CHAPTER Limits and Their Properties Section. A Preview of Calculus................... 305 Section. Finding Limits Graphically and Numerically....... 305 Section.3 Evaluating Limits Analytically............... 309 Section.4 Continuity and One-Sided Limits............. 35 Section.5 Infinite Limits....................... 30 Review Eercises............................. 34 Problem Solving.............................. 37
CHAPTER Limits and Their Properties Section. A Preview of Calculus Solutions to Even-Numbered Eercises. Calculus: velocity is not constant Distance 0 ftsec5 seconds 300 feet 4. Precalculus: rate of change slope 0.08 6. Precalculus: Area 8. Precalculus: Volume 3 6 54 0. (a) Area 5 5 5 3 5 0.47 4 Area 5 5.5 5 5.5 5 3 5 3.5 5 4 4.5 5 9.45 (b) You could improve the approimation by using more rectangles. Section. Finding Limits Graphically and Numerically..9.99.999.00.0. f 0.564 0.506 0.50 0.499 0.494 0.439 0.5 4 Actual it is 4. 4. 3. 3.0 3.00.999.99.9 f 0.485 0.498 0.500 0.500 0.50 0.56 0.5 3 3 Actual it is 4. 6. 3.9 3.99 3.999 4.00 4.0 4. f 0.0408 0.040 0.0400 0.0400 0.0399 0.039 45 0.04 4 4 Actual it is 5. 8. 0. 0.0 0.00 0.00 0.0 0. f 0.0500 0.0050 0.0005 0.0005 0.0050 0.0500 cos 0.0000 (Actual it is 0.) (Make sure you use radian mode.) 305
306 Chapter Limits and Their Properties 0. 3. f 3 4. does not eist since the 3 3 function increases and decreases without bound as approaches 3. 6. sec 8. sin 0 0. Ct 0.35 0.t (a) 0 0 5 (b) t 3 3.3 3.4 3.5 3.6 3.7 4 Ct 0.59 0.7 0.7 0.7 0.7 0.7 0.7 (c) Ct 0.7 t 3.5 t 3.5.9 3 3. 3.5 4 Ct Ct t 3.5 0.47 0.59 0.59 0.59 0.7 0.7 0.7 does not eist. The values of C jump from 0.59 to 0.7 at t 3.. You need to find such that 0 < < implies That is, So take Then 0. < 4 0. < 3.8 < 3.8 < 3.8 < f 3 3 4 < 0.. 0 < < 4 4. 0.0494. < 0. < 4 0. < 4. < 4. < 4. implies 4. < < 4. 3.8 < < 4.. Using the first series of equivalent inequalities, you obtain f 3 4 < 0.. 4. 4 4 4 < 0.0 Hence, if < 0.0 < 0.0 4 0 < 4 < 0.0 0 < 4 < < 0.0 4 < 0.0 4 < 0.0 f L < 0.0 0.0, you have
Section. Finding Limits Graphically and Numerically 307 6. 5 4 9 8. 4 9 < 0.0 5 < 0.0 5 5 < 0.0 If we assume 4 < < 6, then Hence, if 5 < 0.0 5 0 < 5 < 5 < 0.0 < 5 5 < 0.0 5 < 0.0 4 9 < 0.0 f L < 0.0 0.0 0.0 0.0009. you have, 5 0.0 5 3 Given Hence, let Hence, if > 0: 5 < 6 < 3 < 3 <. 0 < 3 < 3 < 6 < 5 < f L < you have, 30. 3 9 3 9 9 3 3. Given Given 3 < < 3 Hence, let Hence, if 3 > 0: 3 9 9 3. 0 < < < 3 3 3 < 3 3 < 9 9 3 < 3 < f L < 3, you have > 0: Hence, any Hence, for any > 0 < < f L < will work. > 0, you have 0 < 34. 4 36. 4 Given > 0: Assuming < < 9, you can choose Then, 0 < 4 < < < 4 < 3. 3 4 < <. 3 3 0 Given Hence, let Hence for > 0: 3 0 < 3 <. 0 < 3 < 3 < 3 0 < f L <, you have
308 Chapter Limits and Their Properties 38. 3 3 0 3 40. f 4 3 Given 3 0 < 3 < If we assume 4 < <, then Hence for > 0: 3 < 0 < 3 < 3 < < 4 3 < 3 0 < f L < 4. you have 4, 3 f 3 5 The domain is all, 3. The graphing utility does not show the hole at 3,. 4 4 4. f 3 9 3 f 6 9 3 3 44. (a) No. The fact that f 4 has no bearing on the eistence of the it of f as approaches. (b) No. The fact that f 4 has no bearing on the value of f at. 3 The domain is all ± 3. The graphing utility does not show the hole at 3, 6. 46. Let p be the atmospheric pressure in a plane at 48. altitude (in feet). p 4.7 lbin 0.00 (.999, 0.00) (.00, 0.00) 50. True 5. False; let.998.00 0 Using the zoom and trace feature, 0.00. 4 0 < < 0.00, 4 < 0.00. f 4, 0, 4 4. f and f 4 0 0 4 4 4 0 That is, for 54. 7 4 4 n 4 0. n f 4 0. n n 4. 7. 4.0 7.0 3 4.00 7.00 4 4.000 7.000 4 0. n 3.9 6.9 f 4 0. n 3.99 6.99 3 3.999 6.999 4 3.9999 6.9999
Section.3 Evaluating Limits Analytically 309 56. f m b, m 0. Let > 0 be given. Take m. 58. g L, L > 0. Let There eists c such that 0 < 0 implies If 0 < c < < then That is, m, m c < m mc < m b mc b < which shows that m b mc b. L < g L < L L < g < 3 L L. g L < Hence for in the interval c, c, c, g > L > 0. > 0 L. Section.3 Evaluating Limits Analytically. 0 (a) g.4 4. 0 (a) 0 0 (b) 4 g 4 5 0 (b) f t 0 t 4 f t 5 t 5 g 3 9 0 f t t t 4 6. 3 3 8 8. 3 33 7 3 0. 0 4. 3 3. 3 3 4 3 3 4 5 3 3 3 6. 3 5 3 5 3 8 3 8. 3 4 3 4 3 0. 4 4 3 4 4. 3 0 3 4. (a) (b) (c) f 3 7 4 3 g 4 4 6 g f g4 6 3 6. (a) f 4 4 34 (b) g 3 6 3 (c) g f g 3 4 8. tan tan 0 30. sin sin 3. cos 3 cos 3 5 34. cos cos 53 3 36. 7 sec 6 7 3 sec 6 3
30 Chapter Limits and Their Properties 38. (a) (b) (c) 4f 4 f 4 3 6 f g f g 3 f g f g 3 3 4 f (d) g f 3 g 3 40. (a) (b) (c) 3 f 3 f 7 3 3 f 8 f 7 8 8 3 f f 7 79 (d) f 3 f 3 7 3 9 4. and h 3 f 3 agree ecept at 0. 44. g and f agree ecept at 0. (a) h f 5 (a) f does not eist. (b) h f 3 (b) f 46. f 3 and g 3 agree ecept at. f g 5 4 48. f 3 and g agree ecept at. f g 3 7 8 4 4 4 8 50. 4 5. 5 4 4 4 8 4 4 4 4 3 6 54. 4 3 56. 3 3 3 3 3 3 3 4 58. 4 4 4 4 4 4 4 4 6 60.
Section.3 Evaluating Limits Analytically 3 6. 3 3 3 3 3 3 3 3 3 3 3 3 64. f 4 6 5.9 5.99 5.999 6 6.00 6.0 6. f Analytically,.5.5.5?.5.5.48 4 6 6 6 4 4 4. 6 4 8 0 0 It appears that the it is 0.5. 66. 5 3 80 00.9.99.999.9999.0.000.00.0. f 7.39 79.0 79.9 79.99? 80.0 80.08 80.80 88.4 4 3 5 Analytically, 5 3 4 3 4 8 6 4 3 4 8 6 80. (Hint: Use long division to factor 5 3. ) 3 cos cos cos tan sin 68. 30 0 70. 0 0 3 tan sin 7. cos 0 0 sin sin cos 74. sec 76. 4 tan sin cos 4 4 4 cos sin sin cos cos sin cos cos sin cos cos sec 4 sin 78. sin 3 sin 3 3 sin 3 3 3
3 Chapter Limits and Their Properties 80. f h cos h h 0. 0.0 0.00 0 0.00 0.0 0. f h.98.9998?.9998.98 Analytically, cos h cos0. h 0 4 5 5 4 The it appear to equal. 8. f sin 3 0. 0.0 0.00 0 0.00 0.0 0. f 0.5 0.0464 0.0? 0.0 0.0464 0.5 sin 3 Analytically,. 3 sin 0 0 3 3 The it appear to equal 0. 84. f h f h h 0 h h 0 h h h 0 h h h h h 0 h h h 0 h 86. 88. h 0 f h f h 4 h 4 h h 4 4h 4 h h 0 h h 0 h Therefore, a f b. h h 4 h 4 4 h 0 h h 0 b a f b a a a a 90. b a f b f sin 6 sin 0 9. f cos 94. h cos 6 0.5 0.5 0.5 6 cos 0 0.5 cos 0
Section.3 Evaluating Limits Analytically 33 96. f and g agree at all points ecept. 98. If a function f is squeezed between two functions h and g, h f g, and h and g have the same it L as c, then f eists and equals L. 00. h sin f, g sin, g 3 3 h f When you are close to 0 the magnitude of g is smaller than the magnitude of f and the magnitude of g is approaching zero faster than the magnitude of f. Thus, when is close to 0 g f 0 0. when t 000 50 st 6t 000 0 seconds 6 s 50 t 50 50 st t 0 6t 000 t 50 t 50 6 t 5 t 50 50 t 6 t 50 t 50 t 50 50 t 6 t 50 t 50 800 ftsec 53 ftsec 04. when t 50 seconds. 4.9 500 4.9t 50 0 5.53 49 The velocity at time t a is sa st 4.9a 50 4.9t 50 4.9a ta t t a a t t a a t t a a t t a 4.9a t a4.9 9.8a msec. Hence, if a 50049, the velocity is 9.850049 54. msec. 06. Suppose, on the contrary, that g eists. Then, since f eists, so would f g, which is a contradiction. Hence, g does not eist. 08. Given f n, n is a positive integer, then n n n c n c n cc n3... c n. 0. Given f 0: > 0, > 0 f 0 < c For every there eists such that whenever f 0 f f 0 < 0 < c < f Now for <. Therefore,. 0.
34 Chapter Limits and Their Properties. (a) If 0, then f 0. f f f f f f f Therefore, f 0. (b) Given f L: 0 f 0 > 0, > 0 f c L < 0 < < f L c <. For every there eists such that whenever Since < for, then f L f L. 4. True. 3 0 3 0 6. False. Let Then f but f. f 3, c 8. False. Let f and g. Then f < g 0. for all 0. But f g 0. cos cos cos cos cos sin cos 0 cos sin sin cos sin 0 0 sin cos. sec f (a) The domain of f is all 0, n. (b) 3 3 The domain is not obvious. The hole at 0 is not apparent. (c) (d) f sec sec sec sec sec sec Hence, tan sec cos sin sec sec cos sin sec. 4. The calculator was set in degree mode, instead of radian mode.
Section.4 Continuity and One-Sided Limits 35 Section.4 Continuity and One-Sided Limits. (a) (b) (c) f f f The function is continuous at. 4. (a) f (b) f (c) f The function is NOT continuous at. 6. (a) f 0 (b) f (c) f does not eist. The function is NOT continuous at. 8. 4 4 0. 4 4 4 4 4 4 4 4 4. 4. 0 6. f 4 f 4 6 f 8. f 0 0. sec. does not eist since sec and sec do not eist. 4. f 6. has a discontinuity at since f is not defined. 8. f has discontinuity at since f f., <,, > 30. f t 3 9 t is continuous on 3, 3. 3. g is not defined. g is continuous on,.
36 Chapter Limits and Their Properties 34. f is continuous for all real. 36. f cos is continuous for all real. 38. f has nonremovable discontinuities at and since f and f do not eist. 3 40. f has a nonremovable discontinuity at since does not eist, and has a removable discontinuity 3 f 9 3 at 3 since f. 3 3 3 6 4. f has a nonremovable discontinuity at since f does not eist, and has a removable discontinuity at since 44. f 3 3 has a nonremovable discontinuity at 3 since f 3 does not eist. f 3. 46. f 3,, < has a possible discontinuity at.. f. 3. f 3 f f f f f is continuous at, therefore, f is continuous for all real. 48. f, has a possible discontinuity at. 4, >. f 4 f 4 does not eist.. f f 4 3 Therefore, f has a nonremovable discontinuity at. csc csc 50. 6, 5 f 6, has possible discontinuities at, 5.,, < or > 5.. 3. f csc f 6 3 3 > f 5 csc 5 6 f 5 f f f 5 f 5 f is continuous at and 5, therefore, f is continuous for all real.
Section.4 Continuity and One-Sided Limits 37 5. f tan has nonremovable discontinuities at each k, k is an integer. 54. f 3 has nonremovable discontinuities at each integer k. 56. 0 f 0 58. f is not continuous at 4 8 8 0 4 sin g( 4 g a a Let a 4. 60. g a a a a a a a Find a such that a 8 a 4. 6. f g Nonremovable discontinuity at. Continuous for all >. Because f g is not defined for <, it is better to say that f g is discontinuous from the right at. 64. f g sin 66. Continuous for all real h Nonremovable discontinuity at and. 3 4 68. cos, f 5, < 0 0 3 f 0 50 0 7 cos f 0 3 f 5 0 Therefore, f 0 f 0 and f is continuous on the entire real line. ( 0 was the only possible discontinuity.) 70. f 3 7. Continuous on 3, f Continuous on 0,
38 Chapter Limits and Their Properties 74. f 3 8 4 76. f 3 3 is continuous on 0,. f 0 and f By the Intermediate Value Theorem, f 0 for at least one value of c between 0 and. 4 4 0 The graph appears to be continuous on the interval 4, 4. Since f is not defined, we know that f has a discontinuity at. This discontinuity is removable so it does not show up on the graph. 78. f 4 tan is continuous on, 3. 8 and f 3 4 3 f 4 tan tan 8 < 0 3 8 > 0. By the Intermediate Value Theorem, f 0 for at least one value of c between and 3. 80. f 3 3 f is continuous on 0,. f 0 and f By the Intermediate Value Theorem, f 0 for at least one value of c between 0 and. Using a graphing utility, we find that 0.596. 8. h 3 tan 84. f 6 8 h is continuous on 0,. f is continuous on 0, 3. h0 > 0 and h.67 < 0. f 0 8 and f 3 By the Intermediate Value Theorem, h 0 for at least one value between 0 and. Using a graphing utility, we find that 0.4503. < 0 < 8 The Intermediate Value Theorem applies. 6 8 0 4 0 or 4 c ( 4 is not in the interval.) Thus, f 0. 86. f f is continuous on 5, 4. The nonremovable discontinuity,, lies outside the interval. f 5 35 6 and f 4 0 3 The Intermediate Value Theorem applies. 6 6 6 5 6 0 35 6 < 6 < 0 3 3 0 or 3 c 3 ( is not in the interval.) Thus, f3 6.
Section.4 Continuity and One-Sided Limits 39 88. A discontinuity at c is removable if you can define (or redefine) the function at c in such a way that the new function is continuous at c. Answers will vary. (a) f (b) f sin (c) f, 0,, 0, 3 y if if < < if if < 3 3 3 90. If f and g are continuous for all real, then so is f g (Theorem., part ). However, fg might not be continuous if g 0. For eample, let f and g. Then f and g are continuous for all real, but fg is not continuous at ±. 9. C.04,.04 0.36t,.04 0.36t, 0 < t t >, t is not an integer t >, t is an integer You can also write C as C.04,.04 0.36 t, 0 < t. t > Nonremovable discontinuity at each integer greater than. C 4 3 3 4 t 94. Let st be the position function for the run up to the campsite. s0 0 ( t 0 corresponds to 8:00 A.M., s0 k (distance to campsite)). Let rt be the position function for the run back down the mountain: r0 k, r0 0. Let f t st rt. When t 0 (8:00 A.M.), f 0 s0 r0 0 k < 0. When t 0 (8:0 A.M.), f 0 s0 r0 > 0. Since f 0 < 0 and f 0 > 0, then there must be a value t in the interval 0, 0 such that f t 0. If f t 0, then st rt 0, which gives us st rt. Therefore, at some time t, where 0 t 0, the position functions for the run up and the run down are equal. 96. Suppose there eists in a, b such that f > 0 and there eists in a, b such that f < 0. Then by the Intermediate Value Theorem, f must equal zero for some value of in, (or, if < ). Thus, f would have a zero in a, b, which is a contradiction. Therefore, f > 0 for all in a, b or f < 0 for all in a, b. 98. If 0, then f 0 0 and f 0. Hence, f is continuous at 0. If 0, then f t 0 for rational, whereas t f t kt k 0 for irrational. Hence, f is not t t continuous for all 0. 00. True. f c L is defined.. f L eists. 3. f c f All of the conditions for continuity are met.
30 Chapter Limits and Their Properties 0. False; a rational function can be written as PQ where P and Q are polynomials of degree m and n, respectively. It can have, at most, n discontinuities. 04. (a) S 60 50 40 30 0 0 5 0 5 0 5 30 t (b) There appears to be a iting speed and a possible cause is air resistance. 06. Let y be a real number. If y 0, then 0. If y > 0, then let 0 < 0 < such that M tan 0 > y (this is possible since the tangent function increases without bound on 0, ). By the Intermediate Value Theorem, f tan is continuous on 0, 0 and 0 < y < M, which implies that there eists between 0 and 0 such that tan y. The argument is similar if y < 0. 08.. f c is defined.. f f c f c eists. Let c. As c, 0 3. f f c. Therefore, f is continuous at c. 0. Define f f f. Since and are continuous on a, b, so is f. f a f a f a > 0 and f b f b f b < 0. By the Intermediate Value Theorem, there eists c in f c f c f c 0 f c f c f f a, b such that f c 0. Section.5 Infinite Limits. 4. sec 4 sec 4 6. f 9 3.5 3. 3.0 3.00.999.99.9.5 f.077 5.08 50.08 500. 499.9 49.9 4.95 0.909 f 3 f 3
Section.5 Infinite Limits 3 8. f sec 6 3.5 3. 3.0 3.00.999.99.9.5 f 3.864 9. 9.0 90 90 9.0 9. 3.864 f 3 f 3 0. 4 3. 4 3 Therefore, is a vertical asymptote. Therefore, 0 is a vertical asymptote. Therefore, is a vertical asymptote. 4. No vertical asymptote since the denominator is never zero. 6. hs and hs. s 5 s 5 Therefore, s 5is a vertical asymptote. hs and hs. s 5 s 5 Therefore, s 5 is a vertical asymptote. 8. f sec has vertical asymptotes at cos n, n any integer. 0. g 3 4 3 6 4 6,, 4 6 8 8 No vertical asymptotes. The graph has holes at and 4.. 4 f 6 4 3 3 9 8 9 4 3, 3, Vertical asymptotes at 0 and 3. The graph has holes at 3and. 4. h 4 3 6. ht tt t t t 4 t t t 4, t has no vertical asymptote since h 4 5. Vertical asymptote at t. The graph has a hole at t.
3 Chapter Limits and Their Properties 8. g tan sin has vertical asymptotes at cos n n, n any integer. 30. 6 7 7 8 3 3 There is no vertical asymptote at 0 since 0 tan. Removable discontinuity at 3. sin Removable discontinuity at 3 3 34. 36. 4 6 38. 6 4 4 3 3 3 5 8 40. 3 9 4. 44. cos 46. cot tan 0 48. and tan. tan 50. Therefore, does not eist. tan f 3 f 0 4 8 8 4 5. f sec 6 f 3 54. The line c is a vertical asymptote if the graph of f approaches ± as approaches c. 6 9 9 6 56. No. For eample, f has no vertical asymptote. 58. P k V k V 0 V k (In this case we know that k > 0.)
Section.5 Infinite Limits 33 00 60. (a) r 50 sec ftsec 6. m 6 3 (b) r 50 sec 3 00 ftsec (c) 50 sec m 0 v c m m 0 v c v c v c 64. (a) Average speed 50 50y 50 y 5 5 y Domain: > 5 Total distance Total time d d dy 50 y y 50 y 50y 50 y 5 (b) 30 40 50 60 y 50 66.667 50 4.857 5 (c) 5 5 As gets close to 5 mph, y becomes larger and larger. 66. (a) (c) A bh r 00 tan 0 50 tan 50 Domain: 00 0, (b) (d) f A 0.3 0.6 0.9..5 0.47 4. 8.0 68.6 630. 0 0.5 68. False; for instance, let 70. True f. The graph of f has a hole at,, not a vertical asymptote. 7. Let f and g and c 0. 4, 74. Given f g, let g. then c f 0 c by Theorem.5. and but 4, 4 4 0.