Solutions for Thomas Calculus Early Transcendentals 14th Us Edition by Hass by 6alsm

CHAPTER 2 LIMITS AND CONTINUITY

2.1 RATES OF CHANGE AND TANGENTS TO CURVES

1. (a) (3)(2) 289 321 19 fff x  

2. (a) (3)(1)() 3 1 312 2 ggg x

3. (a)

3 44 3 442 114

4. (a) ()(0)(21)(21) 2 00 ggg t  

5. (2)(0) 811 31 2022 1 RRR 

6. (2)(1)(81610)(145) 211 22 0 PPP

7. (a) 22 2 ((2)5)(25) 4451yhhh xhh

(1)(1) 20 1(1)2 1 fff x

(4)(2) 88 4(2)6 0 ggg x

26 263 0333

()()(21)(21) ()2 0 ggg

at (2,1) P the slope is 4. (b) (1)4(2)148

8. (a) 22 2 (7(2))(72) 7443yhhh

9. (a)

22 ((2)2(2)3)(22(2)3)yhh xh

13. (a) 33 (1)12(1)(112(1))yhh xh

As 0, h  2 93hh  9  at (1,11) P the slope is 9 (b) (11)(9)(1)119 yxyx

14. (a) 3232 (2)3(2)4(23(2)4)yhh xh

15. (a) 11 22 2(2) 11 2(2)2(2)

11111 24242 ((2)) yxyx

16. (a)

(4) 4 2(4)24 42(2) 4 21111 21222 .

17. (a) (4)4 444242 1 42(42)42

(b) 111 444 2(4)211

18. (a) 7(2)7(2) (9)9 939393 1 93(93)93 h yh

As 0, h  111 6 9393 , h    at (2,3) P the slope is 1 6 (b) 8 1111 3((2))366363 yxyxyx

19. (a) Q Slope of p t PQ   

1 (10,225) Q 650225 2010 42.5m/sec  2 (14,375) Q 650375 2014 45.83m/sec 

3 (16.5,475) Q 650475 2016.5 50.00m/sec  4 (18,550) Q 650550 2018 50.00m/sec  (b) At 20, t  the sportscar was traveling approximately 50 m/sec or 180 km/h.

20. (a) Q Slope of p t PQ   

1 (5,20) Q 8020 105 12m/sec  2 (7,39) Q 8039 107 13.7m/sec 

3 (8.5,58) Q 8058 108.5 14.7m/sec 

4 (9.5,72) Q 8072 109.5 16m/sec 

(b) Approximately 16 m/sec

21. (a)

(b) 17462 112 201420122 56 p t

thousand dollars per year

thousand dollars per year.

The average rate of change from 2012 to 2013 is 11162 20132012 49

thousand dollars per year.

So, the rate at which profits were changing in 2012 is approximately 1 2 (3549)42  thousand dollars per year.

22. (a) ()(2)/(2)Fxxx

(b) The rate of change of ()Fx at 1 x  is 4

23. (a) (2)(1) 21 2121 0.414213 ggg x

(1)(1) 11 (1)1 gghgh xhh

(b) () gxx 

(d) The calculator gives 11 1 2 0 lim. h hh

24. (a) i) 111 326 (3)(2) 1 32116 ff

(d)

NOTE: Answers will vary in Exercises 25 and 26. 25. (a) 150

(b) At  1 2 ,7.5: P Since the portion of the graph from 0 t  to 1 t  is nearly linear, the instantaneous rate of change will be almost the same as the average rate of change, thus the instantaneous speed at 1 2 t  is 157.5 10.5  15mi/hr. At (2,20): P Since the portion of the graph from 2 t  to 2.5 t  is nearly linear, the instantaneous rate of change will be nearly the same as the average rate of change, thus 2020 2.52 0mi/hr. v  For values of t less than 2, we have

Q Slope of s t PQ   

1 (1,15) Q 1520 12 5mi/hr 

2 (1.5,19) Q 1920 1.52 2mi/hr 

3 (1.9,19.9) Q 19.920 1.92 1mi/hr 

Thus, it appears that the instantaneous speed at 2 t  is 0 mi/hr. At (3,22) : P

Q Slope of s t PQ   

1 (4,35) Q 3522 43 13mi/hr 

2 (3.5,30) Q 3022 3.53 16mi/hr 

3 (3.1,23) Q 2322 3.13 10mi/hr 

Thus, it appears that the instantaneous speed at 3 t  is about 7 mi/hr.

Q Slope of s t PQ   

1 (4,35) Q 3530 43.5 10mi/hr 

2 (3.75,34) Q 3430 3.753.5 16mi/hr 

3 (3.6,32) Q 3230 3.63.5 20mi/hr 

26. (a) gal 1015 30day [0,3]:1.67; A

(b) At (1,14): P Q Slope of A t PQ

Slope of s t PQ    1 (3,22) Q 2230 33.5 16mi/hr 

2 (3.25,25) Q 2530 3.253.5 20mi/hr  3 (3.4,28) Q 2830 3.43.5 20mi/hr 

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

(1.1,13.85)

Thus, it appears that the instantaneous rate of consumption at 1 t  is about 1.45gal/day. At (4,6): P Q Slope of A t PQ    1 (5,3.9) Q

Q Slope of A t PQ    1 (0,15) Q 1514 01 1gal/day 2 (0.5,14.6) Q 14.614 0.51 1.2gal/day 3 (0.9,14.86) Q 14.8614 0.91 1.4gal/day Q Slope of A t PQ    1 (3,10) Q 106 34 4gal/day

2 (3.5,7.8) Q 7.86 3.54 3.6gal/day 3 (3.9,6.3) Q 6.36 3.94 3gal/day

Thus, it appears that the instantaneous rate of consumption at 1 t  is 3gal/day.

(solutioncontinuesonnextpage) Q Slope of s t PQ    1 (2,20) Q 2022 23 2mi/hr  2 (2.5,20) Q 2022 2.53 4mi/hr  3 (2.9,21.6) Q 21.622 2.93 4mi/hr 

At (8,1) : P

1 (9,0.5) Q

Slope of A t PQ

1 (7,1.4) Q 1.41 78 0.6gal/day

2 (7.5,1.3) Q 1.31 7.58 0.6gal/day

3 (7.9,1.04) Q 1.041 7.98 0.6gal/day

Thus, it appears that the instantaneous rate of consumption at 1 t  is 0.55gal/day. (c) It appears that the curve (the consumption) is decreasing the fastest at 3.5. t  Thus for (3.5,7.8) P



Slope of s t PQ 

1 (2.5,11.2) Q 11.27.8 2.53.5 3.4gal/day 

2 (3,10) Q 107.8 33.5 4.4gal/day

3 (3.4,8.2) Q 8.27.8 3.43.5 4gal/day

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

2.2 LIMIT OF A FUNCTION AND LIMIT LAWS

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

2. (a) 0 (b) 1

(c) Does not exist. As t approaches 0 from the left, ()ft approaches 1. As t approaches 0 from the right, ()ft approaches 1. There is no single number L that ()ft gets arbitrarily close to as 0 t  (d) 1

3. (a) True

(b) True (c) False (d) False (e) False (f) True (g) True (h) False (i) True (j) True (k) False

4. (a) False

(b) False (c) True (d) True (e) True (f) True (g) False (h) True (i) False

5. || 0 lim x xx  does not exist because || 1 xx xx  if 0 x  and || 1 xx xx  if 0. x  As x approaches 0 from the left, || x x approaches 1. As x approaches 0 from the right, || x x approaches 1. There is no single number L that all the function values get arbitrarily close to as 0 x 

6. As x approaches 1 from the left, the values of 1 1 x become increasingly large and negative. As x approaches 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 1, x  so 1 1 1 lim xx  does not exist.

7. Nothing can be said about ()fx because the existence of a limit as 0xx  does not depend on how the function is defined at 0x In order for a limit to exist, ()fx must be arbitrarily close to a single real number L when x is close enough to 0x That is, the existence of a limit depends on the values of ()fx for x near 0 , x not on the definition of ()fx at 0x itself.

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

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

10. No, because the existence of a limit depends on the values of ()fx when x is near 1, not on (1) f itself. If 1 lim() x fx  exists, its value may be some number other than (1)5. f  We can conclude nothing about 1 lim(), x fx  whether it exists or what its value is if it does exist, from knowing the value of (1) f alone. 11. 22 3

13. 6 lim8(5)(7)8(65)(67)8

15. 33 2(2)5 25 1111(2) 2

2241 3333 2/3

25. 2 (5)(2) 310 55 555 limlimlim xxxx xxxxx

26. 2 (5)(2) 710 22 222

223

28. 2 2 (2)(1) 322 (2)(1)2 2 111

29. 3222 2(2)

30. 322 42222 58(58)58 316(316)316 000

32 (1)(1)(1) 1 1(1)(1) 111 limlimlim uuuu uuuuuuu

34. 2 3 42 (2)(24) 8 16(2)(2)(4) 222 limlimlim vvvv vvvvvvv

2 2 243 44412 (4)(8)328 (2)(4)

35. 33 111 96 999(3)(3)393 limlimlim xx xxxxxxx

36. 2 (4)(2)(2) 4 222 444 limlimlimxxxxxxx xxxx xx

37.

(1)32 1 32 3232 111 limlimlim xx x xxxxxx

22 2 2 8383 83 1 11 (1)83

43. 0 lim(2sin1)2sin01011 x x

45. 111 coscos01 00 limseclim1 xx

47. 1sin10sin0100 1 3cos3cos033 0 lim xx xx

48. 22 0 lim(1)(2cos)(01)(2cos0) x xx 

49. lim4cos()lim4lim xxx xxx

50. 22 00 lim7seclim(7sec) xx xx

51. (a) quotient rule

2 222 00 limsinlimsin(sin0)00 xx xx

(b) difference and power rules (c) sum and constant multiple rules

52. (a) quotient rule

(b) power and product rules (c) difference and constant multiple rules

53. (a) lim()()lim()lim() xcxcxc fxgxfxgx

(b) lim2()()2lim()lim() xcxcxc fxgxfxgx

(d) lim() () 55 ()()lim()lim()5(2)7 lim

54. (a) 444 lim[()3]lim()lim3330 xxx gxgx 

(b) 444 lim()limlim()(4)(0)0 xxx xfxxfx

(d) 4 44 lim() () 3 ()1lim()lim101 4 lim3 x xx gxgx xfxfx

55. (a) lim[()()]lim()lim() xbxbxb fxgxfxgx

(b) lim()()lim()lim() xbxbxb fxgxfxgx 

58. 22 2 (2)(2) 444 000 limlimlim hhh hh hhh

59. [3(2)4][3(2)4] 3 00 limlim3 hh hhhh

60.

61.

11 2 22 2 1 2(2) 00022(2) limlimlimhhh hhhhhhh

7777 77 77 00 limlimhhh hhhhh

63. 22 0 lim5252(0)5 x x

64. 2 0 lim(2)202 x x 

65. (a)

and 0 lim x

and 0 lim2cos x x

2 0 66 0 lim11 1 x x

by the sandwich theorem, 0 lim()5 x fx

 2(1)2;  by the sandwich theorem, 0 lim()2 x gx

and 0 lim11; x   by the sandwich theorem, sin 22cos 0 lim1 xx xx  

(b) For 0,(sin)/(22cos) xyxxx   lies between the other two graphs in the figure, and the graphs converge as 0 x 

66. (a)   22 1111 00022422422 limlimlim0 xx

(b) For all 0, x   the graph of 2 ()(1cos)/ fxxx  lies between the line 1 2 y  and the parabola 2 1 2 /24, yx  and the graphs converge as 0. x 

67. (a) 2 ()(9)/(3) fxxx

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

68. (a)  2 ()(2)/2 gxxx x

70. (a) 22 ()(23)/(43) hxxxxx  x 2.9

(b) (c) 2 2 (3)(1) 231 (3)(1)1 43 () xxxxx xxxx x hx

74. (a) 2 ()(1cos)/ Gttt 

Gt  

(b)

75. (a) 1/(1) () fxxx 

x fx  

(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).

76. (a) ()(31)/ x fxx 

77. lim() xc fx  exists at those points c where 4

78. Nothing can be concluded about the values of ,,fg and h at 2 x  Yes, (2) f could be 0. Since the conditions of the sandwich theorem are satisfied, 2 lim()50. x fx

79. 444 44 lim()lim5lim()5 ()5

80. (a) 22 22 2 lim()lim() () 4 lim 2 1 lim xx x fxfxfx xxx

83. (a) 1 0 limsin0 xx x  

(b)

1sin1

84. (a)

by the sandwich theorem; 11

by the sandwich theorem.

(b)

85–90. Example CAS commands: Maple:

f:x-(x^416)/(x2);  x0:2; 

plot(f(x),xx0-1..x01,colorblack, 

title"Section 2.2,85(a); #" ) 

limit(f(x),xx; 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(x1,3)1)/x. 

Mathematica: (assigned function and values for x0 and h may vary)

Clear[f,x]

322 f[x_]:(xx5x3)/(x1)  x01;h0.1; 

Plot[f[x],{x,x0h,x0h}] 

Limit[f[x],xx0] 

2.3 THE PRECISE DEFINITION OF A LIMIT

Step 1: 55xx 

55 x

Step 2: 572,    or 514.  

The value of δ which assures 51 x     7 x  is the smaller value, 2.  

Step 1: 22xx      22 x   

Step 2: 211,   or 275.   

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

Step 1: (3)3xx     33 x

Step 2: 7 1 22 3,

 or 1 2 3

The value of  which assures (3) x

Step

The

Step 1: 33xx

Step 2: 32.75910.2409,   or 33.23910.2391.   

The value of  which assures 3 x    2.75913.2391 x  is the smaller value, 0.2391.  

7. Step 1: 55xx     55 x  

Step 2: From the graph, 54.90.1,   or 55.10.1;    thus 0.1   in either case.

8. Step 1: (3)3xx     33 x

Step 2: From the graph, 33.10.1,

or 32.90.1;

thus 0.1.

9. Step 1: 11xx

Step 2: From the graph, 97 1616 1,

or 259 1616 1;

thus 7 16

10. Step 1: 33xx

Step 2: From the graph, 32.610.39,

or 33.410.41;

thus 0.39.

11. Step 1: 22xx

Step 2: From the graph,

12. Step 1: (1)1xx

Step 2: From the graph, 5 2 1

13. Step 1: (1)1xx

Step

14. Step

Step 2: From the graph, 111 22.012

15. Step 1: (1)50.0140.01 xx

Step 2: 44

16. Step 1: (22)(6)0.02240.02

Step 2: (2)2

17. Step 1:

18. Step 1:

Step 2: 11 44xx

Then 1 4 0.160.09   1 4 or0.360.11;thus0.09.

19. Step 1: 193111931 xx 

219441916 xx     41916153 xx     or 3 15 x 

Step 2: 1010xx       1010. x  

Then 1037,   or 1015   5;    thus 5.  

20. Step 1: 7411741xx    3759725 xx 

Step 2: 2323xx

23x23.

21. Step 1: 1111440.050.050.05xx

Step 2: 44xx

22. Step 1: 2230.10.130.1

Step 2: 33xx

23. Step 1: 2240.50.540.5

4.53.5, x   for x near 2

Step 2: (2)2xx   

24. Step 1: 11(1)0.10.110.1 xx

Step 2: (1)1xx 

Then 10 1 99 1,

25. Step 1: 22 (5)111161

Step 2: 44xx     44. x  

Then 415415

or 1010 911 x

0.1270, or 417174

0.1231; thus 1740.12.  

26. Step 1: 120120120 511514 xxx      11 41206 63020 xx      or 20 30. x  

Step 2: 2424xx     2424. x  

Then 24204,   or 24   306;    thus 4.   

27. Step 1: 20.030.0320.03mxmmxm    0.0320.032 mmxm   0.03 22 m x   0.03 m

Step 2: 22xx     22. x  

Then 0.030.03 22, mm    or 0.030.03 22. mm      In either case, 0.03 . m  

28. Step 1: 33 mxmccmxmc     33 cmmxcm  33cc mm x  

Step 2: 33xx     33. x  

Then 33, cc mm    or 3   3 cc mm     In either case, c m  

29. Step 1: 22 () mmmxbbccmx    22 mmccmxc    11 22 cc mm x  

Step 2: 11 22xx     11 22 x  

Then 11 22 , cc mm    or 1 2   1 2 cc mm     In either case, c m  

30. Step 1: 0.050.05 ()()0.050.050.050.050.05 11. mm mxbmbmxmmmxm x  

Step 2: 11xx

Then 0.050.05 11, mm

Step 1: (32)(3)0.020.0262 xx

Step 2: 033 xx

32. 1 lim(32)(3)(1)21 x x  

Step 1: (32)10.030.0333 xx

Step 2: (1)1xx

Step 1:

Step 2: 22xx

34. 2 (x5)(x1) x6x5 x5x5x5(x5) limlim

Step 1:   2 65 5 (4)0.05 xx x  

Step 2: (5)5xx

Then 55.050.05,

35. 3 lim1515(3)164 x x 

Step 1: 1540.50.515 xx

11.25519.253.852.25. xx 

Step 2: (3)3xx 

Then 33.850.85,

36. 44 2 2 lim2 xx  

38. Step 1: (37)239

39. Step 1: 525xx

Then

40. Step 1: 4242 xx

Step 2: 0.

41. Step 1:

Step 2:

Then 1111,

that is, the smaller of the two distances.

42. Step 1: For 22 2,44xxx 

Step 2:

43. Step 1: 1111

48. Step 1: 0:2020xxx      2  0; x   2 0:002. x xx     Step 2: 0.xx    Then 22 ,     or 2     2.    Choose 2   

49. By the figure, 1 sin x xxx for all 0 x  and 1 sinfor0. x xxxx  Since 0 lim() x x   0 lim0, x x   then by the sandwich theorem, in either case, 1 0 limsin0. xx x  

50. By the figure, 222 1 sin x xxx  for all x except possibly at 0. x  Since 2 0 lim() x x   2 0 lim0, x x   then by the sandwich theorem, 2 1 0 limsin0. xx x  

51. As x approaches the value 0, the values of ()gx approach k. Thus for every number 0,   there exists a 0   such that 00 x    ().gxk 

52. Write xhc  Then 0 xc   , xc  ()xchc    , c   , hcch   0 h   00. h   Thus, lim() xc fxL   for any 0,   there exists 0   such that () fxL  whenever 0 xc   () fhcL  whenever 0 00lim(). h hfhcL   

53. Let 2 ().fxx  The function values do get closer to 1 as x approaches 0, but 0 lim()0, x fx   not 1 The function 2 () fxx  never gets arbitrarily close to 1 for x near 0.

54. Let 1 0 2 ()sin,,and0 fxxLx There exists a value of x 6 (namely) x   for which 1 2 sin x   for any given 0.   However, 0 limsin0, x x   1 2 not. The wrong statement does not require x to be arbitrarily close to 0x As another example, let 11 2 ()sin,, x gxL and 0 0. x  We can choose infinitely many values of x near 0 such that 11 2 sin x  as you can see from the accompanying figure. However, 1 0 limsin xx  fails to exist. The wrong statement does not require all values of x arbitrarily close to 0 0 x  to lie within 0of L   1 2 Again you can see from the figure that there are also infinitely many values of x near 0 such that 1 sin0. x  If we choose 1 4   we cannot satisfy the inequality 11 2 sin x   for all values of x sufficiently near 0 0. x  55.

228.999.01 x    or3.3843.387. x  To be safe, the left endpoint was rounded up and the right endpoint was rounded down.

56.

To be safe, the left endpoint was rounded up and the right endpoint was rounded down.

57.

(b) 0111() xxfx

Then ()1(1)1 fxxx 1 x  That is, ()11fx  no matter how small  is taken when 1 1 1lim x x     ()1.fx   (c) 1011(). xxfxx     Then ()1.51.51.5 fxxx  1.510.5.  Also, 011xx     1()1. fxx

Then ()1.5fx  (1)1.50.5 xx

0.510.50.5 x  Thus, no matter how small  is taken, there exists a value of x such that 1 x  but 1 2 ()1.5fx   1 lim()1.5. x fx  

58. (a) For 22()2()42. xhxhx       Thus for 2,()4 hx    whenever 22 x   no matter how small we choose 2 0lim()4. x hx      

(b) For 22()2()31 xhxhx       Thus for 1,()3 hx    whenever 22 x   no matter how small we choose 2 0lim()3. x hx      

(c) For 2 22()xhxx     so 2 ()22.hxx No matter how small 0   is chosen, 2 x is close to 4 when x is near 2 and to the left on the real line 2 2 x  will be close to 2. Thus if 1,()2 hx    whenever 22 x   no matter how small we choose 2 0lim()2. x hx      

59. (a) For 3 3()4.8()40.8.xfxfx       Thus for 0.8,()4 fx    whenever 3 3 x   no matter how small we choose 3 0lim()4. x fx      

(b) For 3 3()3()4.81.8xfxfx       Thus for 1.8,()4.8 fx    whenever 3 3 x   no matter how small we choose 3 0lim()4.8. x fx    

(c) For 3 3()4.8()31.8xfxfx       Again, for 1.8,()3 fx    whenever 3 3 x   no matter how small we choose 3 0lim()3. x fx      

60. (a) No matter how small we choose 0,   for x near 1 satisfying 11, x  the values of ()gx are near 1 ()2gx  is near 1. Then, for 1 2   we have 1 2 ()2gx  for some x satisfying 11, x  or 1 01lim()2. x xgx       (b) Yes, 1 lim()1 x gx   because from the graph we can find a 0   such that ()1gx   if 0(1). x   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),xx0-1..x01,colorblack,  # (a) title"Section 2.3,#61(a)");  L:limit(f(x),xx0);  # (b) epsilon:0.2;  # (c) plot([f(x),L-epsilon,Lepsilon],xx0-0.01..x00.01,  colorblack, linestyle[1,3,3],title"Section 2.3, #61(c)"); 

q:fsolve(abs(f(x)-L )epsilon, xx0-1..x01); #(d)

delta:abs(x0-q);  plot([f(x),L-epsilon,Lepsilon],xx0-delta..x0delta,colorblack,title"Section 2.3,#61(d)");  for eps in [0.1,0.005,0.001]do #(e)

q:fsolve(abs(f(x)-L)eps,xx0-1..x01); 

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,Leps],xx0-delta..x0delta,  colorblack,linestyle[1,3,3],titlehead));  enddo:

Mathematica (assigned function and values for x0, eps and del may vary): Clear[f,x] y1:Leps;y2:Leps;x01;  2 f[x_]:(3x(7x1)Sqrt[x]5)/(x1) 

Plot[f[x],{x,x00.2,x00.2}]  L:Limit[f[x],xx0]  eps0.1;del0.2; 

Plot[{f[x],y1,y2},{x,x0del,x0del}, PlotRange{L2eps, L2eps}]

2.4 ONE-SIDED LIMITS

1. (a) True (b) True (c) False (d) True (e) True (f) True (g) False (h) False (i) False (j) False (k) True (l) False

2. (a) True (b) False (c) False (d) True (e) True (f) True (g) True (h) True (i) True (j) False (k) True

3. (a) 2 2 2 lim()12, x fx    2 lim()321 x fx  

 (c) 4 2 4 lim()13, x fx

(b) No, 2 lim() x fx  does not exist because 22 lim()lim() xx fxfx

lim()13 x fx

(d) Yes, 4 lim()3 x fx   because 44 3 lim()lim() xx fxfx  

4. (a) 2 2 2 lim()1, x fx    2 lim()321,(2)2 x fxf

(b) Yes, 2 lim()1 x fx   because 22 1 lim()lim() xx fxfx

(d) Yes, 1 lim()4 x fx   because 11 4 lim()lim() xx fxfx

5. (a) No, 0 lim() x fx   does not exist since  1 sin x does not approach any single value as x approaches 0

(b) 00 lim()lim00 xx fx  

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

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

7. (a) (b) 11 lim()1lim() xx fxfx   

8. (a) (b) 11 lim()0lim() xx fxfx   

9. (a) domain: 02 x  range: 01 y  and 2 y  (b) lim() xc fx  exists for c belonging to (0,1)(1,2) 

10. (a) domain: x   range: 11 y  (b) lim() xc fx  exists for c belonging to (,1)(1,1)(1,)

18. (a) 2(1)2(1) (1) 1 11 limlimxxxx xx xx  

(b) 2(1)2(1) (1) 1 11 limlimxxxx xx xx

(|1|(1)for1) xxx  1 lim22 x x  

19. (a) If 2 0, x   then sin0, x  so that sin sin sinsin 000 limlimlim11 xx xx xxx   

(b) If 2 0, x   then sin0, x  so that sin sin sinsin 000 limlimlim11 xx xx xxx 

20. (a) If 2 0, x   then cos1, x  so that 1cos1cos1cos (cos1)1cos cos1 0000 limlimlimlim11 xxx xxx xxxx

(b) If 2 0, x   then cos1, x  so that cos1cos1 (cos1) cos1 000 limlimlim11 xx xx

21. (a) 3 3 3 lim1

25.

csc2211111 cos5sin2cos52sin2cos522 0000 limlimlimlim1(1)

31.

coscos 1 sincossincossincossincossin 0000 limlimlimlim xxxxxxxx xxxxxxxxxx xxx

sinsin 111 cos 000 limlimlim(1)(1)12 xx xx x xxx 

32.

33. 22 (1cos)(1cos) 1cos1cossin sin2(2sincos)(1cos)(2sincos)(1cos)(2sincos)(1cos) 0000 limlimlimlim  

sin0 (2cos)(1cos)(2)(2) 0 lim0

34.

xxxxx xx xxxx

(1cos) 1cos 1 1cos 1 2 9 9909 222222 sin3 sin3sin3 2 33 9 0 lim (0) (1cos) cos sin3sin31 0000 lim limlimlimlim 0 xxx x

35. sin(1cos) sin 1cos 00 limlim1since1cos0as 0 t tt

36. sin(sin) sin sin 00 limlim1sincesin0as 0 h hhhh

sinsin2sin2111 sin2sin222sin222 000 limlimlim11

sin5sin5455sin5455 sin4sin45445sin444 000 limlimlim11 xxxxx xxxxxx

40. cos2cos2cos2 1 0000sin22sincos2cos2 limsincot2limsinlimsinlim

41.

tan3sin3sin383 11 sin8cos3sin8cos3sin838 000 limlimlimxxxx xxxxxxx xx

42.

sin3cot5sin3sin4cos5sin3sin4cos5345 000cot4cos4sin5cos4sin5345 limlimlimyyyyyyyyy yyyyyyyyyy yy

sin3sin45cos5 34 1212 34sin5cos4555 0 lim1111 yyyy yyyyy

43. sin cos 222 cos3 sin3 tansinsin3 000cot3coscos3 limlimlim

000 limlimlim1

45. 22 1cos31cos31cos31cos3sin33sin3sin3 221cos32(1cos3)2(1cos3)231cos3 00000 limlimlimlimlimxxxxxxx xxxxxxxxx

3sinsin30 21cos211

46. 22 2 22222 coscoscos1cos(cos1)cos(cos1)cos(cos1)cos(sin) cos1 00000(cos1)(cos1) limlimlimlimlimxxxxxxxxxxx x xxxxxxxxxxxx 

xxx xxx

sinsincos 11 cos1112

47. Yes. If lim()lim(), xaxa fxLfx   then lim(). xa fxL   If lim()lim(), xaxa fxfx    then lim() xa fx  does not exist.

48. Since lim() xc fxL   if and only if lim() xc fxL    and lim(), xc fxL   then lim() xc fx  can be found by calculating lim(). xc fx  

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

50. If f is an even function of x, then ()().fxfx  Given 2 lim()7 x fx 

then 2 lim()7. x fx

 However, nothing can be said about 2 lim() x fx  because we don’t know 2 lim() x fx

51. (5,5)55Ix

52. (4,4)44Ix

Also, 22 555.

Also, 44422

53. As 0 x  the number x is always negative. Thus, (1)10xx xx

which is always true independent of the value of x. Hence we can choose any 0

with

54. Since 2 x   we have 2 x  and 22.xx Then, 22 2 2 110xx xx

which is always true so long as 2 x  Hence we can choose any

and thus 22 x

55. (a)

400 lim400. x x

Just observe that if 400401, x

then 400 x

Thus if we choose 1,   we have for any number 0

that 400400 x

(b) 400 lim399. x

any number 0

Just observe that if 399400 x

then 399. x

that 400400 x

we conclude that 400 lim x x

does not exist. 56. (a) 00 lim()lim00; xx fxx 

for x positive. Choose 2

(b)

0 lim()0. x fx    

2 1 00 lim()limsin0 x xx fxx   by the sandwich theorem since

22 2 1 sin x xxx  for all 0. x

Since 222 00 xxx  whenever , x   we choose    and obtain

2 1 sin0 x x 

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

2.5 CONTINUITY

1. No, discontinuous at 2, x  not defined at 2 x 

2. No, discontinuous at 3, x  3 1 lim()(3)1.5 x gxg  

3. Continuous on [ 1,3]

4. No, discontinuous at 1, x  11 1.5lim()lim()0 xx kxkx 

5. (a) Yes (b) Yes, 1 lim() 0 x fx    (c) Yes (d) Yes

6. (a) Yes, (1)1 f 

(b) Yes, 1 lim()2 x fx   (c) No (d) No

7. (a) No (b) No

8. [ 1,0)(0,1)(1,2)(2,3) 

9. (2)0, f  since 22 lim()2(2)40lim() xx fxfx  

10. (1) f should be changed to 1 2lim() x fx  

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

Removable discontinuity at 0 x  by assigning the number 0 lim()0 x fx   to be the value of (0) f rather than (0)1. f 

12. Nonremovable discontinuity at 1 x  because 1 lim() x fx  fails to exist 11 (lim()2andlim()1). xx fxfx   

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

13. Discontinuous only when 202xx    14. Discontinuous only when 2 (2)02 xx    15. Discontinuous only when 2 430(3)(1)03xxxxx

   or 1 x 

16. Discontinuous only when 2 3100(5)(2)05 xxxxx      or 2 x 

17. Continuous everywhere. (|1|sinxx  defined for all x; limits exist and are equal to function values.)

18. Continuous everywhere. (||10 x   for all x; limits exist and are equal to function values.)

19. Discontinuous only at 0 x 

20. Discontinuous at odd integer multiples of 22 ,i.e.,(21), xn   n an integer, but continuous at all other x

21. Discontinuous when 2x is an integer multiple of ,i.e.,2, xn    n an integer 2 , xn   n an integer, but continuous at all other x.

22. Discontinuous when 2 x is an odd integer multiple of 222 ,i.e.,(21), xn 

n an integer 21,xn   n an integer (i.e., x is an odd integer). Continuous everywhere else.

23. Discontinuous at odd integer multiples of 22 ,i.e.,(21), xn   n an integer, but continuous at all other x

24. Continuous everywhere since 2 1sin10sin1 xx   2 1sin1; x   limits exist and are equal to the function values.

25. Discontinuous when 230 x  or 3 2 x   continuous on the interval  3 2 , .   

26. Discontinuous when 3 10 x  or 1 3 x   continuous on the interval  1 3 ,.   

27. Continuous everywhere: (21)1/3 x is defined for all x; limits exist and are equal to function values.

28. Continuous everywhere: (2)1/5 x is defined for all x; limits exist and are equal to function values.

29. Continuous everywhere since 2 (3)(2) 6 33 333 limlimlim(2)5(3) xxxx xxxxx xg

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

31. Discontinuous at 1; x  2 1 lim(2)3, x x    but 1 lim, x x ee   so that 1 lim() x fx  does not exist while (1); fe  and 00 lim(1)1lim, x xx xe   so that 0 lim()1(0) x fxf  

32. Discontinuous at ln2, x  since 202lnln2ln2 xxx eeex      

33. limsin(sin)sin(sin)sin(0)sin0, x xx   

34.

and function continuous at x 

2222 0 limsin(cos(tan))sin(cos(tan(0)))sincos(0)sin1, t t    and function continuous at 0. t 

35. 22222 111 limsec(sectan1)limsec(secsec)limsec((1)sec) yyy yyyyyyyy 

2 sec((11)sec1)  sec01,  and function continuous at 1. y 

36.

1/3 4444 0 limtancos(sin)tancos(sin(0))tancos(0)tan1, x x 

37. 2 42 193sec2193sec016 0 limcoscoscoscos, tt

38.

and function continuous at 0. t 

6 22 1 66 3 limcsc53tancsc53tan45393, x xx 

39. 

0 222 0 limsinsinsin1, x x ee

40.

 111 2 1 limcoslncosln1cos(0), x

41. 2 (3)(3) 9 3(3) ()3, xxx xx gxx

42. 2 (5)(2) 310 22 ()5, tttt tt htt

43. 2 32 3 (1)(1) 11 (1)(1)1 1 (), ssssss fsssss

44. 2 2 (4)(4) 164 (4)(1)1 34 (), xx xx xxxx x gx

45. As defined, 2 3 lim()(3)18 x fx

For ()fx to be continuous we must have 4 3

46. As defined, 2 lim()2 x gx 

and 2 2 lim()(2)4. x gxbb

For ()gx to be continuous we must have 1 2 4 2.bb

47. As defined, 2 lim()12 x fx

For ()fx to be continuous we must have 2 12223 aaa

48. As defined, 0 11 0 lim() bb bb x gx

49. As defined, 1 lim()2 x fx 

and 1 lim()(1), x fxabab  

and 1 lim()(1) x fxabab   and 1 lim()3. x fx    For ()fx to be continuous we must have 2 ab and 5 2 3 aba    and 1 2 . b 

50. As defined, 0 lim()(0)22 x gxabb   and 2 0 lim()(0)33, x gxabab  

and 2 2 lim()(2)3 x gxab   4 3ab and 0 lim()3(2)51. x gx  

For ()gx to be continuous we must have 23bab  and 4 31 ab   3 2 a  and 3 2 . b 

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

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

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

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

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

56. cos(cos)0. xxxx    If 2 , x 

22 cos0.   If 2 , x  

22 cos0.  Thus cos0 xx for some x between 2  and 2  according to the Intermediate Value Theorem, since the function cos xx is continuous.

57. Let 3 ()151, fxxx which is continuous on [ 4,4]. Then (4)3, f  (1)15, f  (1)13, f  and (4)5. f  By the Intermediate Value Theorem, ()0fx  for some x in each of the intervals 41, x  11, x  and 1 4. x  That is, 3 1510xx 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 ab  Then ()()()22 Fxxaxbx  is continuous for all values of x, so it is continuous on the interval [ ,].ab Moreover () Faa  and (). Fbb  By the Intermediate Value Theorem, since 2 , ab ab   there is a number c between a and b such that 2 ().Fab x  

59. Answers may vary. Note that f is continuous for every value of x. (a) 3 (0)10,(1)18(1)103.ff Since 3 10,   by the Intermediate Value Theorem, there exists a c so that 01 c  and ().fc  

(b) 3 (0)10,(4)(4)8(4)1022.ff Since 22310, by the Intermediate Value Theorem, there exists a c so that 40 c  and ()3.fc 

(c) (0)10, f  3 (1000)(1000)8(1000)10999,992,010. f  Since 105,000,000999,992,010,  by the Intermediate Value Theorem, there exists a c so that 01000 c  and ()5,000,000.fc 

60. All five statements ask for the same information because of the intermediate value property of continuous functions.

(a) A root of 3 ()31 fxxx is a point c where ()0.fc 

(b) The point where 3 yx  crosses 31yx have the same y-coordinate, or 3 31yxx ()fx   3 310.xx

(d) The points where 3 3 yxx  crosses 1 y  have common y-coordinates, or 3 31yxx ()fx   3 310.xx

(e) The solutions of 3 310xx are those points where 3 ()31 fxxx has value 0.

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

62. Answers may vary. For example, 1 1 () gxx   has a discontinuity at 1 x  because 1 lim() x gx  does not exist. 11 lim()andlim(). xx gxgx  

63. (a) Suppose 0x is rational 0 ()1.fx   Choose 1 2   For any 0   there is an irrational number x (actually infinitely many) in the interval 00 (,) xx ()0fx   Then 0 0|| xx   but 0 |()()| fxfx 1 2 1  ,   so 0 lim() xx fx  fails to exist f  is discontinuous at 0x rational.

On the other hand, 0x irrational 0 ()0fx   and there is a rational number x in 00 (,)()1. xxfx    Again 0 lim() xx fx  fails to exist f  is discontinuous at 0x irrational. That is, f is discontinuous at every point.

(b) f is neither right-continuous nor left-continuous at any point 0x because in every interval 00 (,) xx  or 00 (,) xx   there exist both rational and irrational real numbers. Thus neither limits 0 lim() xx fx  and 0 lim() xx fx   exist by the same arguments used in part (a).

64. Yes. Both () fxx  and 1 2 () gxx are continuous on [0,1]. However () () fx gx is undefined at 1 2 x  since  () 1 2() 0 fx ggx   is discontinuous at 1 2 x 

65. No. For instance, if ()0,fx  (),gxx    then

()00hxx

is continuous at 0 x  and ()gx is not.

66. Let 1 1 () fxx  and ()1.gxx Both functions are continuous at 0. x  The composition (())fgfgx  11 (1)1xx   is discontinuous at 0, x  since it is not defined there. Theorem 10 requires that ()fx be continuous at (0), g which is not the case here since (0) 1 g  and f is undefined at 1.

67. Yes, because of the Intermediate Value Theorem. If ()fa and ()fb did have different signs then f would have to equal zero at some point between a and b since f is continuous on [ ,].ab

68. Let ()fx be the new position of point x and let ()() .dxfxx  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, ()0dx  for some point in between. That is, () fxx  for some point x, which is then in its original position.

69. If (0)0 f  or (1)1, f  we are done (i.e.,0or1inthosecases). cc Then let (0)0fa and (1)1fb because 0()1. fx  Define ()() gxfxxg   is continuous on [0,1]. Moreover, (0)(0)00 gfa and (1)(1)110 gfb  by the Intermediate Value Theorem there is a number c in (0, 1) such that ()0()0 gcfcc    or (). fcc 

70. Let () 2 0. fc   Since f is continuous at xc  there is a 0   such that ()() xcfxfc      ()()(). fcfxfc    If ()0,fc  then 3 11 222()()()()()0 fcfcfxfcfx       on the interval (,). cc If ()0,fc  then 3 11 222()()()()() 0 fcfcfxfcfx       on the interval (,). cc

71. By Exercise 52 in Section 2.3, we have 0 lim()lim(). xch fxLfchL   Thus, ()fx is continuous at 0 lim()()lim()(). xch xcfxfcfchfc  

72. By Exercise 71, it suffices to show that 0 limsin()sin h chc   and 0 limcos()cos . h chc   Now  0000 limsin()lim(sin)(cos)(cos)(sin)(sin)limcos(cos)limsin hhhh chchchchch 

. By Example 11 Section 2.2, 0 limcos1 h h   and 0 limsin0. h h   So 0 limsin()sin h chc   and thus ()sin fxx  is continuous at xc  Similarly,

00 limcos()lim(cos)(cos)(sin)(sin) hh chchch   00 (cos)limcos(sin)limsin hh chch 

cos c  Thus, ()cos gxx  is continuous at xc 

73. 1.8794,1.5321,0.3473 x 

75. 1.7549 x 

77. 3.5156 x 

79. 0.7391 x  74. 1.4516,0.8547,0.4030 x  76. 1.5596 x  78. 3.9058,3.8392,0.0667 x  80. 1.8955,0,1.8955 x 

2.6 LIMITS INVOLVING INFINITY; ASYMPTOTES OF GRAPHS

1. (a)

2 lim()0 x fx   (c) 3 lim()2 x fx

0 lim()1 x fx

(g) 0 lim()doesnotexist x fx

(i) lim()0 x fx 

2. (a) 4 lim()2 x fx

(e) 3 lim() x fx

lim() x fx

(b) 3 lim()2 x fx 

(h) lim()1 x fx

Note: In these exercises we use the result / 1 lim0whenever0.

15. (a)

16. (a)

17. (a) 3 3239 2 77 1 36 limlim7 x x x xxx

18. (a) 1 4 3 4256 1 234 9 99 2 2 256 limlim x xxx xx xxx xx

19. (a) 1031 1 54 26 6 1031 1

0 (same process as part (a))

7 (same process as part (a))

21.

22.

(b) 72413 323 35135 673673

(a) 83325 554 529529 3434

(b) 83325 554 529529 3434

53. (a)

(c)

and 0 lim(2cot),

(1) 0 lim xxx

1/3 2 3 0 lim xx 

1/5 2 0 lim xx

2/31/32 11 () 00 limlim xxx x 

2 limsec x x 

so the limit does not exist

2 111 (2)(2)positivepositive 4 22 limlim xxx

2 111 (2)(2)positivenegative 4 22 limlim xxxxx

2 111 (2)(2)negativenegative 4 22 limlim xxxxx

(b) 2 3222 (2)(1) 321 1 4 2(2) 222 limlimlim, xx xxx xxxxxx xx

(d) 2 3222 (2)(1) 321 1 4 2(2) 222 limlimlim, xx xxx xxxxxx xx

2 322 (2)(1)negativenegative 32 positivenegative 002(2) limlimxxxx xxxxx x

(a) 2 3 (2)(1)(1)

(2)(1)(1)negative

(d) 2 3 (2)(1)(1) 320 (2)(2)(2)(1)(3) 4 111 limlimlim0 xxxxx xxxxxxxx xx

(a)

61. (a) 2/32/3 12 (1) 0 lim xxx

62. (a) 1/34/3 11 (1) 0 lim xxx

(b) 1/3 3 0 lim2 tt 

(b) 3/5 1 0 lim7 tt 

2/32/3 12 (1) 0 lim xxx

2/32/3 12 (1) 1 lim xxx

(b) 1/34/3 11 (1) 0 lim xxx

70. domain(,1)(1, 1)(1, );

77. Here is one possibility.

Here is

Here is one possibility. 79. Here is one possibility.

81. Here is one possibility.

82. Here is one possibility.

83. Yes. If () () lim2 fx xgx   then the ratio the polynomials’ leading coefficients is 2, so () () lim2 fx xgx   as well.

84. Yes, it can have a horizontal or oblique asymptote.

85. At most 1 horizontal asymptote: If () () lim, fx xgx L   then the ratio of the polynomials’ leading coefficients is L, so () () lim fx xgx L   as well.

86.

93.

94. For any 0,   take 1. N  Then for all yN  we have that ()0.

95. For every real number 0, B  we must find a 0   such that for all x, 2 1 00. x xB     Now, 22 2 1111 00. xBB x BBxx

Choose 1 , B   then 1 0 B xx 

2 1 x B   so that 2 1 0 lim. xx  

96. For every real number 0, B  we must find a 0   such that for all x, 1 00. x xB     Now, 11 0 xB Bx

so that 1 0 lim. xx  

97. For every real number 0, B  we must find a 0   such that for all x, 2 2 (3) 03. x xB     Now, 2 2 (3) 0 x B

2 2 (3) 21 2 (3) 0 x xB B  2 22 (3)03. xBB  X Choose 2 , B   then 2 2 (3) 030 x xB     so that 2 2 (3) 3 lim . xx  

98. For every real number 0, B  we must find a 0   such that for all x, 2 1 (5) 0(5) x xB      Now, 2 2 111 (5) 0(5)5. xBB Bxx   Choose 1 . B   Then 0(5) x   2 11 (5) 5 Bx xB      so that 2 1 (5) 5 lim xx   

99. (a) We say that ()fx approaches infinity as x approaches 0x from the left, and write 0 lim(), xx fx   if for every positive number B, there exists a corresponding number 0   such that for all x, 00 () .xxxfxB    

(b) We say that ()fx approaches minus infinity as x approaches 0x from the right, and write 0 lim(), xx fx    if for every positive number B (or negative number B) there exists a corresponding number 0   such that for all x, 00 (). xxxfxB    

(c) We say that ()fx approaches minus infinity as x approaches 0x from the left, and write 0 lim(), xx fx   if for every positive number B (or negative number B) there exists a corresponding number 0   such that for all x, 00 (). xxxfxB    

100. For 11 0,0. BxB Bx  Choose 1 B   Then 11 00xxBx       B so that 1 0 lim. x x   

101.

103.

115. (a) y   (see accompanying graph)

(b) y   (see accompanying graph)

116. (a) 0 y  and a cusp at 0 x  (see the accompanying graph)

(b) 3 2 y  (see accompanying graph)

3 3 24 1, (see accompanying graph)

CHAPTER 2 PRACTICE EXERCISES

1. At 11 1:lim ()lim()1 xx xfxfx

1 lim ()1(1) x fxf

is continuous at 1 fx

xfxfx

At 00 0:lim ()lim ()0

lim ()0. x fx

But 0 (0)1lim () x ffx   is discontinuous at 0. fx

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

At 11 1:lim ()1andlim ()1 xx xfxfx

1 lim()does not exist x fx

is discontinuous at 1. fx 

2. At 11 1:lim ()0 and lim ()1 xx xfxfx 

1 lim () does not exist x fx

is discontinuous at 1 fx

At 00 0:lim ()andlim () xx xfxfx  

0 lim()does not exist x fx 

is discontinuous at 0. fx

At 1 11 1:lim ()lim ()1lim ()1. xx x xfxfxfx 

1 But(1)0lim () x ffx 

is discontinuous at 1. fx

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

(a) 00 lim(3())3 lim ()3(7)21 tttt ftft 

(b) 00 2 lim(())lim22()(7)49 tttt ftft 

(d) 00 0 000 lim ()lim () () 7 ()7lim (()7)lim ()lim707 lim1 tttt tttttt ftftft tgtgtgt

(e) 00 limcos(())coslim ()cos01 tttt gtgt 

(f) 00 lim|()|lim ()|7|7 tttt ftft 

(g) 000 lim(()())lim ()lim ()707 tttttt ftgtftgt

(h) 

0 0 1111 ()lim ()77 lim tt tftft

4. (a) 00 lim()lim ()2 xx gxgx 

5. Since 0 lim 0 x x   we must have that x0 lim  (4())0. gx  Otherwise, if 0 lim (4()) x gx  is a finite positive number, we would have 4()

so the limit could not equal 1 as 0 x  Similar reasoning holds if 0 lim (4()) x gx  is a finite negative number. We conclude that 0 lim () 4 x gx   .

6. 4044 2limlim ()limlim xxxx xgxx

(since 0 lim x  ()gx is a constant) 21 42 0 lim (). x gx

7. (a) 1/31/3 lim ()lim () xcxc fxxcfc   for every real number cf  is continuous on (,). 

(b) 3/43/4 lim ()lim () xcxc gxxcgc 

for every nonnegative real number cg  is continuous on [0,). 

(d) 1/6 1/6 1 lim()lim () xc cxc kxxkc   for every positive real number ck  is continuous on (0,) 

8. (a)    11 22,, nI nn    where I  the set of all integers.

(b) (, (1)), nI nn    where I  the set of all integers.

9. (a) 2 32 44 (2)(2) 2 (7)(2)(7) 514 000 lim limlim ,2; xxxxx xxxxxxxxx xx x      the limit does not exist because 2 (7) 0 lim x xx x    and 2 (7) 0 lim x xx x

(b) 2 32 44 (2)(2) 2 (7)(2)(7) 514 222 limlimlim, 2, xxxxx xxxxxxxxx xx x

10. (a) 2 54332 (1) 002(21) limlimxxxx xxxxxxx x

and 20 (7)2(9) 2 lim0 x

11. 11 11 12 (1)(1)1 111 limlimlim xx xxxxx xx

12. 22 22 442222222 () 11 ()()2 limlimlim xaxa xxaxaxaxaa axaxa 

13. ()(2)22222 000 limlimlim (2)2 xhxxhxhx hh hhh xhx

14. ()(2)22222 000 limlimlim (2) xhxxhxhx xhhxx xhh

15. 11 22 2(2) 11 0002(2)424 limlimlim xx xxxxxxx

16. 332 (2)8(6128)8 2 000 limlimlim (612)12 xxxx xxxxx xx

17. 1/32/31/3 1/3 2/31/3 (1)(1)(1) 1 1(1) (1)(1) 11 limlimxxxxx xxxxxx

18. 1/31/3 2/3 (4)(4) 16 646488 limlimxxx xxx x

2/31/32/31/3 (1)(1) 1 112 1113 (1)(1)1 11 limlimxxx xxxxxx x

1/31/32/31/3 2/31/3 (4)(4)(416)(8) 8 (8)(416) 64 lim xxxxx xxxxx

1/31/3 2/31/32/31/3 (64)(4)(8)(4)(8)(44)(88) 8 1616163 (64)(416)416 6464 limlimxxxxx xxxxxx x

19. tan2cos sin 2 tancos 2sin 00 limlimxx x xxxx x  

20. 1 sin limcsc lim x xx x  

222 limsinsinsinsinsin1 x

22. limcos(tan)cos(tan)2222cos()(1)1 x xx

23. sin 888 3sin 3(1)1 31 00 limlim4 x x x

24.

cos 21cos 21cos 2 1 sinsincos 21 00 limlimxxx xxxx x

2 4sincos 4(0)(1) cos 2111 0 lim 0 xx

25. Let x = t 3  30 limln(3)limln tx tx

26.

2 1 limln2ln10 t tt 

cos22 21sin2 sin(cos 21)sin (cos 21) 00 limlimxx xxxxx x

30. 11 55()2 lim2lim(()) xxgxxxgx

31. 2 2 31 () 111 limlim ()0 since lim (31)4 x xgxxxgxx

32. 2 2 5 () 222 lim0lim () since lim (5)1 x xgxxx gxx

33. (a) (1)1and(2)5hasarootbetween fff

1 and 2 by the Intermediate Value Theorem. (b), (c) root is 1.32471795724

34. (a) (2)2and(0)2hasarootbetween fff

2 and 0 by the Intermediate Value Theorem. (b), (c) root is 1.76929235424

35. At

2 (1) 1 11 limlim1,and xx x xx

lim() x fx

 ()fx does not exist, the function f cannot be extended to a continuous function at 1. x

does not exist so

cannot be extended to a continuous function at 1 x

either. 36. The discontinuity at 0 x  of  1 ()sin fxx  is nonremovable because 1 0 limsin xx  does not exist.

37.

38.

39.

40.

47. sin sin 1 limlim0 since as lim0.xx xxxxxx xx

48. cos 1cos1 2 limlim0lim0

50. 2/315/3

51.

52.

1/0 1 limcoscos(0)111 x xx ee

1 limln1ln10 xt

53. 1 2 limtan x x

54.

311 1 limsin0sin(0)000 t tt e

55. (a) 222 444 333 33 is undefined at 3:lim and lim xxx xxx xx yx 

(b) 22 22 22 2121 1 is undefined at 1:lim xxxx

, thus 3 is a vertical asymptote.

57. domain[4,2)(2,

58. Since

CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES

2. (a)

(b)

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

4. (a)

6. We want to know in what interval to hold values of h to make V satisfy the inequality |1000||361000| Vh   10.  To find out, we solve the inequality: |361000|101036100010 hh

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 8.98.80.1  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 2 11 lim ()lim (7)6(1). xx fxxf

Step 1: 22 |(7)6|1

Step 2: |1|111. xxx

Then 11 or 11

Choose

min11,11,

then 0|1| x    2 |(7)6| x   and 1 lim()6 x fx   By the continuity text, ()fx is continuous at 1. x 

8. Show

11 44 11 24 lim ()lim 2. x xx gxg  

Step 1: 11111 2224242 2222. xxx x      

Step 2: 1111 4444 . xx   

  X Then 1111 4424424(2) ,

Choose 4(2) ,   

or 1111 4424244(2)

 the smaller of the two values. Then 11 42 02 x

By the continuity test, ()gx is continuous at 1 4 . x 

9. Show 22 lim ()lim231(2). xx hxxh

Step 1: 231231

Step 2: |2|2or22.

10. Show 55 lim()lim9 2(5).

Step 2: 0|5|555 xxx

Then 222 59(2)(2)42,  

or 222 59(2)4(2)2

Choose 2 2,   the smaller of the two values. Then, 0|5|92, xx

 so 5 lim92. x x  

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

11. Suppose 1L and 2L are two different limits. Without loss of generality assume 21.LL  Let 1 21 3 (). LL  Since 0 1 lim() xx fxL

1 () fxL

  1 11 33211211212 ()()()43()2. LLLfxLLLLLfxLL  

21211221 23()443()2. LLfxLLLLfxLL

If 12 min{,}    both inequalities must hold for 0 0|| : xx   1212 1212 1221 4 3() 2 5()0 43()2 LLfxLLLLLL LLfxLL

That is, 12 0 LL and 12 0, LL a contradiction.

12. Suppose lim() xc fxL   If 0, k  then lim()lim000 xcxc kfx 

lim() xc fx  and we are done. If 0, k  then given any 0,   there is a 0   so that || 0|||()||||()| k xcfxLkfxL

kfxL

Thus lim()lim(). xcxc kfxkLkfx

Since

(b) Since 333

0,01()0lim()

(d) Since 4224 0,1001()0 xxxxxx 

24 0 lim() as in part (c). x fxxA

14. (a) True, because if lim(()()) xa fxgx   exists then lim(()())lim()lim xaxaxa fxgxfx   [(()())()] fxgxfx lim() xa gx  exists, contrary to assumption.

(b) False; for example take 1 () fxx  and 1 ().gxx  Then neither 0 lim() x fx  nor 0 lim() x gx  exists, but  11 000 lim(()())limlim00 xxxxx fxgx   exists.

(c) True, because ()|| gxx  is continuous (())|()| gfxfx   is continuous (it is the composite of continuous functions).

(d) False; for example let 1,0 ()() is discontinuous at 0. However |()|1 1,0 x fxfxxfx x

is continuous at 0. x

15. Show 2 (1)(1) 1 1111(1) lim()limlim2,1. xxx xxx xx fxx

Define the continuous extension of ()fx as 2 1 1 ,1 ().We now prove the limit of () as 1

exists and has the correct value.

Step 1: 2 (1)(1) 1 1(1)(2)2(1) xxx

Step 2: |(1)|111 xxx

Then 11,or11 

Since the conditions of the continuity test are met by (),Fx then ()fx has a continuous extension to ()Fx at 1.

Define the continuous extension of ()gx as

exists and has the correct value.

Step 1: 2 (3)(1) 23 262(3)22 xxxx xx 

Step 2: |3|333. xxx

Then, 3 322,

or 3322

We now prove the limit of ()gx as 3

    Choose 2.    Then 0|3| x   2 (3)(1) 23 262(3) 3 2lim2.xxxx xx x        Since the conditions of the continuity test hold for (),()Gxgx can be continuously extended to ()Gx at 3. x 

17. (a) Let 0   be given. If x is rational, then ()|()0||0||0|; fxxfxxx    i.e., choose .    Then |0||()0| xfx      for x rational. If x is irrational, then ()0|()0|0 fxfx    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|0||()0| xfx      Therefore, f is continuous at 0. x 

(b) Choose 0. xc Then within any interval (,) cc there are both rational and irrational numbers. If c is rational, pick 2 c   No matter how small we choose 0   there is an irrational number x in 2 (,)|()()||0|. ccfxfcccc     That is, f is not continuous at any rational 0. c  On the other hand, suppose c is irrational ()0.fc   Again pick 2 c   No matter how small we choose 0   there is a rational number x in (,) cc with 3 222 ||. cccxcx   Then |()()||0| fxfcx 2 ||xc    f is not continuous at any irrational 0. c  If 0, xc repeat the argument picking || 22 . cc  Therefore f fails to be continuous at any nonzero value .xc  18. (a) Let m n c  be a rational number in [0,1] reduced to lowest terms 1 ().fcn   Pick 1 2 . n   No matter how small 0   is taken, there is an irrational number x in the interval (,)|()()| ccfxfc   111 2 0. nnn   Therefore f is discontinuous at ,xc  a rational number.

(b) Now suppose c is an irrational number ()0.fc   Let 0   be given. Notice that 1 2 is the only rational number reduced to lowest terms with denominator 2 and belonging to [0,1]; 1 3 and 2 3 the only rationals with denominator 3 belonging to [0,1]; 1 4 and 3 4 with denominator 4 in [0,1]; 3 12 555 ,, and 4 5 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 12,,,. prrr  Let min{||:1,,}. i crip   Then the interval (,) cc contains no rational numbers with denominator . N  Thus, 0|||()()| xcfxfc     |()0| fx  1 |()|fxN   f is continuous at xc  irrational.

(c) The graph looks like the markings on a typical ruler when the points (,()) xfx on the graph of ()fx are connected to the -axis x 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 1x is a point on the equator “just after” noon 1 xR    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, 11 ()()0. TxTxR   At exactly the same moment in time pick 2x to be a point just before midnight 2 xR    is just before noon. Then 22 ()()0 TxTxR   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 ()()0;TcTcR   i.e., there is always a pair of antipodal points on the earth’s equator where the temperatures are the same. 20.

(c)

(d)

22. ()2 cos (0)02 cos 02 0 fxxxf    and ()2 cos()20. f   Since ()fx is continuous on [,0],  by the Intermediate Value Theorem, ()fx must take on every value between [2,2].  Thus there is some number c in [,0]  such that ()0;fc  i.e., c is a solution to 2 cos 0. xx

23. (a) The function f is bounded on D if () fxM  and () fxN  for all x in D. This means () MfxN  for all x in D. Choose B to be max {||, ||}.MN Then |()|. fxB  On the other hand, if |()|, fxB  then ()() BfxBfxB   and ()() fxBfx   is bounded on D with NB  an upper bound and MB  a lower bound. (b) Assume () fxN  for all x and that . LN  Let 2 . LN   Since 0 lim () xx fxL   there is a 0   such that 0 0|||()| xxfxL      222 ()()()LLNLNLNfxLLfxLfx    3 2 LN  But 2 () LLNNNNfx       contrary to the boundedness assumption (). fxN  This contradiction proves . LN  (c) Assume ()Mfx  for all x and that LM  Let 2 ML   As in part (b), 0 2 0||xxLML    3 222 ()(), MLM fLML xLfxM    a contradiction.

24. (a) If ,ab  then 0||max {,}abababab     || 2 22222 ab abababaa   If ,ab  then 0||() abababba    || 2 22222 max{,} ab ababbab abb

(b) Let || 22 min{,}. ababab

25. 2 sin(1cos) sin(1cos )sin(1cos ) 1cos 1cos 1cos 1cos1cos 1cos(1cos ) 0000 limlimlimlim xxxxxx xxxxxxxxxxx

2 sinsin sin 0 (1cos )1cos 2 00 1limlim10. xxx xxxxxx

26.

sin sin sin 1 sinsin 0000 limlim1limlim1100. x x x xxx x xxx xxxx x 

27. sin(sin )sin(sin )sin(sin ) sin sin sinsin 0000 limlimlimlim111. xxxxx xxxxxx xxx 

28. 222 22 sin()sin()sin() 0000 limlim(1)limlim(1)111 xxxxxx xxxxxxxxx xx 

29. 222 22 sin(4)sin(4)sin(4) 2 44 2222 limlim(2)limlim(2)14 4 xxx xxxxxxx xx 

30. sin(3)sin(3)sin(3)1111 96699993333 limlimlimlim1 xxx xxxxxx xxx 

31. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is an oblique asymptote. 3/2 2233 11 2, xx xx yx    thus the oblique asymptote is 2.yx 

32. As

111 ,0sin01sin1, xxx x      thus as

11 ,sin1sin; xx xyxxxx  thus the oblique asymptote is yx 

33. As 2222 ,11; xxxxx    as 2 ,, xxx  and as 2 ,; xxx  thus the oblique asymptotes are yx  and .yx 

34. As 2 ,22 xxxxx    (2);22 as ,, xxxxxx  and as 2 ,; xxx  asymptotes are yx  and yx 

35. Assume 1 ab and 2 1 ()() a xxbxaxbxxbx    2 ()()()0; fxaxbxxbx  f is continuous for all x-values and (0)0,fab 22 ()()() fabaabaab  22 () () ()0.aaabab     

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

36. (a) 1"1" 0 0 lim, abxa x x

 so 10 1, aa

2 11 0 lim24 bb xbx

  then (1)1 1111 0011(11) limlimbxbxbx xbxxbxxx 

Chapter 2 Limits and Continuity