x 1 x ( x 1) lim 2 1   and lim 2 1  . x 1 x ( x 1) x 1 x ( x 1)

10. (a) lim

(b)

exist because

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Chapter 2 Practice Exercises 1 x x 1 1 x

1 x x 1 (1 x )(1 x )

x2 a2 4 4 x a x  a

2 2 2 x a ( x  a )( x  a )

 lim

( x  h)2  x 2 h x 0

 lim

14. lim

1 1 2 x 2

x 0

x1/3 1 x 1 x 1

17. lim

 lim (2 x  h)  h

tan 2 x x 0 tan  x

19. lim

( x3  6 x 2 12 x 8) 8 x x 0

  14

 lim ( x 2  6 x  12)  12 x 0

 lim

( x 1)( x 1)

2/3 1/3 x 1 ( x 1)( x  x 1)

x 1 2/3 1/3 x 1 x  x 1

 lim

 11111 

2 3

( x1/3  4)( x1/3  4) ( x1/3  4)( x1/3  4) ( x 2/3  4 x1/3 16)( x 8)  lim  x 8 x 8 ( x 8)( x 2/3  4 x1/3 16) x 64 x 64 1/3 1/3 ( x 64) ( x  4) ( x 8) ( x  4) ( x 8) (4  4) (88) lim  lim 2/3 1/3  161616  83 2/3 1/3 x 64 ( x 64) ( x  4 x 16) x 64 x  4 x 16

 lim

sin 2 x  cos  x x 0 cos 2 x sin  x

lim csc x  lim

x 

1 x 0 4  2 x

 lim

x  

21. lim sin

x 0

( x1/3 1) ( x 2/3  x1/3 1)( x 1)  2/3 1/3 x 1 ( x 1) ( x 1)( x  x 1)

x 2/3 16 x 8 x 64

( x 2  2 hx  h 2 )  x 2 h x 0

 lim

lim

 12

h 0

 lim



20.

1 2 2 xa x  a

 lim

 lim (2 x  h)  2 x

2 (2  x ) x 0 2 x (2 x )

(2  x )3 8 x x 0

 12

( x 2  2 hx  h 2 )  x 2 h h 0

 lim

16. lim

18.

( x  h)2  x 2 h h 0

13. lim

lim

 lim

( x2 a2 )

 lim

12. lim

15.

1 x 1 1 x

 lim

11. lim

1 sin x

 lim

x 0

cos  x x 2 x  1 1 1  2  2  sin2 x2 x  cos   2 x   sin  x    x 



 2x  sin x   sin  2  sin    sin  2   1

22. lim cos 2 ( x  tan x)  cos 2 (  tan  )  cos 2 ( )  (1)2  1 x 

8x x 0 3sin x  x

23. lim

24. lim  x 0

8 sin x x 0 3 x 1

 lim

cos 2 x 1 sin x



8 3(1) 1

4

 sin 2 x 2 x 1  lim cos 2 x 1  lim  cossin2xx1  cos cos 2 x 1  x 0 sin x (cos 2 x 1) x 0 sin x (cos 2 x 1) x 0 2

 lim

4sin x cos 2 x x 0 cos 2 x 1

 lim



4(0)(1) 2 11

0

25. Let x = t  3  lim ln(t  3)  lim ln x   t 3



x 0 



26. lim t 2 ln 2  t  ln1  0 t 1

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107

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108

Chapter 2 Limits and Continuity

 

27. 1  cos   1  e1  ecos( / )  e1   e 1   ecos( / )   e  lim

 0 

 ecos( / )  0 by the

Sandwich Theorem 28.

lim

2e1/ z

z 0 e1/ z

1

 lim

 120  2

1/ z

z  1 e

1/3

29.

  lim [4 g ( x)]1/3  2   lim 4 g ( x)    x 0  x 0 

30.

1 x  g ( x) x 5 3 x 2 1 x 1 g ( x )

32.

x 0 

 2  lim ( x  g ( x))  12  5  lim g ( x) 

lim

31. lim

 2  lim 4 g ( x)  8, since 23  8. Then lim g ( x )  2.

x 5

1 2

 lim g ( x)  12  5 x 5

   lim g ( x)  0 since lim (3 x 2  1)  4 x 1

5 x 2 x 2 g ( x )

x 1

 0  lim g ( x)   since lim (5  x 2 )  1

lim

x 2

33. (a) f (1)  1 and f (2)  5  f has a root between  1 and 2 by the Intermediate Value Theorem. (b), (c) root is 1.32471795724 34. (a) f (2)  2 and f (0)  2  f has a root between  2 and 0 by the Intermediate Value Theorem. (b), (c) root is 1.76929235424 35. At x  1:

lim

 lim

x 1 x ( x 2 1) x 2 1

x 1

lim



x 1

x ( x 2 1)

f ( x)  lim

2 x 1 | x 1|

 lim x  1, and x 1

f ( x)  lim

x ( x 2 1) 

x 1

| x 2 1|

 lim

 lim (  x)  ( 1)  1. Since lim x 1

lim

x 1

x ( x 2 1)

2 x 1 ( x 1)

x 1



f ( x) 

f ( x)  lim f ( x) does not exist, the x1

function f cannot be extended to a continuous function at x  1. At x  1:

lim f ( x)  lim

x 1

x ( x 2 1) 2

x 1 | x 1| x ( x 2 1)

 lim

x 1

x 2 1

 lim

x ( x 2 1) 2

x 1  ( x 1)

 lim ( x)  1, and lim f ( x)  lim x 1

x 1

x ( x 2 1)

2 x 1 | x 1|

 lim x  1. x 1

Again lim f ( x) does not exist so f cannot be extended to a continuous function at x  1 either. x 1

36. The discontinuity at x  0 of f ( x)  sin

lim sin 1x does not exist.  1x  is nonremovable because x 0

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Chapter 2 Practice Exercises 37. Yes, f does have a continuous extension at a  1: define f (1)  lim x 41  43 . x 1 x  x

38. Yes, g does have a continuous extension at a  2 :

 

5 cos  g 2  lim 4  2   54 .   2

39. From the graph we see that lim h(t )  lim h(t ) t 0 

t 0 

so h cannot be extended to a continuous function at a  0.

40. From the graph we see that lim k ( x)  lim k ( x) x 0 

x 0 

so k cannot be extended to a continuous function at a  0.

41.

43.

2 lim 52xx73  lim 7x  5200  52 x  x  5 x 2 lim x  43x 8  lim

x 



1 x  3 x

42.

2 lim 2 x2 3  lim

x  5 x  7

x 

2

20

x2 5  72 x

 5  0  52



 42  8 3  0  0  0  0 3x

44.

45.

lim 2 1 x  x 7 x 1

x2 7 1  1 x  x 2 x

 lim

 100 0  0

2 lim x x71 x  lim x 17   x  x  1 x

46.

x 4  x3 3 x  12 x 128

lim

x 1 128 x  12  3

 lim



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110 47. 48.

Chapter 2 Limits and Continuity lim sin x x   x 

 lim 1  0 since  x    as x    lim sin x  0. x   x  x   x 

cos  1

lim



 

 lim 2  0  lim cos 1  0.     sin x

2 x

49.

1 x  x2 x lim x sin  lim sin x x x  sin x  x  1 x

50.

  x 2/3  x 1  lim  1 x 5/3   1 0  1 lim 2/3 2 2 x  x  cos x x   1 cos x  1 0  x2/3 

51. 52.

53.

54.

 1100 0  1



lim e1/ x cos 1x  e0  cos(0)  1 1  1

x 

 

lim ln 1  1t  ln1  0

x

lim tan 1 x   2

x 



lim e3t sin 1 1t  0  sin 1 (0)  0  0  0

t 

2 2 is undefined at x  3 : lim xx 34   and lim xx 34   , thus x  3 is a vertical asymptote.   x 3 x 3 2 2 2 (b) y  x2  x  2 is undefined at x  1: lim x2  x  2   and lim x2  x  2  , thus x  1 is a vertical

55. (a) y 

x2  4 x 3

x  2 x 1

x 1 x  2 x 1

x 1 x  2 x 1

asymptote. 2 (c) y  x2  x 6 is undefined at x  2 and  4: lim

x 2  x 6 2 x 2 x  2 x 8

2  lim xx34  56 ; lim x2  x 6  lim xx34    x  2 x 8 x2 x 4 x  2 x 8 x 4 2 x  x  6 x  3 lim 2  lim x  4  . Thus x  4 is a vertical asymptote.

x 4 x  2 x 8

56. (a)

x 4

1 x 2 x 1 x  x 2 1

y  12 x : lim

 lim

1

1 x  1 2 x



1 x 2 2 x  x 1

1 1

 1 and lim

 lim

x  1

1 1 x2



1 1

 1, thus y  1 is a

horizontal asymptote. (b)

y

x 4 : x4 x2  4 x

lim

x 

lim

x 

x 4 x4

 lim

x2  4 x

1

4 x

 lim

x 

 lim

x 

1 0 1 0



x  1 4x

1



1

 1, thus y  1 is a horizontal asymptote.

1 0 1

 1 and lim

x 

x x

 lim

x 

1

1



1 0 1

x2 4 x



1 1

 lim

1

x 

x x2

 1,

thus y  1 and y  1 are horizontal asymptotes.

(d) y 

x 2 9 : 9 x 2 1

lim

x 

x 2 9 9 x 2 1

 lim

x 

1 9

thus y  13 is a horizontal asymptote.

x2 1 x2



1 0 9 0



1 3

and lim

x 

x 2 9 9 x 2 1

 lim

x 

1

x2 9  12 x



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1 0 9 0

 13 ,

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Chapter 2 Additional and Advanced Exercises 57. domain  [4, 2)  (2, 4]; y in range and y  16  x 2 x2

lim

x 2



ax 2  4 x b

x b 

x x  x b

a

x  x  x b

 lim

CHAPTER 2

1. (a) x x

lim

x2



16  x 2 x 2

 , and

4 x2

a

ax 2  4 x  x b

   vertical asymptote is x  b; lim  a  horizontal asymptote is y  a ,

4 x2

x  x  x b

 lim

a

4 x2

lim

x 

x  x  x b

 lim

a

4 x2

ax 2  4 x b

  a  horizontal asymptote is y   a

ADDITIONAL AND ADVANCED EXERCISES 0.1

, if x  4, then y  0,

 ,  range  (, )

58. Since lim  lim

16 x 2 x 2

0.01

0.001

0.0001 0.00001

0.7943 0.9550 0.9931 0.9991 0.9999

Apparently, lim x x  1 x 0 

(b)

2. (a) x

 1x 

1/(ln x )

1000

0.3679 0.3679 0.3679

1 x  x

Apparently, lim (b)

100

1/(ln x )

 0.3678  1e

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111

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112

Chapter 2 Limits and Continuity

v2 c2

lim L  lim L0 1 

v c 

 L0 1 

lim v 2

v c 2

 L0 1  c 2  0 c

The left-hand limit was needed because the function L is undefined if v  c (the rocket cannot move faster than the speed of light). 4. (a)

x 2

 1  0.2  0.2  2x  1  0.2  0.8  2x  1.2  1.6  x  2.4  2.56  x  5.76.

(b)

x 2

 1  0.1  0.1  2x  1  0.1  0.9  2x  1.1  1.8  x  2.2  3.24  x  4.84.

5. |10  (t  70)  104  10|  0.0005  |(t  70)  104 |  0.0005  0.0005  (t  70)  104  0.0005  5  t  70  5  65  t  75  Within 5 F. 6. We want to know in what interval to hold values of h to make V satisfy the inequality |V  1000|  |36 h  1000|  10. To find out, we solve the inequality: 990  h  1010  8.8  h  8.9 |36 h  1000|  10  10  36 h  1000  10  990  36 h  1010  36  36 where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe. The interval in which we should hold h is about 8.9  8.8  0.1 cm wide (1 mm). With stripes 1 mm wide, we can expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking. 7. Show lim f ( x)  lim ( x 2  7)  6  f (1). x 1

x 1

Step 1: |( x 2  7)  6|      x 2  1    1    x 2  1    1    x  1   . Step 2: | x  1|      x  1      1  x    1. Then   1  1   or   1  1   . Choose   min 1  1   , 1    1 , then 0  | x  1|   





|( x 2  7)  6|   and lim f ( x)  6. By the continuity text, f ( x ) is continuous at x  1. x 1

1 x  14 2 x

8. Show lim g ( x)  lim Step

x  14 1: 21x



 2  g 14 .

 2      21x  2    2    21x  2    412  x  412 .

 14      x  14      14  x    14 . Then   14  412    14  412  4(2 ) , or   14  412   

Step 2:

1 4  2

 14  4(2 ) . Choose   4(2 ) , the smaller of the two values. Then 0  x  14    21x  2   and lim 21x  2. x 1 4

By the continuity test, g ( x) is continuous at x  14 . 9. Show lim h( x)  lim 2 x  3  1  h(2). x 2

Step 1:

x2

2 x  3  1      2 x  3  1    1    2 x  3  1   

Step 2: | x  2|      x  2   or    2  x    2.

(1 )2 3 2

x

(1 )2 3 . 2

2 2 2 (1 ) 2 3 (1 ) 2 3 1(1 )2  2 , or   2  (1 ) 3    (1 ) 3  2  (1 ) 1    2      2 2 2 2 2 2 2 2 2   2 . Choose     2 , the smaller of the two values. Then, 0  | x  2|    2 x  3  1  ,

Then   2 

so lim 2 x  3  1. By the continuity test, h( x) is continuous at x  2. x2

10. Show lim F ( x)  lim 9  x  2  F (5). x 5

Step 1:

x 5

9  x  2      9  x  2    9  (2   )2  x  9  (2   ) 2 .

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

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Chapter 2 Additional and Advanced Exercises

113

Step 2: 0  | x  5|      x  5      5  x    5. Then   5  9  (2   )2    (2   )2  4   2  2, or   5  9  (2   ) 2    4  (2   ) 2   2  2. Choose    2  2, the smaller of the two values. Then, 0  | x  5|    9  x  2  , so lim 9  x  2. x 5

By the continuity test, F ( x) is continuous at x  5.

11. Suppose L1 and L2 are two different limits. Without loss of generality assume L2  L1. Let   13 ( L2  L1 ). Since lim f ( x)  L1 there is a 1  0 such that 0  | x  x0 |  1  | f ( x )  L1 |      f ( x )  L1   x  x0

  13 ( L2  L1 )  L1  f ( x)  13 ( L2  L1 )  L1  4 L1  L2  3 f ( x)  2 L1  L2 . Likewise, lim f ( x)  L2 so x  x0

there is a  2 such that 0  | x  x0 |   2  | f ( x )  L2 |      f ( x)  L2     13 ( L2  L1 )  L2  f ( x)  13 ( L2  L1 )  L2  2 L2  L1  3 f ( x)  4 L2  L1  L1  4 L2  3 f ( x)  2 L2  L1. If   min{1 ,  2 } both inequalities must 4 L  L  3 f ( x)  2 L1  L2  hold for 0  | x  x0 |   : 1 2   5( L1  L2 )  0  L1  L2 . That is, L1  L2  0 and L1  4 L2  3 f ( x)  2 L2  L1  L1  L2  0, a contradiction. 12. Suppose lim f ( x)  L. If k  0, then lim k f ( x)  lim 0  0  0  lim f ( x) and we are done. If k  0, then given x c

x c

any   0, there is a   0 so that 0  | x  c |     | f ( x)  L |  | k |  | k || f ( x )  L |    | k ( f ( x)  L)|  

 |(kf ( x))  (kL)|  . Thus lim k f ( x)  kL  k  lim f ( x)  . x c  x c 

13. (a) Since x  0 , 0  x3  x  1  ( x3  x)  0  lim f ( x3  x)  lim f ( y )  B where y  x3  x. 

x 0 



y 0

(b) Since x  0 , 1  x  x  0  ( x  x)  0  lim f ( x  x)  lim f ( y )  A where y  x3  x. 3



x 0 



y 0



x 0  4

y 0 2



(d) Since x  0 ,  1  x  0  0  x  x  1  ( x  x )  0  lim f ( x  x 4 )  A as in part (c). 4

x 0 

14. (a) True, because if lim ( f ( x)  g ( x)) exists then lim ( f ( x)  g ( x))  lim f ( x)  lim [( f ( x)  g ( x))  f ( x)]  x a

lim g ( x) exists, contrary to assumption.

x a

(b) False; for example take f ( x) 





1 x

x a

and g ( x)   1x . Then neither lim f ( x) nor lim g ( x) exists, but x 0

x 0

lim ( f ( x)  g ( x))  lim 1x  1x  lim 0  0 exists. x 0 x 0 x 0 (c) True, because g ( x) | x | is continuous  g ( f ( x))  | f ( x)| is continuous (it is the composite of continuous functions). 1, x  0 (d) False; for example let f ( x)    f ( x) is discontinuous at x  0. However | f ( x)| 1 is  1, x  0 continuous at x  0. x 2 1 x 1 x 1

15. Show lim f ( x )  lim x 1

 lim

x 1

( x 1)( x 1) ( x 1)

 2, x  1.

 x 2 1 , x  1 . We now prove the limit of f ( x) as x  1 Define the continuous extension of f ( x) as F ( x )   x 1  2 , x  1

exists and has the correct value. Step 1:

x 2 1  ( 2) x 1

    

( x 1)( x 1) ( x 1)

 2      ( x  1)  2  , x  1    1  x    1.

Step 2: | x  (1)|     x  1      1  x    1. Copyright  2018 Pearson Education, Inc.

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114

Chapter 2 Limits and Continuity Then   1    1    , or   1    1    . Choose   . Then 0  | x  (1)|   2  xx 11  (2)    lim F ( x)  2. Since the conditions of the continuity test are met by F ( x), then f ( x) has x 1 a continuous extension to F ( x) at x  1.

x 2  2 x 3 x 3 2 x  6

16. Show lim g ( x)  lim x 3

( x 3)( x 1) x 3 2( x 3)

 lim

 2, x  3.

 x 2  2 x 3 , x  3 . We now prove the limit of g ( x) as x  3 Define the continuous extension of g ( x) as G ( x)   2 x 6 , x3 2 exists and has the correct value.

Step 1:

x 2  2 x 3  2 2 x 6

    

( x 3)( x 1) 2( x 3)

 2      x21  2  , x  3  3  2  x  3  2.

Step 2: | x  3|      x  3    3    x    3. Then, 3    3  2    2, or   3  3  2    2. Choose   2. Then 0  | x  3|  

2 ( x 3)( x 1)  x 2x2x63  2    lim 2( x 3)  2. Since the conditions of the continuity test hold for G ( x), g ( x) can be x 3 continuously extended to G ( x ) at x  3.

17. (a) Let   0 be given. If x is rational, then f ( x)  x  | f ( x)  0|  | x  0|   | x  0|  ; i.e., choose   . Then | x  0|    | f ( x)  0|   for x rational. If x is irrational, then f ( x)  0  | f ( x)  0|    0   which is true no matter how close irrational x is to 0, so again we can choose   . In either case, given   0 there is a     0 such that 0  | x  0|    | f ( x)  0|  . Therefore, f is continuous at x  0. (b) Choose x  c  0. Then within any interval (c   , c   ) there are both rational and irrational numbers. If c is rational, pick   2c . No matter how small we choose   0 there is an irrational number x in (c   , c   )  | f ( x)  f (c)|  |0  c |  c  2c  . That is, f is not continuous at any rational c  0. On the other hand, suppose c is irrational  f (c)  0. Again pick   2c . No matter how small we choose   0 there is a rational number x in (c   , c   ) with | x  c |  2c    2c  x  32c . Then | f ( x)  f (c)|  | x  0|  | x| 

c 2

   f is not continuous at any irrational c  0.

If x  c  0, repeat the argument picking   value x  c.

|c| 2

 2c . Therefore f fails to be continuous at any nonzero

18. (a) Let c  mn be a rational number in [0, 1] reduced to lowest terms  f (c)  1n . Pick   21n . No matter how small   0 is taken, there is an irrational number x in the interval (c   , c   )  | f ( x)  f (c)|  0  1n  1n  21n  . Therefore f is discontinuous at x  c, a rational number.

(b) Now suppose c is an irrational number  f (c)  0. Let   0 be given. Notice that 12 is the only rational number reduced to lowest terms with denominator 2 and belonging to [0, 1]; 13 and 23 the only rationals with denominator 3 belonging to [0, 1]; 14 and 34 with denominator 4 in [0, 1]; 15 , 52 , 53 and 54 with denominator 5 in [0, 1]; etc. In general, choose N so that

1 N

   there exist only finitely many rationals in [0, 1] having

denominator  N , say r1 , r2 , , rp . Let   min {| c  ri |: i  1,  , p}. Then the interval (c   , c   ) contains no rational numbers with denominator  N . Thus, 0  | x  c |    | f ( x)  f (c)|  | f ( x )  0|  | f ( x)|  N1    f is continuous at x  c irrational.

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Chapter 2 Additional and Advanced Exercises

115

(c) The graph looks like the markings on a typical ruler when the points ( x, f ( x)) on the graph of f ( x ) are connected to the x-axis with vertical lines.

19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero point, 0, on the equator  0   R represents the midnight point (at the same exact time). Suppose x1 is a point on the equator “just after” noon  x1   R is simultaneously “just after” midnight. It seems reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T ( x1 )  T ( x1   R )  0. At exactly the same moment in time pick x2 to be a point just before midnight  x2   R is just before noon. Then T ( x2 )  T ( x2   R )  0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c between 0 (noon) and  R (simultaneously midnight) such that T (c)  T (c   R )  0; i.e., there is always a pair of antipodal points on the earth’s equator where the temperatures are the same. 1  ( f ( x )  g ( x )) 2 x c 4  1 (32  ( 1) 2 )  2. 4

20. lim f ( x ) g ( x)  lim x c



 ( f ( x)  g ( x))2   



(b) At x





( x )  g ( x))    lim ( f ( x)  g ( x))    x c 



2

 

1(1 a ) 1 1 a  lim 1 a1 a 1 1 a  lim  1  12 a    1 1 a a  1 1 0 ( 1  1 a ) a 0 a 0 a 0 a 0 1(1 a )  a  1  1: lim r (a )  lim  lim  1 1 0 a 1 a 1 a ( 1 1 a ) a 1 a ( 1  1 a ) 1 (1 a ) a  0: lim r (a )  lim 1 a1 a  lim 1 a1 a 1 1 a  lim  lim     a ( 1 1 a )    a 1 1 a 0 a 0 a 0 a 0 a 0 a ( 1 1 a ) 1  lim   (because the denominator is always negative); lim r (a) a 0 1 1 a a 0   1  lim   (because the denominator is always positive). a 0 1 1 a

21. (a) At x  0: lim r (a )  lim At x



1  lim ( f 4  x c







Therefore, lim r (a ) does not exist. a 0

At x  1: lim r (a )  lim a 1

a 1

1 1 a a

1 a 1 1 1 a

 lim

1

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116

Chapter 2 Limits and Continuity (c)

(d)

22. f ( x)  x  2 cos x  f (0)  0  2 cos 0  2  0 and f ( )    2 cos( )    2  0. Since f ( x ) is continuous on [ , 0], by the Intermediate Value Theorem, f ( x ) must take on every value between [  2, 2]. Thus there is some number c in [ , 0] such that f (c)  0; i.e., c is a solution to x  2 cos x  0. 23. (a) The function f is bounded on D if f ( x)  M and f ( x)  N for all x in D. This means M  f ( x)  N for all x in D. Choose B to be max {| M |, | N |}. Then | f ( x)|  B. On the other hand, if | f ( x)|  B, then  B  f ( x)  B  f ( x)   B and f ( x)  B  f ( x) is bounded on D with N  B an upper bound and M   B a lower bound. (b) Assume f ( x)  N for all x and that L  N . Let   L 2N . Since lim f ( x)  L there is a   0 such that x  x0

0  | x  x0 |    | f ( x)  L |    L    f ( x )  L    L  L 2N  f ( x)  L  L 2N  L 2N  f ( x)  3 L2 N . But L  N  L 2N  N  N  f ( x) contrary to the boundedness assumption f ( x)  N . This contradiction proves L  N . (c) Assume M  f ( x) for all x and that L  M . Let   M2 L . As in part (b), 0  | x  x0 |    L  M2 L  f ( x)  L  M2 L  3L 2 M  f ( x) 

M L 2

 M , a contradiction.

25.

lim 

x 0

sin(1cos x ) x

sin(1cos x ) 1cos x 1 cos x  x  1 cos x x 0 1cos x

 lim



a b 2

| a b |

 2  a 2 b  a 2 b  22a  a. |a b| If a  b, then a  b  0  | a  b |  (a  b)  b  a  max {a, b}  a 2 b  2  a 2 b  b 2 a  22b  b. | a b | (b) Let min {a, b}  a 2 b  2 .

24. (a) If a  b, then a  b  0  | a  b |  a  b  max {a, b} 

2 sin(1cos x ) x  lim x1(1cos  x x) 1 cos cos x 0 x 0

 lim

2 x  lim sin x  sin x  1  0  0.  1  lim x (1sin x ) x 0 x cos 1 cos x 2  x 0

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Chapter 2 Additional and Advanced Exercises 26.

lim sin x x 0 sin x

 lim

x 0

sin(sin x ) x

 lim

sin( x 2  x ) x x 0

 lim

sin( x 2  4) x2 x 2

 lim

sin( x 3) x 9 x 9

 lim

27. lim

x 0

x sin x  x sin x

x 0

28. lim

29. lim

x x

sin(sin x ) sin x  x sin x sin( x 2  x ) 2

x x

x 0

30. lim



sin( x 2  4) 2

x 4

x 2

 1  lim

x 0 

 lim

x 0



 lim

x  1 1  0  0.

x 0 

sin(sin x )  lim sinx x sin x x 0

 ( x  1)  lim

sin( x 2  x )

x2  x

x 0

sin( x 2  4)

 ( x  2)  lim

x2 4

x2

sin( x 3)  1 x 3 x 3 x 9



1 sin x x

 1 1  1.

 lim ( x  1)  1 1  1. x 0

 lim ( x  2)  1  4  4. x2

sin( x 3)  lim x 3 x 9 x 9

 lim

117

1 x 3

 1  16  16 .

31. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is 3/ 2 an oblique asymptote. y  2 x  2 x 3  2 x  3 , thus the oblique asymptote is y  2 x. x 1

32. As x  ,

1 x

x 1



 



   x; thus

 0  sin 1x  0  1  sin 1x  1, thus as x  , y  x  x sin 1x  x 1  sin 1x

the oblique asymptote is y  x.

33. As x   , x 2  1  x 2  x 2  1  x 2 ; as x  , x 2   x, and as x   , x 2  x; thus the oblique asymptotes are y  x and y   x. 34. As x   , x  2  x  x 2  2 x  x( x  2)  x 2 ; as x  , x 2   x, and as x   , x 2  x; asymptotes are y  x and y   x. 35. Assume 1  a  b and

a x

x

1 x b

 a ( x  b)  x 2 ( x  b)  x  f ( x)  a( x  b)  x 2 ( x  b)  x  0; f is

continuous for all x-values and f (0)  ab  0, f (a  b)  a 2  (a  b)2 a  (a  b)  a2 a  (a  b) 2 a  b  0.    ()

()

Thus, by the Intermediate Value Theorem there is at least one number c, 0  c  a  b, so that f (c)  0  a(c  b)  c 2 (c  b)  c  0  ac  c  c 1b .

36. (a)

lim

x 0

a bx 1 x

 " a01" , so

1bx 1 1bx 1  x 1bx 1 x 0

a  1  0  a  1, then lim

b  b 2b4 x 0 1 bx 1 2 tan( ax  a ) b  2 "tan 0 b  2" "b  2" lim   0 , x 1 0 x 1

(1bx ) 1 ( 1bx 1) x x 0

 lim

(b)

tan( ax  a ) x 1 x 1

so b  2  0  b  2, then lim

sin a ( x 1)

 lim cos aa( x 1)  a ( x 1)  cosa 0 1  a  3 x 1

37.

tan a ( x 1)

 lim a  a ( x 1) x 1

1/6 4 4 1/6 1/6 1/6 2 1/6 1/3 2/3 1  lim ( x ) 1  lim ( x 1)( x 1)(( x ) 1)  lim ( x 1)( x 1)  (2)(2)  4 lim x 1/2 1/6 1/6 1/6 2 1/6 1/3 3 1/6 3 3 3 x 1 1 x x 1 1 ( x ) x 1 (1 x )(1 ( x )  ( x ) ) x 1 1 x  x

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118 38.

Chapter 2 Limits and Continuity lim

3x 4  x 4 x

lim

4x x

x 0





 lim x 0



(3 x  4)  x  4 x

3x  4  x 4 x x 0

 4  lim

3x  4  x 4 (3 x  4) (  x )  4  lim 22x  2; assume x  43  lim  lim x x    x 0 x 0 x 0

does not exist.

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