CHAPTER 2 Limits and Their Properties

Transcription

1 CHAPTER Limits and Their Properties Section. A Preview of Calculus...5 Section. Finding Limits Graphically and Numerically...5 Section. Section. Evaluating Limits Analytically...5 Continuity and One-Sided Limits...57 Section.5 Infinite Limits...6 Review Eercises...67 Problem Solving Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2 CHAPTER Limits and Their Properties Section. A Preview of Calculus. Precalculus: ( 0 ft/sec)( 5 sec) 00 ft. Calculus required: Slope of the tangent line at is the rate of change, and equals about (a) Precalculus: bh (b) Calculus required: Area Area 5 0 sq. units bh.5 5 sq. units 7. f 6 y (a) 0 P (b) slope m, ( 6 ) 8 ( )( ) For, m For For (c) At P.5, m , m.5.5, 8, the slope is. You can improve your approimation by considering values of close to. 9. (a) Area Area (b) You could improve the approimation by using more rectangles. D (a) (b) D ( ) ( ) ( ) (c) Increase the number of line segments. Section. Finding Limits Graphically and Numerically f () Actual it is Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3 6 Chapter Limits and Their Properties f () ( ) Actual it is f () sin.0000 Actual it is. Make sure you use radian mode f () e.0000 Actual it is f () ( ) ln.0000 Actual it is f () Actual it is f () Actual it is f () sin.0000 Actual it is. Make sure you use radian mode sin π 0 9. does not eist. For values of to the left of, ( ) for values of to the right of, ( ), whereas.. ln 0 5. cos( ) 0 does not eist because the function oscillates between and as approaches 0. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4 Section. Finding Limits Graphically and Numerically 7 7. (a) f ( ) eists. The black dot at (, ) indicates that f ( ). (b) f does not eist. As approaches from the left, f () approaches.5, whereas as approaches from the right, f () approaches. (c) f ( ) does not eist. The hollow circle at (, ) indicates that f is not defined at. (d) f eists. As approaches, f approaches : f. 9. c f eists for all c.. 5 f c eists for all c.. One possible answer is: y y f f 5 5. Ct ( t ) (a) (b) (c) t C t.5 Ct.6 t C t Ct does not eist because the values of C approach different values as t approaches from both sides. f < 0.. So, take δ 0.. If 0 < < 0., then 7. You need f < 0., as desired. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5 8 Chapter Limits and Their Properties 9. You need to find δ such that 0 < < δ implies f < 0.. That is, 0. < < < < 0. 9 < < > > > > 9 > >. 9 So take δ. Then 0 < < δ implies < < < <. 9 Using the first series of equivalent inequalities, you obtain f < 0... ( ) 8 L ( ) 8 < < 0.0 < < < δ 0.0 So, if 0 < < δ, you have < < < 0.0 ( ) f L < ( ) L ( ) < 0.0 ( )( ) < 0.0 < 0.0 < 0.0 < 0.0 If you assume < <, then δ So, if 0 < < δ 0.00, you have < < < 0.0 ( ) < 0.0 < 0.0 f 5. ( ) ( ) f L < Given ε > 0: ( ) 5 < ε < ε So, let δ ε. So, if 0 < < δ ε, you have < ε 5 < ε L < ε. 7. ( ) Given ε > 0: ( ) ( ) < ε < ε ( ) < ε < ε So, let δ ε. So, if ( ) 0 < < δ ε, you have < ε < ε < ε f L < ε. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6 Section. Finding Limits Graphically and Numerically Given ε > 0: < ε 0 < ε So, any δ > 0 will work. So, for any δ > 0, you have < ε f L < ε. 0 0 Given ε > 0: So, let δ ε. 0 < ε < ε < ε So, for 0 < 0 < δ ε, you have f < ε < ε 0 < ε L < ε. 5. Given ε > 0: < ε < So, let δ ε. So, for ( ) ε ( 0) ε < < 0 < < δ ε, you have < ε < ε ( ) ( ) < ε ( ) < ε because < 0 f L < ε. 55. ( ) Given ε > 0: ( ) < ε < ε ( )( ) < ε ε < If you assume 0 < <, then δ ε. ε So, for 0 < < δ, you have ( ) < ε < ε < ε < ε f 57. f < ε. π 59. f f π The domain is [ 5, ) (, ). The graphing utility does not show the hole at,. 6 9 f 6 6. f The domain is all 0 ecept 9. The graphing utility does not show the hole at ( 9, 6 ). 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7 50 Chapter Limits and Their Properties 6. f 5 means that the values of f approach 5 as 8 gets closer and closer to (i) The values of f approach different numbers as approaches c from different sides of c: y 69. f ( ) 0 e.788 y 7 (0,.78) 5 (ii) The values of f increase or decrease without bound as approaches c: y f () f () (iii) The values of f oscillate between two fied numbers as approaches c: y (.999, 0.00) (.00, 0.00) 67. (a) C π r C 6 r cm π π π 5.5 (b) When C 5.5: r cm π 6.5 When C 6.5: r.05 cm π So, < r <.05. πr 6; ε 0.5; δ (c) π Using the zoom and trace feature, δ So, δ, δ.999,.00. Note: for. 7. False. The eistence or noneistence of f at c has no bearing on the eistence of the it f as c. of 75. False. Let, f 0, f 0 f ( ) f 0.5 is true. 0.5 As approaches 0.5 from either side, f approaches Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8 Section. Finding Limits Graphically and Numerically Using a graphing utility, you see that sin sin, etc. 0 0 So, sin n n If and f L c f L, then for every ε > 0, there eists δ > 0 and δ > 0 such that c < δ < ε and δ c f L < < ε. Let δ equal the smaller of c f L δ and δ. Then for c δ, L L L f f L L f f L < ε ε. So, L L < ε. Because ε > 0 is arbitrary, it follows that L L. < you have 8. f L 0 c means that for every ε > 0 there eists δ > 0 such that if 0 < c < δ, then ( f L) 0 < ε. This means the same as f L < ε when 0 < c < δ. So, f L. c 85. Answers will vary. 87. The radius OP has a length equal to the altitude z of the h h triangle plus. So, z. h Area triangle b Area rectangle bh Because these are equal, h b bh h h 5 h P h. 5 h O b Section. Evaluating Limits Analytically h (a) h 0 (b) h 5 f cos (a) f 0 (b) f π π 6 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9 5 Chapter Limits and Their Properties ( ) ( ) sec sec 0 0 5π sin sin 5π 6 6 π π tan tan π sin sin π π π cos cos 0 9. e cos e cos 0 0 ln e ln e.. (a) f 5 g 6 (b) (c) g( f ) g( f) g 6 5. (a) f (b) g (c) g( f ) g 7. (a) g g c c (b) f g f g 5 c c c 6 c c c (c) f( g ) f g (d) f cg f c g c 6 c c f f 9. (a) f f (b) c c (c) f f c c 8 c c (d) f f. f and g agree ecept at 0. g f 0 (a) 0 0 (b) g f agree ecept at. g f. f ( ) and g (a) (b) g f 5. f 0 and g. f g agree ecept at 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10 Section. Evaluating Limits Analytically 5 8 and g agree ecept at. f g 7. f 9. f ln 6 ln 6 and g 6 agree ecept at. ln f g ( )( ) ( ) ( )( 5 ) ( ) ( ) ( ) ( ) ( ) sin sin 67. () ( ) sin cos sin cos 0 0 () cos sin cot π π e 0 ( e ) e e e e e e e e sin sin 7. sin () sin ( h) h 0 h 0 cos cos h cos h h ( h) sin t sin t 79. () t 0 t t 0 t 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

11 5 Chapter Limits and Their Properties 8. f f () ? It appears that the it is 0.5. The graph has a hole at 0. Analytically, 0 0 ( ) ( ) 8. f f () ? It appears that the it is The graph has a hole at 0. ( ) Analytically, f() t sin t t t f (t) ? It appears that the it is. The graph has a hole at t 0. sin t sin t Analytically, (). t 0 t t 0 t 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

12 Section. Evaluating Limits Analytically f sin f () ? It appears that the it is 0. The graph has a hole at 0. sin sin Analytically, 0() f ln f () It appears that the it is. ln Analytically, f f f f ( ) ( ) ( )( ) ( )( ) ( )( ) ( ) ( ) f ( ) So, f 0 0 f. 97. f cos 99. f sin sin ( ) cos 0 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

13 56 Chapter Limits and Their Properties 0. f sin 07. f, g sin, h sin 0.5 g h f sin You say that two functions f and g agree at all but one point (on an open interval) if f g for all in the interval ecept for c, where c is in the interval. 05. An indeterminant form is obtained when evaluating a it using direct substitution produces a meaningless fractional epression such as 00.That is, f cg for which f g 0. c c When the -values are "close to" 0 the magnitude of f is approimately equal to the magnitude of g. So, g f when is "close to" st t s s( t) ( 6t 500) t t t t 6 6t 500 t t 6( t ) t t 6( t )( t ) t t 6 6 ft/sec t ( t ) The wrench is falling at about 6 feet/second.. st ().9t 00 s () s() t.9() 00 (.9t 00) t t t t t ( t ).9 9 t t.9 t t ( t )( t ) ( t ).9 9. m/sec The object is falling about 9. m/sec.. Let f and g /. f 0 g do not eist. However, and 0 f g [] and therefore does eist. ε > 5. Given f b, show that for every 0 a δ > 0 such that f b < ε whenever there eists c < δ. Because f b b b 0 < ε for every ε > 0, any value of δ > 0 will work. 7. If b 0, the property is true because both sides are equal to 0. If b 0, let ε > 0 be given. Because f L, there eists δ > 0 such that c f L < ε b whenever 0 < c < δ. So, whenever 0 < c < δ, we have ε or which implies that bf b f L < bf bl < ε bl. c 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

14 Section. Continuity and One-Sided Limits M f f g M f ( M f ) f g M f c c c So, ( 0) ( 0) c 0 f g 0 M f g M c f g 0. c. Let, if 0 f, if < 0 f f does not eist because for < 0, f and for f. False. The it does not eist because the function approaches from the right side of 0 and approaches from the left side of True 7. False. The it does not eist because f ( ) approaches from the left side of and approaches 0 from the right side of. 0, cos cos cos cos cos sin 0 0 sin sin 0 cos sin sin 0 0 cos () ( cos ) ( cos ) sec f (a) The domain of f is all 0, π / nπ. (b) The domain is not obvious. The hole at 0 is not apparent. (c) f 0 sec sec sec sec (d) sec ( sec ) tan sin ( sec ) cos sec sec sin So, 0 0 cos sec ().. The graphing utility was set in degree mode, instead of radian mode. Section. Continuity and One-Sided Limits. (a) f (b) f (c) f The function is continuous at and is continuous on (, ).. (a) f 0 (b) f 0 (c) f 0 The function is NOT continuous at. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

15 58 Chapter Limits and Their Properties 5. (a) f 7. (b) f (c) f does not eist. The function is NOT continuous at does not eist because 9 decreases without bound as ( 0) ( ) 5 7. f 9. f ( ) ( ) f f. cot does not eist because π cot and cot do not eist. π π. ( ) ( for < ) 5. ( ) ( ) does not eist because 5 and 6. ( ) 7. ( ) ln ln 0 does not eist. 9. ln ln ln has discontinuities at and because f are not defined.. f f ( ) and. f has discontinuities at each integer k f f. because k k 5. g 9 is continuous on [ 7, 7 ]. 7. f f. f is continuous on [, ] f. f. f 6 has a nonremovable discontinuity at 0. 9 is continuous for all real. has nonremovable discontinuities at and f do not eist. ± because f 5. f cos is continuous for all real. is not continuous at 0,. Because for 0, 0 is a removable discontinuity, whereas is a nonremovable discontinuity. 7. f 9. f is continuous for all real. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

16 Section. Continuity and One-Sided Limits f ( )( 5) discontinuity at 5 has a nonremovable because f 5 does not eist, and has a removable discontinuity at because f f because f has a nonremovable discontinuity at does not eist., f, > has a possible discontinuity at f ( ). f f f. f ( ) f f is continuous at, so, f is continuous for all real. 57. f,, > has a possible discontinuity at. f f f does not eist. f ( ).. So, f has a nonremovable discontinuity at. 59. f π tan, <, π tan, < <, or has possible discontinuities at,.. f f. f f( ). f ( ) f f( ) f f is continuous at ±, so, f is continuous for all real. 6. f ( ) ln, 0, < 0 has a possible discontinuity at 0.. f ( 0) ln( 0 ) ln f 0 f f 0 0 does not eist. So, f has a nonremovable discontinuity at f csc has nonremovable discontinuities at integer multiples of π f has nonremovable discontinuities at each integer k. 67. f f 0 f is not continuous at. 69. f ( a ) Find a so that 7. f () a a 7. 8 Find a so that a 8 a. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

17 60 Chapter Limits and Their Properties 7. Find a and b such that ( a b) a b and a b a b. a b a b, f a, < < a, b 75. f ( ) arctan( ) Find a such that ( ae ) ae a a. 77. f( g ) ( ) Continuous for all real 79. f( g ) ( 5) 6 Nonremovable discontinuities at ± 8. y Nonremovable discontinuity at each integer f sin The graph appears to be continuous on the interval [, ]. Because f ( 0) is not defined, you know that f has a discontinuity at 0. This discontinuity is removable so it does not show up on the graph. 9. f ( ) ln 8. g, > 5, Nonremovable discontinuity at f.5 Continuous on (, ) π 87. f( ) sec 8 The graph appears to be continuous on the interval [, ]. Because f ( 0) is not defined, you know that f has a discontinuity at 0. This discontinuity is removable so it does not show up on the graph. f is continuous on the interval 9. 7 [, ]. f ( ) and f ( 8 ). By the Intermediate Value Theorem, there eists a number c in [, ] such that f( c ) 0. π 95. h is continuous on the interval 0,. and h π 0.9 > 0. By the Intermediate Value π Theorem, there eists a number c in 0, such that hc 0. h ( 0) < 0 Continuous on: ( ) ( ), 6,,,,, 6, 6,0, 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

18 Section. Continuity and One-Sided Limits f f is continuous on [ 0, ]. f ( 0) and f () By the Intermediate Value Theorem, f() c 0 for at least one value of c between 0 and. Using a graphing utility to zoom in on the graph of f, you find that Using the root feature, you find that g() t cost t g is continuous on [ 0, ]. g( 0) > 0and g ().9 < 0. By the Intermediate Value Theorem, gc () 0 for at least one value of c between 0 and. Using a graphing utility to zoom in on the graph of g(), t you find that t Using the root feature, you find that t f e f is continuous on [ 0, ]. 0 f e f e e 0 < 0 and > 0. By the Intermediate Value Theorem, f( c ) 0 for at least one value of c between 0 and. Using a graphing utility to zoom in on the graph of f, you find that Using the root feature, you find that f f is continuous on [ 0, 5 ]. f f 0 and 5 9 < < 9 The Intermediate Value Theorem applies. 0 ( )( ) 0 or c is not in the interval. So, f (). 05. f f is continuous on [ 0, ]. f f 0 and 9 < < 9 The Intermediate Value Theorem applies ( )( ) ( has no real solution. ) c So, f. 07. (a) The it does not eist at c. (b) The function is not defined at c. (c) The it eists at c, but it is not equal to the value of the function at c. (d) The it does not eist at c. 09. If f and g are continuous for all real, then so is f (Theorem., part ). However, f g might not be continuous if g 0. For eample, let f g and g. Then f and g are continuous for all real, but f g is not continuous at ±.. True. f ( c) L is defined.. f ( ) Leists. c. f ( c) f( ) c All of the conditions for continuity are met.. False. A rational function can be written as P Q where P and Q are polynomials of degree m and n, respectively. It can have, at most, n discontinuities. 5. f( t) 8 t t () f t 56 At the end of day, the amount of chlorine in the pool has decreased to about 8 oz. At the beginning of day, more chlorine was added, and the amount is now about 56 oz. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

19 6 Chapter Limits and Their Properties 0.0, 0 < t 0 Ct t 9, t > 0, tnot an integer t 0, t > 0, t an integer 7. () C There is a nonremovable discontinuity at each integer greater than or equal to 0. Note: You could also epress C as 0.0, 0 < t 0 Ct () t, t > 0 9. Let s() t be the position function for the run up to the s 0 0 t 0corresponds to 8:00 A.M., campsite. ( ( 0 ) s k (distance to campsite)). Let rt () be the position function for the run back down the mountain: r 0 k, r 0 0. Let f ( t) s( t) r( t ). When t 0 (8:00 A.M.), f 0 s 0 r 0 0 k < 0. When 0 t (8:0 A.M.), f s r > 0. Because 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. t s t r t So, at some time t, where 0 t 0, the position functions for the run up and the run down are equal. in [ a, b] such that f( ) > 0 and there eists in [ a, b] such that f must equal zero for some value of in [, ] ( or [, ] if < ). [ a, b], which is a contradiction. Therefore, f > 0. Suppose there eists f < 0. Then by the Intermediate Value Theorem, So, f would have a zero in all in [ a, b] or f < 0 for all in [ a, b ].. If 0, 5. (a) then f ( 0) 0 and f 0. So, f is 0 continuous at 0. If 0, f t 0 for rational, whereas then t f t kt k 0 for irrational. So, f is not t t continuous for all f S (b) There appears to be a iting speed and a possible cause is air resistance., c, > c f is continuous for < c and for > c. At c, you need c c. Solving c c 0, you obtain ± ± 5 c. t for 9. f c c, c > 0 c c and 0, c, 0) ( 0, ) Domain: 0 c c c c c c c c c c c c c c c Define f ( 0) ( c). h 5 to make f continuous at 0. h has nonremovable discontinuities at ±, ±, ±,. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

20 Section.5 Infinite Limits 6. f( ) (a) Domain: all 0 (b) P() P P P P P() P P P 5 5 Continuing this pattern, you see that P infinitely many values of. So, the finite degree polynomial must be constant: P for all. for (c) f f, (d) For near 0 and negative, 0 For near 0 and positive, 0.. Section.5 Infinite Limits. f As approaches from the left, is a small negative number. So, f. As approaches from the right, is a small positive number. So, f.. f ( ) As approaches from the left or right, ( ) small positive number. So, f f. 5. is a 7. π tan π tan 9. f f ( ) f f Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

21 6 Chapter Limits and Their Properties. f f ( ) f f So, 0 is a vertical asymptote. 5. and So, is a vertical asymptote. and So, is a vertical asymptote. 7. No vertical asymptote because the denominator is never zero. 9. ( )( ) ( )( ) So, is a vertical asymptote. ( )( ) So, is a vertical asymptote t t t t So, t 0 is a vertical asymptote.. f Vertical asymptotes at and. 5. f ( )( ) No vertical asymptote because f. The graph has a hole at. 7. f e is a vertical asymptote. ( t ) ln 9. ht () t t is a vertical asymptote.. f e 0 is a vertical asymptote.. f sin π tan π has vertical asymptotes at cos π n, n any integer. t 5. st () has vertical asymptotes at t nπ, n a sin t nonzero integer. There is no vertical asymptote at t 0 because t. t 0sin t 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

22 Section.5 Infinite Limits ( ) sin e ( 8) 9. Removable discontinuity at π ln cos ln cos ln 0 ( π ) sec π ( ) and sec π. ( ) So, sec π does not eist. ( ) 6. f f 8 Vertical asymptote at 5. f ( e ) e ( e )( e ) e e, f 5 f Removable discontinuity at ( ) ( 6) ( )( ) A it in which f increases or decreases without bound as approaches c is called an infinite it. is not a number. Rather, the symbol f ( ) c says how the it fails to eist. 69. One answer is f. 6 y ( )( ) 0 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

23 66 Chapter Limits and Their Properties m0 7. m v ( c ) m0 m v c v c ( v c ) π 00π 75. (a) r 50π sec ft sec 6 π (b) r 50π sec 00π ft sec (c) 50π sec θ θ ( π ) 77. (a) (b) Total distance Average speed Total time d 50 ( d ) ( d y) 50 y y 50y 50 y 50 y 50y 50 y y Domain: > y (c) A bh r θ 0 0 tan θ 0 θ 50 tan θ 50θ 79. (a) π Domain: 0, (b) θ f ( θ ) As gets close to 5 mi/h, y becomes larger and larger (c) θ π A 8. False. For instance, let f or. g 8. False. The graphs of y tan, y cot, y sec and y csc have vertical asymptotes. 85. Let f and g and but, , and c Given, let. f g Then c g 0 by Theorem.5. c f 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

24 Review Eercises for Chapter 67 < < δ. Equivalently, < whenever < δ, >. M 89. f is defined for all >. Let M > 0 be given. You need 0 So take δ. Then for > and M Review Eercises for Chapter < δ, > M 8 δ > such that f and so f > M. > M whenever. Calculus required. Using a graphing utility, you can estimate the length to be 8.. Or, the length is slightly longer than the distance between the two points, approimately f f () f f () f 0 7. h ( ), 0 (a) h 0 0 (b) h 5 ( t ) ln 9. f() t t (a) f ( t) t 0 (b) f( t) 0 t does not eist. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

25 68 Chapter Limits and Their Properties. ( ) 5 Let ε > 0 be given. Choose δ ε. Then for 0 < < δ ε, you have < ε 5 < ε ( ) f. ( ) L < ε. Let ε > 0 be given. You need < ε < ε <. ε ε Assuming < <, you can choose δ. 5 ε So, for 0 < < δ, you have 5 ε ε < < 5 < ε ( ) ( ) f < ε < ε < ε L < ε. 5. ( ) t 6.5 t t 9. t t t t. ( ) ( )( ). 5. ( ) ( ) ( ) ( 5)( 5 5) ( ) cos cos 0 0 sin sin 7. () 0 0 π 0 π 9. e sin e sin. ( π ) ( π ) ( π ) sin 6 sin 6 cos cos 6 sin ( cos ) sin () Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

26 Review Eercises for Chapter 69 c. f g 7 c 5. f g 7. f (a) f() Actual it is. (b) The graph has a hole at. f (c) ( ) ( )( ) s s t 9. v t t t t 50 t t ( t ).9 6 t t.9 t t ( t )( t ) ( t ).9 9. m/sec The object is falling at about 9. m/sec.. ( ). f 0 5. ht t does not eist because ht () and ht. t t 7. f 7 Continuous on (, ) 9. f k k where k is an integer. k where k is an integer. k Nonremovable discontinuity at each integer k Continuous on ( k, k ) for all integers k 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

27 70 Chapter Limits and Their Properties 5. f ( ) ( )( ) f 5 Removable discontinuity at Continuous on (,) (, ) 5. f ( ) ( ) Nonremovable discontinuity at Continuous on (,) (, ) f 55. f f Nonremovable discontinuity at Continuous on (, ) (, ) 57. csc f π Nonremovable discontinuities at each even integer. Continuous on ( k, k ) for all integers k. 59. g e is continuous on all intervals ( n n ),, where n is an integer. g has nonremovable discontinuities at each n. 6. f 5 Find c so that ( c ) c 6 5 c c f is continuous on [, ]. f () < 0 and f > 0. So by the Intermediate Value Theorem, there is at least one value c in (, ) such that c t 65. A Nonremovable discontinuity every 6 months 67. g Vertical asymptote at f 8 ( 0) Vertical asymptote at 0 7. g ln( 5 ) 7. Vertical asymptotes at 5 and sin sin csc 0 0 sin 8. ln ( sin ) 0 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

28 Problem Solving for Chapter f (a) tan tan 0 (b) Yes, define f tan, 0., 0 Now f is continuous at 0. Problem Solving for Chapter. (a) Perimeter PAO y y Perimeter PBO y y (b) r f() (c) r Perimeter Perimeter PAO PBO r Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

29 7 Chapter Limits and Their Properties. (a) There are 6 triangles, each with a central angle of 60 π. So, Area heagon 6 6 π bh () sin.598. h sin h sin θ θ Error Area (Circle) Area (Heagon) π 0.55 (b) There are n triangles, each with central angle of θ π n. So, π n sin( π n) An n bh n () sin. n (c) n (d) As n gets larger and larger, π n approaches 0. Letting π n, 5. (a) (b) A n which approaches Slope 5 π π. 5 Slope of tangent line is. 5 y ( 5) y 5 69, (c) Q (, y) (, 69 ) A n ( π ) ( π ) ( π ) sin n sin n sin π π n n m 5 69 (d) m ( ) ( 5)( 69 ) 5 ( 5)( 69 ) 5 ( ) This is the same slope as part (b). 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

30 Problem Solving for Chapter 7 7. (a) 0 (b) 7 Domain: 7, or [ 7, ) (, ) (c) f (d) f ( )( ) ( )( )( ) 9. (a) ( )( ) ( )( ) f : g, g (b) f continuous at : g (c) f : g, g, g 0.. (a) y. (a) f f ( 0) 0 f 0 0 f.7 (b) f y f f (c) f is continuous for all real numbers ecept 0, ±, ±, ±,. (b) (i) P a ab, (ii) P 0 a ab, (iii) P 0 b ab, (iv) P b a ab, b (c) Pab, is continuous for all positive real numbers ecept a, b. (d) The area under the graph of U, and above the -ais, is. 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.