CHAPTER 1 Limits and Their Properties

Transcription

1 CHAPTER Limits and Their Properties Section. A Preview of Calculus Section. Finding Limits Graphically and Numerically Section.3 Evaluating Limits Analytically Section.4 Continuity and One-Sided Limits Section.5 Infinite Limits Review Eercises Problem Solving

2 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 Precalculus: Area 8. Precalculus: Volume (a) Area Area (b) You could improve the approimation by using more rectangles. Section. Finding Limits Graphically and Numerically f Actual it is f Actual it is f Actual it is f cos (Actual it is 0.) (Make sure you use radian mode.) 305

3 306 Chapter Limits and Their Properties f 3 4. does not eist since the 3 3 function increases and decreases without bound as approaches sec 8. sin 0 0. Ct t (a) (b) t Ct (c) Ct 0.7 t 3.5 t Ct Ct t 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 < < < < 0. < 4 0. < 4. < 4. < 4. implies 4. < < < < 4.. Using the first series of equivalent inequalities, you obtain f 3 4 < < 0.0 Hence, if < 0.0 < < 4 < < 4 < < < < 0.0 f L < , you have

4 Section. Finding Limits Graphically and Numerically < < < 0.0 If we assume 4 < < 6, then Hence, if 5 < < 5 < 5 < 0.0 < 5 5 < < < 0.0 f L < you have, Given Hence, let Hence, if > 0: 5 < 6 < 3 < 3 <. 0 < 3 < 3 < 6 < 5 < f L < you have, Given Given 3 < < 3 Hence, let Hence, if 3 > 0: < < < < 3 3 < < 3 < f L < 3, you have > 0: Hence, any Hence, for any > 0 < < f L < will work. > 0, you have 0 < Given > 0: Assuming < < 9, you can choose Then, 0 < 4 < < < 4 < < < Given Hence, let Hence for > 0: 3 0 < 3 <. 0 < 3 < 3 < 3 0 < f L <, you have

5 308 Chapter Limits and Their Properties 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, f f (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, 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 Using the zoom and trace feature, < < 0.00, 4 < f 4, 0, 4 4. f and f That is, for n 4 0. n f 4 0. n n n f 4 0. n

6 Section.3 Evaluating Limits Analytically 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 (a) 0 0 (b) 4 g (b) f t 0 t 4 f t 5 t 5 g f t t t (a) (b) (c) f g g f g (a) f (b) g (c) g f g tan tan sin sin 3. cos 3 cos cos cos sec sec 6 3

7 30 Chapter Limits and Their Properties 38. (a) (b) (c) 4f 4 f f g f g 3 f g f g f (d) g f 3 g (a) (b) (c) 3 f 3 f f 8 f f f 7 79 (d) f 3 f and h 3 f 3 agree ecept at 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 f 3 and g agree ecept at. f g

8 Section.3 Evaluating Limits Analytically f f Analytically,.5.5.5? It appears that the it is f ? Analytically, (Hint: Use long division to factor 5 3. ) 3 cos cos cos tan sin tan sin 7. cos 0 0 sin sin cos 74. sec tan sin cos cos sin sin cos cos sin cos cos sin cos cos sec 4 sin 78. sin 3 sin 3 3 sin 3 3 3

9 3 Chapter Limits and Their Properties 80. f h cos h h f h ? Analytically, cos h cos0. h The it appear to equal. 8. f sin f ? sin 3 Analytically,. 3 sin The it appear to equal f h f h h 0 h h 0 h h h 0 h h h h h 0 h h h 0 h 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 cos cos 0

10 Section.3 Evaluating Limits Analytically 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 st 6t seconds 6 s 50 t st t 0 6t 000 t 50 t 50 6 t 5 t t 6 t 50 t 50 t t 6 t 50 t ftsec 53 ftsec 04. when t 50 seconds t The velocity at time t a is sa st 4.9a t a ta t t a a t t a a t t a a t t a 4.9a t a a msec. Hence, if a 50049, the velocity is 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.

11 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 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.

12 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 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, g is not defined. g is continuous on,.

13 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 f has a nonremovable discontinuity at since does not eist, and has a removable discontinuity 3 f 9 3 at 3 since f 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 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 > f csc f > 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.

14 Section.4 Continuity and One-Sided Limits f tan has nonremovable discontinuities at each k, k is an integer. 54. f 3 has nonremovable discontinuities at each integer k f f is not continuous at sin g( 4 g a a Let a g a a a a a a a Find a such that a 8 a 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 cos, f 5, < f 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,

15 38 Chapter Limits and Their Properties 74. f 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 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 f 4 tan tan 8 < > 0. By the Intermediate Value Theorem, f 0 for at least one value of c between and 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 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 < 8 The Intermediate Value Theorem applies or 4 c ( 4 is not in the interval.) Thus, f f f is continuous on 5, 4. The nonremovable discontinuity,, lies outside the interval. f and f The Intermediate Value Theorem applies < 6 < or 3 c 3 ( is not in the interval.) Thus, f3 6.

16 Section.4 Continuity and One-Sided Limits 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 < 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, t, t, 0 < t t >, t is not an integer t >, t is an integer You can also write C as C.04, t, 0 < t. t > Nonremovable discontinuity at each integer greater than. C 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 True. f c L is defined.. f L eists. 3. f c f All of the conditions for continuity are met.

17 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 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 < 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 f f 3 f 3

18 Section.5 Infinite Limits 3 8. f sec f f 3 f 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 ,, No vertical asymptotes. The graph has holes at and f , 3, Vertical asymptotes at 0 and 3. The graph has holes at 3and. 4. h 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.

19 3 Chapter Limits and Their Properties 8. g tan sin has vertical asymptotes at cos n n, n any integer There is no vertical asymptote at 0 since 0 tan. Removable discontinuity at 3. sin Removable discontinuity at cos 46. cot tan and tan. tan 50. Therefore, does not eist. tan f 3 f f sec 6 f The line c is a vertical asymptote if the graph of f approaches ± as approaches c 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.)

20 Section.5 Infinite Limits (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) y (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 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,