CHAPTER 2 LIMITS AND CONTINUITY

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CHAPTER LIMITS AND CONTINUITY RATES OF CHANGE AND LIMITS (a) Does not eist As approaches from the right, g() approaches 0 As approaches from the left, g() approaches There is no single number L that all the values g() get arbitrarily close to as Ä (c) 0 (a) 0 (c) Does not eist As t approaches 0 from the left, f(t) approaches As t approaches 0 from the right, f(t) approaches There is no single number L that f(t) gets arbitrarily close to as t Ä 0 (a) True True (c) False (d) False (e) False (f) True (a) False False (c) True (d) True (e) True 5 lim does not eist because if 0 and if 0 As approaches 0 from the left, Ä 0 kk kk kk kkapproaches As approaches 0 from the right, kkapproaches There is no single number L that all the function values get arbitrarily close to as Ä 0 6 As approaches from the left, the values of become increasingly large and As approaches from the right, the values become increasingly large and positive There is no one number L that all the function values get arbitrarily close to as Ä, so lim does not eist Ä 7 Nothing can be said about f() because the eistence of a limit as Ä! does not depend on how the function is defined at! In order for a limit to eist, f() must be arbitrarily close to a single real number L when is close enough to! That is, the eistence of a limit depends on the values of f() for near!, not on the definition of f() at! itself 8 Nothing can be said In order for lim f() to eist, f() must close to a single value for near 0 regardless of Ä 0 the value f(0) itself 9 No, the definition does not require that f be defined at in order for a limiting value to eist there If f() is defined, it can be any real number, so we can conclude nothing about f() from lim f() 5 Ä 0 No, because the eistence of a limit depends on the values of f() when is near, not on f() itself If lim f() eists, its value may be some number other than f() 5 We can conclude nothing about lim f(), Ä Ä whether it eists or what its value is if it does eist, from knowing the value of f() alone

58 Chapter Limits and Continuity (a) f() a * b/( ) 0 00 000 0000 00000 f() 6 60 600 6000 60000 600000 9 99 999 9999 99999 999999 f() 59 599 5999 59999 599999 5999999 The estimate is lim f() 6 Ä$ 9 ( )( ) (c) f() if Á, and lim ( ) 6 Ä$ (a) g() a b/ Š È g() 8 8 88 88 88 886 Š ÈŠ È (c) g() È if Á È, and lim È Š È È È È Š È Ä È (a) G() ( 6)/ a b 59 599 5999 59999 599999 5999999 G() 658 556 5056 5005 5000 50000 6 60 600 6000 60000 600000 G() 56 8 98 998 999 999

Section Rates of Change and Limits 59 6 6 (c) G() a b (6)() if Á6, and lim ' 8 05 Ä' (a) h() a b/ a b 9 99 999 9999 99999 999999 h() 056 00505 000500 000050 000005 0000005 0 00 000 0000 00000 h() 9580 9950 999500 999950 999995 999999 ( )( ) (c) h() ()() if Á, and lim Ä$ 5 (a) f() a b/ akkb 0 00 000 0000 00000 f() 0 00 000 0000 00000 9 99 999 9999 99999 999999 f() 9 99 999 9999 99999 999999

60 Chapter Limits and Continuity ( )( ) (c) f(), 0 and Á kk ( )( ), and lim ( ) ( ), 0 and Á Ä ( ) 6 (a) F() a b/ a kkb 0 00 000 0000 00000 F() 0 00 000 0000 00000 9 99 999 9999 99999 999999 F() 9 99 999 9999 99999 999999 ( )( ), 0 (c) F() kk ( )( ), and lim ( ), 0 and Á Ä 7 (a) g( )) (sin ))/) ) 0 00 000 0000 00000 g( )) 998 99998 999999 999999 999999 999999 ) 0 00 000 0000 00000 g( )) 998 99998 999999 999999 999999 999999 lim g( )) ) Ä! 8 (a) G(t) ( cos t)/t t 0 00 000 0000 00000 G(t) 9958 99995 99999 5 5 5 t 0 00 000 0000 00000 G(t) 9958 99995 99999 5 5 5 lim G(t) 05 t Ä!

Section Rates of Change and Limits 6 Graph is NOT TO SCALE 9 (a) f() 9 99 999 9999 99999 999999 f() 8678 660 67695 6786 67877 67879 0 00 000 0000 00000 f() 855 697 6806 67897 6788 67878 lim f() 06788 Ä Graph is NOT TO SCALE Also the intersection of the aes is not the origin: the aes intersect at the point (ß 780) 0 (a) f() a b/ 0 00 000 0000 00000 f() 6 0669 0995 09867 09868 0986 0 00 000 0000 00000 f() 005 09599 098009 09855 098606 0986 lim f() 0986 Ä! lim () lim (0) 0 Ä Ä! lim ( ) ˆ 0 lim Ä Ä () $

6 Chapter Limits and Continuity ( ) 5 lim ( ) ( )(( ) ) 9 6 lim Ä Ä ( ) 7 lim sin sin 8 lim Ä Ä cos cos? f f() f() 8 9? f f() f( ) 0?? ( ) 9 (a) 9? g g() g( )? g g(0) g( ) 0? ( )? 0 ( ) 0 (a) 0? h hˆ hˆ? h hˆ hˆ 0È È? t? t 6 6 (a)? g g( ) g(0) ( ) ( )? g g( ) g( ) ( ) ( )? t 0 0? t ( ) (a) 0? R R() R(0) È8È?) 0? P P() P() (8 6 0) (%&)?) 0 5 (a) Q Slope of PQ 650 5? p? t Q (0ß5) 0 0 5 m/sec 650 75 Q (ß75) 0 58 m/sec 650 75 Q $ (65ß75) 0 65 5000 m/sec 650 550 Q %(8ß550) 0 8 5000 m/sec At t 0, the Cobra was traveling approimately 50 m/sec or 80 km/h 6 (a) Q Slope of PQ 7 (a) 80 0? p? t Q (5ß0) 0 5 m/sec 80 9 Q (7ß9) 0 7 7 m/sec 80 58 Q $ (85ß58) 0 85 7 m/sec 80 7 Q %(95ß7) 0 95 6 m/sec Approimately 6 m/sec? p 7 6? t 99 99 56 thousand dollars per year (c) The average rate of change from 99 to 99 is The average rate of change from 99 to 99 is? p 6 7? t 99 99? p 6? t 99 99 5 thousand dollars per year 9 thousand dollars per year So, the rate at which profits were changing in 99 is approimatley a 5 9 b thousand dollars per year

8 (a) F() ( )/( ) 0 00 000 F() 0 0 00 000? F 0 ( )? F ( )??? F 0 ( )? F 00 ( )? 0? 00? F!!!% ( )? 000!!!%; 50; ; 0;!!%; The rate of change of F() at is? g g() g() È? g g(5) g() È5?? 5 05? h) g() Èh? (h) h Section Rates of Change and Limits 6 9 (a) 0 0989 g() È h 0 00 000 0000 00000 È h 0880 00987 000998 000099 000005 0000005 Š È h /h 0880 0987 0998 099 05 05 (c) The rate of change of g() at is 05 Èh (d) The calculator gives lim h h Ä! 0 (a) i) c f() f() 6 6 T f(t) f() T T T T T T T T T(T) T( T) T ii), T Á T 0 00 000 0000 00000 f(t) 07690 0975 099750 0999750 099997 099999 af(t) f() b/ at b 08 088 0500 0500 0500 0500 (c) The table indicates the rate of change is 05 at t (d) lim T Ĉ T -6 Eample CAS commands: Maple: f := -> (^ 6)/( ); 0 := ; plot( f(), 0-0+, color black, title Section, (a) ); limit( f(), 0 ); In Eercise, note that the standard cube root, ^(/), is not defined for <0 in many CASs This can be overcome in Maple by entering the function as f := -> (surd(+, ) )/ Mathematica: (assigned function and values for 0 and h may vary) Clear[f, ] f[_]:=( 5 )/( ) 0= ; h = 0; Plot[f[],{, 0 h, 0 h}] Limit[f[], Ä 0]

6 Chapter Limits and Continuity CALCULATING LIMITS USING THE LIMIT LAWS lim ( 5) ( 7) 5 5 9 lim (0 ) 0 () 0 6 6 Ä( Ä lim a 5 b () 5() 0 Ä $ $ lim a 8 b ( ) ( ) ( ) 8 8888 6 Ä 5 lim 8(t 5)(t 7) 8(6 5)(6 7) 8 6 lim s(s ) ˆ ˆ ˆ t Ä' s Ä 5 7 lim 8 lim Ä 6 6 8 Ä& 7 57 $ 9 lim y Ä& y ( 5) 5 5 5y 5 ( 5) 0 0 lim y Ä y y 5y6 () 5( ) 6 0 6 0 5 lim ( ) (( ) ) ( ) 7 Ä *)% *)% *)% lim ( ) ( ) ( ) Ä% %Î$ %Î$ %Î$ lim (5 y) [5 ( )] (8) ˆ Î$ (8) % 6 y Ä$ Î$ Î$ Î$ lim (z 8) ((0) 8) ( 8) z Ä! % 5 lim h Ä! 6 lim h Ä! È h È (0) È 5 5 5 5 È 5h È 5(0) È Èh Èh Èh ah b h h h h Èh 7 lim lim lim lim lim h Ä 0 h Ä 0 È h Ä 0 hš Èh h Ä 0 hš Èh h Ä 0 $ È È5h È5h È5h a5hb 5h 5 h h 5h È5h 8 lim lim lim lim lim h Ä 0 h Ä 0 È h Ä 0 hš È5h h Ä 0 hš È5h h Ä 0 5 5 È 5 5 5 ( 5)( 5) 5 5 5 0 9 lim lim lim Ä& Ä& Ä& ( )( ) 0 lim lim lim Ä$ Ä$ Ä$ 0 ( 5)( ) lim lim lim ( ) 7 Ä& 5 Ä& 5 & Ä&

7 0 ( 5)( ) lim lim lim ( 5) 5 Ä Ä Ä t t (t )(t ) t lim lim lim t Ä t t Ä (t )(t ) t Ä t t t (t )(t ) t t t (t )(t ) t lim lim lim t Ä t Ä t Ä ( ) ( ) 5 lim $ lim lim Ä Ä Ä 5y 8y y (5y 8) 5y 8 $ 8 6 lim lim lim y Ä 0 y% 6y y Ä! y y 6 y Ä! y 6 6 a b Section Calculating Limits Using the Limit Laws 65 u au b(u)(u) au b(u) ( )( ) u u u (u ) u u % 7 lim $ lim lim u Ä u Ä a b u Ä v 8 (v ) av v b v v (v )(v ) v (v ) v ()(8) 8 $ 8 lim lim lim v Ä v% 6 v Ä a b v Ä a b È È 9 6 9 lim lim lim Ä* Ä* ˆ È ˆ È Ä* È È 9 ( ) ˆ È ˆ È 0 lim lim lim lim ( ) 6 Ä% È Ä% È Ä% È ˆ È Ä% ˆ È ˆ È () () lim lim lim lim Ä È Ä È È Ä ( ) Š È ˆ ˆ Ä È È Š È Š È 8 8 8 $ $ a 8b* ( ) Š 8 $ ( ) Š È 8 $ lim lim lim Ä Ä È Ä ( )( ) lim lim Ä ( ) Š È )$ Ä È )$ È Š È Š È a b6 ( ) Š ( ) Š È lim lim lim Ä Ä È Ä ( )( ) 6 lim lim Ä ( ) Š È Ä È È abš È 5 abš È 5 lim lim lim Ä È 5 Ä Š Š Ä È 5 È 5 a 5b9 abš È 5 È 5 È9 ( )( ) lim lim Ä Ä È 5 Š È 5Š È 5 a 5b ( ) Š 5 ( ) Š È 5 5 lim lim lim Ä Ä È Ä 9 ( )( ) 6 5 lim lim lim Ä ( ) 5 Ä ( ) 5 Ä Š È Š È È È

66 Chapter Limits and Continuity abš 5È 9 abš 5È 9 6 lim lim lim Ä 5È 9 Ä Š Š Ä 5 È 9 5 È 9 5 a 9 b abš 5È 9 abš 5È 9 5È 9 5 È5 5 6 ( )( ) 8 lim lim lim Ä Ä Ä 7 (a) quotient rule difference and power rules (c) sum and constant multiple rules 8 (a) quotient rule power and product rules (c) difference and constant multiple rules 9 (a) lim f() g() lim f() lim g() (5)( ) 0 Ä c Ä c Ä c lim f() g() lim f() lim g() (5)( ) 0 Ä c Ä c Ä c (c) lim [f() g()] lim f() lim g() 5 ( ) Ä c Ä c Ä c f() (d) lim lim f() 5 5 Ä c f() g() Äc lim f() lim g() 5 ( ) 7 Äc Äc 0 (a) lim [g() ] lim g() lim $$! Ä% Ä% Ä% lim f() lim lim f() ()(0) 0 Ä% Ä% Ä% (c) lim [g()] lim g() [ ] 9 Ä% Ä% g() lim g() (d) lim Ä% f() Ä% lim f() lim 0 Ä% Ä% (a) lim [f() g()] lim f() lim g() 7 ( ) Ä b Ä b Ä b lim f() g() lim f() lim g() (7)( ) Ä b Ä b Ä b (c) lim g() lim lim g() ()( ) Ä b Ä b Ä b 7 7 (d) lim f()/g() lim f()/ lim g() Ä b Ä b Ä b (a) lim [p() r() s()] lim p() lim r() lim s() 0 ( ) Ä Ä Ä Ä lim p() r() s() lim p() lim r() lim s() ()(0)( ) 0 Ä Ä Ä Ä (c) lim [ p() 5r()]/s() lim p() 5 lim r() lim s() [ () 5(0)]/ Ä Ä Ä Ä ( h) hh h( h) lim lim lim lim ( h) h Ä! h h Ä! h h Ä! h h Ä! ( h) ( ) hh h(h) lim lim lim lim (h ) h Ä! h h Ä! h h Ä! h h Ä! [( h) ] [() ] 5 lim lim h Ä! h h Ä! h h ˆ ˆ c ( h) h h h h( h) h( h) c b h c cb h 6 lim lim lim lim h Ä! h Ä! h Ä! h Ä! 6

È È Š È È Š È È 7h 7 7h 7 7h 7 (7 h) 7 hš 7h 7 hš È7hÈ7 7 lim lim lim h Ä! h h Ä! È È h Ä! h h È7 h È7 È7h È7 È7 lim lim h Ä! Š h Ä! Section Calculating Limits Using the Limit Laws 67 È È Š È Š È (0 h) (0) h h (h ) hš h hš Èh 8 lim lim lim h Ä! h h Ä! È h Ä! h lim lim h Ä! hš Èh h Ä! Èh 9 lim È5 È5 (0) È5 and lim È5 È5 (0) È5; by the sandwich theorem, Ä! Ä! lim f() È 5 Ä! 50 lim a b 0 and lim cos () ; by the sandwich theorem, lim g() Ä! Ä! Ä! 0 sin 5 (a) lim Š and lim ; by the sandwich theorem, lim Ä! 6 6 Ä! Ä! cos For Á 0, y ( sin )/( cos ) lies between the other two graphs in the figure, and the graphs converge as Ä 0 5 (a) lim Š lim lim 0 and lim ; by the sandwich theorem, Ä! Ä! Ä! Ä! cos lim Ä! For all Á 0, the graph of f() ( cos )/ lies between the line y and the parabola y /, and the graphs converge as Ä 0 % % 5 lim f() eists at those points c where lim lim Thus, c c Ê c a c b 0 Ä c Ä c Ä c Ê c 0,, or Moreover, lim f() lim 0 and lim f() lim f() Ä! Ä! Ä Ä 5 Nothing can be concluded about the values of f, g, and h at Yes, f() could be 0 Since the conditions of the sandwich theorem are satisfied, lim f() 5Á 0 Ä lim f() lim 5 lim f() 5 f() 5 55 lim Ä% Ä% Ä% Ê lim f() 5 () Ê lim f() 5 7 Ä% lim lim % Ä% Ä% Ä% Ä%

68 Chapter Limits and Continuity f() lim f() lim f() Äc Äc 56 (a) lim Ê lim f() Ä lim % Ä Äc f() f() f() f() lim lim lim lim Ê lim Ä Ä Ä ˆ Ä Ä f() 5 f() 5 57 (a) 0 0 lim lim ( ) lim Š ( ) lim [f() 5] lim f() 5 Ä Ä Ä Ä Ä Ê lim f() 5 Ä f() 5 0 0 lim lim ( ) Ê lim f() 5 as in part (a) Ä Ä Ä f() f() f() 58 (a) 0 0 lim lim lim lim lim lim f() That is, lim f( Ä! Ä! Ä! Ä! Ä! ) 0 Ä! Ä! f() f() f() f() 0 0 lim lim lim lim That is, lim 0 Ä! Ä! Ä! Ä! Ä! 59 (a) lim sin 0 Ä! Ÿ sin Ÿ for Á 0: 0 Ê Ÿ sin Ÿ Ê lim sin 0 by the sandwich theorem; Ä! 0 Ê sin Ê lim sin 0 by the sandwich theorem Ä! 60 (a) lim cos ˆ 0 Ä! $ Ÿ cos ˆ Ÿ for Á 0 Ê Ÿ cos ˆ Ÿ Ê lim cos ˆ $ $ $ 0 by the sandwich Ä! theorem since lim 0 Ä! THE PRECISE DEFINITION OF A LIMIT Step : k5 k $ Ê $ 5 $ Ê $ 5 $ 5 Step : $ 5 7 Ê $, or $ 5 Ê $ The value of $ which assures k 5 k $ Ê 7 is the smaller value, $

Section The Precise Definition of a Limit 69 Step : k k $ Ê $ $ Ê $ $ Step : $ Ê $, or $ 7 Ê $ 5 The value of $ which assures k k $ Ê 7 is the smaller value, $ Step : k ( ) k $ Ê $ $ $ Ê $ $ 7 5 Step : $ Ê $, or $ $ Ê $ 7 $ k k $ Ê $ The value of which assures ( ) is the smaller value, Step : ˆ $ Ê $ $ Ê $ $ 7 ˆ 7 Step : $ Ê $, or $ Ê $ The value of $ which assures $ Ê is the smaller value, $ 5 Step : $ Ê $ $ Ê $ $ 9 8 7 $ $ Ê 9 7 $ 8 Step : $ Ê $, or $ Ê $ The value of which assures is the smaller value, 6 Step : k k $ Ê $ $ Ê $ $ Step : $ $ 759 Ê $ 009, or $ $ 9 Ê $ 09 The value of $ which assures k k $ Ê 759 9 is the smaller value, $ 09 7 Step : k 5 k $ Ê $ 5 $ Ê $ 5 $ 5 Step : From the graph, $ 5 9 Ê $ 0, or $ 5 5 Ê $ 0; thus $ 0 in either case 8 Step : k ( ) k $ Ê $ $ Ê $ $ Step : From the graph, $ Ê $ 0, or $ 9 Ê $ 0; thus $ 0 9 Step : k k $ Ê $ $ Ê $ $ 9 7 5 9 7 6 6 6 6 6 Step : From the graph, $ Ê $, or $ Ê $ ; thus $ 0 Step : k k $ Ê $ $ Ê $ $ Step : From the graph, $ 6 Ê $ 09, or $ Ê $ 0; thus $ 09 Step : k k $ Ê $ $ Ê $ $ Step : From the graph, $ È Ê $ È 0679, or $ È5 Ê $ È5 06; thus $ È5

70 Chapter Limits and Continuity Step : k ( ) k $ Ê $ $ Ê $ $ È 5 È 5 È È Step : From the graph, $ Ê $ 080, or $ Ê $ 00; thus $ È 5 Step : k ( ) k $ Ê $ $ Ê $ $ 6 7 6 9 9 9 9 5 5 5 Step : From the graph, $ Ê $ 077, or $ Ê 06; thus $ 06 Step : Ê Ê $ $ $ $ $ Step : From the graph, $ 0 Ê $ 0 0008, or $ 99 Ê $ 99 0005; thus $ 0008 5 Step : k( ) 5k 00 Ê k k 00 Ê 00 00 Ê 99 0 Step : k k $ Ê $ $ Ê $ $ Ê $ 00 6 Step : k( ) ( 6) k 00 Ê k k 00 Ê 00 00 Ê 0 98 Ê 0 99 Step : k ( ) k $ Ê $ $ Ê $ $ Ê $ 00 7 Step : ¹ È ¹ 0 Ê 0 È 0 Ê 09 È Ê 08 Ê 09 0 Step : k 0 k $ Ê $ $ Then, $!Þ* Ê $!Þ* or $!Þ; thus, $ 09 8 Step : È 0 0 È 0 0 È Ê Ê 06 Ê 06 06 Step : Ê Ê $ $ $ $ $ Then, $ 06 Ê $ 009 or $ 06 Ê $ 0; thus $ 009 9 Step : ¹ È9 $ ¹ Ê È9 $ Ê È9 % Ê 9 6 Ê % 9 6 Ê 5 or 5 Step : k 0 k $ Ê $ 0 $ Ê $ 0 $ 0 Then $ 0 Ê $ 7, or $ 0 5 Ê $ 5; thus $ 5 0 Step : ¹ È 7 ¹ Ê È 7 % Ê È 7 5 Ê 9 7 5 Ê 6 Step : k k $ Ê $ $ Ê $ $ Then $ 6 Ê $ 7, or $ Ê $ 9; thus $ 7 Step : 0 0 0 005 Ê 005 005 Ê 0 0 Ê or 5 Step : k k $ Ê $ $ Ê $ $ 0 Then $ % or $, or $ 5 or $ ; thus $ Step : k k! Ê 0 0 Ê 9 Ê È9 È Step : ¹ È ¹ $ Ê $ È $ Ê $ È $ È Then $ È È9 Ê $ È È9 009, or $ È È Ê $ È È 0086; thus $ 0086

Section The Precise Definition of a Limit 7 Step : k k 05 Ê 05 05 Ê 5 5 Ê È 5 kk È 5 Ê È 5 È 5, for near Step : k ( ) k $ Ê $ $ Ê $ $ Then $ È5 Ê $ È5 0, or $ È5 Ê $ È5 09; thus $ È5 0 Step : ( ) 9 0 0 0 0 0 Ê 0 0 Ê 0 0 Ê 9 or 9 Step : k ( ) k $ Ê $ $ Ê $ $ 0 0 9 9 Then $ Ê $, or $ Ê $ ; thus $ 5 Step : ka 5b k Ê k 6k Ê 6 Ê 5 7 Ê È5 È7 Step : k k $ Ê $ $ Ê $ % $ % Then $ %È5 Ê $ %È5 070, or $ %È7 Ê $ È7 % 0; thus $ È7 0 6 Step : 0 5 0 0 Ê & Ê 6 Ê 0 6 Ê 0 0 or 0 0 Step : k k $ Ê $ $ Ê $ $ Then $ 0 Ê $, or $ 0 Ê $ 6; thus Ê $ 7 Step : km mk 00 Ê 00 m m 00 Ê 00 m m 00 m Ê 00 00 m m Step : k k $ Ê $ $ Ê $ $ 00 00 00 00 00 m m m m m Then $ Ê $, or $ Ê $ In either case, $ 8 Step : km mk c Ê c m m c Ê c m m c m Ê m Step : k k $ Ê $ $ Ê $ $ B $ $ c Then $ $ $ Ê $ c c, or $ $ $ Ê $ c In either case, $ c m m m m m 9 Step : (m b) ˆ m b m m m c c - Ê c m c Ê c m c Ê m m Step : Ê Ê $ $ $ $ $ c Then $ Ê $ c c, or $ Ê $ c In either case, $ c m m m m m 0 Step : k(m b) (m b) k 005 Ê 005 m m 005 Ê 005 m m 005 m 005 005 Ê m m Step : k k $ Ê $ $ Ê $ $ 005 005 005 005 005 m m m m m Then $ Ê $, or $ Ê $ In either case, $ lim ( ) () Ä Step : ka b( ) k 00 Ê 00 6 00 Ê 60 598 Ê 0 99 or 99 0 Step : 0 k k $ Ê $ $ Ê $ $ $ $ Then $ $ 99 Ê $ 00, or $ $ 0 Ê $ 00; thus $ 00 lim ( ) ( )( ) Ä Step : k( ) k 00 Ê 00 00 Ê 00 00 Ê 0 099 c c m

7 Chapter Limits and Continuity Step : k ( ) k $ Ê $ $ Ê $ $ Then $ 0 Ê $ 00, or $ 099 Ê $ 00; thus $ 00 ( )( ) lim lim lim ( ), Ä Ä () Á Ä ( )( ) Step : ¹ Š ¹ 005 Ê 005 () % 005 Ê 95 05, Á Ê 95 05, Á Step : k k $ Ê $ $ Ê $ $ Then $ 95 Ê $ 005, or $ 05 Ê $ 005; thus $ 005 65 ( 5)( ) lim lim lim ( ), 5 Ä& 5 Ä& (5) Á Ä& 65 ( 5)( ) Step : ¹ Š 5 ( ) ¹ 005 Ê 005 (5) 005 Ê 05 95, Á 5 Ê 505 95, Á 5 Step : k ( 5) k $ Ê $ 5 $ Ê $ & $ & Then $ &505 Ê $ 005, or $ &95 Ê $ 005; thus $ 005 5 lim È 5 È 5( ) È6 Ä$ Step : ¹ È 5 ¹ 05 Ê 05 È 5 05 Ê 5 È 5 5 Ê 5 5 05 Ê 5 5 95 Ê 85 5 Step : k ( ) k $ Ê $ $ Ê $ $ $ $ Then $ $85 Ê $ 085, or $ $5 Ê 075; thus $ 075 6 lim Ä Step : 0 0 0 0 5 5 0 Ê 0 0 Ê 6 Ê 6 Ê 6 or Step : k k $ Ê $ $ Ê $ $ 5 Then $ Ê 5 $, or $ Ê $ ; thus $ 7 Step : k(9 ) 5k % Ê % % Ê % % Ê % % % Ê %% % Step : k k $ Ê $ $ Ê $ % $ % Then $ % Ê $ %, or $ % % % Ê $ % Thus choose $ % 8 Step : % % k( 7) k % Ê % 9 % Ê 9 % *% Ê Step : k k $ Ê $ $ Ê $ $ % Then $ $ Ê % % $, or $ Ê % % $ Thus choose $ 9 Step : ¹ È 5 ¹ % Ê % È 5 % Ê % È 5 % Ê (%) 5 (%) Ê (%) & (%) 5 Step : k 9 k $ Ê $ 9 $ Ê $ 9 $ 9 Then $ *% % % * Ê $ % % %, or $ *% % % * Ê $ % % % Thus choose the smaller distance, $ % % % 0 Step : ¹ È ¹ % Ê % È % Ê % È % Ê (%) % (%) Ê ( %) ( %) Ê ( %) % ( %) % Step : k 0 k $ Ê $ $

Section The Precise Definition of a Limit 7 Then $ ( %) % % % Ê $ % % %, or $ ( %) % % Thus choose the smaller distance, $ % % Step : For Á, k k % Ê % % Ê % % Ê È % kk È % Ê È% È % near B Step : k k $ Ê $ $ Ê $ $ Then $ È % Ê $ È %, or $ È% Ê $ È% Choose $ min š È % ßÈ %, that is, the smaller of the two distances Step : For Á, k k % Ê % % Ê % % Ê È % kk È % Ê È % È % near B Step : k ( ) k $ Ê $ $ Ê $ $ Then $ È%% Ê $ È%%, or $ È%% Ê $ È%% Choose $ min š È%% ß È%% Step : % Ê % % Ê % % Ê % % Step : k k $ Ê $ $ Ê $ $ Then $ Ê $ %, or $ Ê $ % Choose $ % % % % % % % %, the smaller of the two distances Step : % $ % % Ê % % Ê % % Ê Ê $% $% Ê É k k É, or É É for near È$ $ % $% $% $% Step : ¹ È ¹ $ Ê $ È $ Ê È$ È $ Then È È, or È $ É Ê $ É $ É Ê $ É È $% $% $% $% Choose $ min š È É ßÉ È $% $% * 5 Step : ¹ Š ( 6) ¹ % Ê % ( ) 6 %, Á Ê % % Ê % $ % $ Step : k ( ) k $ Ê $ $ Ê $ $ $ Then $ $% $ Ê $ %, or $ $% $ Ê $ % Choose $ % 6 Step : ¹ Š ¹ % Ê % ( ) %, Á Ê % % Step : k k $ Ê $ $ Ê $ $ Then $ % Ê $ %, or $ % Ê $ % Choose $ % % 7 Step : : l( ) l % Ê! % since Þ Thus,!; % : l(6 ) l % Ê! Ÿ 6 6 % since Thus, Ÿ 6 Step : k k $ Ê $ $ Ê $ $ % % % % % Then $ Ê $, or $ Ê $ Choose $ 6 6 6 % 8 Step :!: k 0 k % Ê %! Ê 0; 0:! % Ê! Ÿ %

7 Chapter Limits and Continuity Step : k 0 k $ Ê $ $ % % % Then $ Ê $, or $ % Ê $ % Choose $ 9 By the figure, Ÿ sin Ÿ for all 0 and sin for 0 Since lim ( ) lim 0, Ä! Ä! then by the sandwich theorem, in either case, lim sin 0 Ä! 50 By the figure, Ÿ sin Ÿ for all ecept possibly at 0 Since lim a b lim 0, then Ä! Ä! by the sandwich theorem, lim sin 0 Ä! 5 As approaches the value 0, the values of g() approach k Thus for every number % 0, there eists a $! such that! k 0 k $ Ê kg() k k % 5 Write h c Then!lcl$ Í$ c $, Ác Í$ ah cbc $, h c Ác Í$ h $, h Á!Í!lh!l$ Thus, limf a b L Ífor any %!, there eists $! such that lf a b L l % whenever!lcl $ Äc ÍlfahcbL l % whenever!lh!l$ Ílim fah cb L 5 Let f() The function values do get closer to as approaches 0, but lim f() 0, not The Ä! function f() never gets arbitrarily close to for near 0 hä! 5 Let f() sin, L, and 0 There eists a value of (namely, ) for which sin % for any! 6 % Ä!!! Ä! given 0 However, lim sin 0, not The wrong statement does not require to be arbitrarily close to As another eample, let g() sin, L, and 0 We can choose infinitely many values of near 0 such that sin as you can see from the accompanying figure However, lim sin fails to eist The wrong statement does not require all values of arbitrarily close to! 0 to lie within % 0 of L Again you can see from the figure that there are also infinitely many values of near 0 such that sin 0 If we choose % we cannot satisfy the inequality sin % for all values of sufficiently near 0! 55 ka * k Ÿ 00 Ê 00 Ÿ ˆ 9 Ÿ 00 Ê 899 Ÿ Ÿ 90 Ê (899) Ÿ Ÿ (90) 899 90 Ê É Ÿ Ÿ É or 8 Ÿ Ÿ 87 To be safe, the left endpoint was rounded up and the right endpoint was rounded down V 56 V RI Ê I Ê V 5 0 0 0 R 0 Ÿ 0 Ê 0 Ÿ 5 Ÿ 0 Ê 9 Ÿ Ÿ 5 Ê Ê R R R R 9 0 5 (0)(0) R (0)(0) 5 Ÿ Ÿ 9 Ê 5 Ÿ R Ÿ 8

To be safe, the left endpoint was rounded up and the right endpoint was rounded down Section The Precise Definition of a Limit 75 57 (a) $ 0 Ê $ Ê f() Then kf() k k k That is, kf() k no matter how small $ is taken when $ Ê lim f() Á Ä 0 $ Ê $ Ê f() Then kf() k k( ) k kk That is, kf() k no matter how small $ is taken when $ Ê lim f() Á Ä (c) $! Ê $ Ê f() Then kf() 5k k 5k 5 5 05 Also,! $ Ê $ Ê f() Then kf() 5 k k( ) 5k k 05k 05 05 05 Thus, no matter how small $ is taken, there eists a value of such that $ $ but kf() 5 k Ê lim f() Á 5 Ä 58 (a) For $ Ê h() Ê kh() k Thus for %, kh() k % whenever $ no matter how small we choose $ 0 Ê lim h() Á Ä For $ Ê h() Ê kh() k Thus for %, kh() k % whenever $ no matter how small we choose $ 0 Ê lim h() Á Ä (c) For $ Ê h() so kh() k k k No matter how small $ 0 is chosen, is close to when is near and to the left on the real line Ê k kwill be close to Thus if %, kh() k % whenever $ no mater how small we choose $ 0 Ê lim h() Á Ä 59 (a) For $ Ê f() 8 Ê kf() k 08 Thus for % 08, kf() k % whenever $ no matter how small we choose $ 0 Ê lim f() Á Ä$ For $ Ê f() Ê kf() 8k 8 Thus for % 8, kf() 8 k % whenever $ no matter how small we choose $ 0 Ê lim f() Á 8 Ä$ (c) For $ Ê f() 8 Ê kf() k 8 Again, for % 8, kf() k % whenever $$ no matter how small we choose $ 0 Ê lim f() Á Ä$ 60 (a) No matter how small we choose $ 0, for near satisfying $ $, the values of g() are near Ê kg() kis near Then, for % we have kg() k for some satisfying $ $, or! k k $ Ê lim g() Á Ä Yes, lim g() because from the graph we can find a $! such that kg() k % if! k ( ) k $ Ä 6-66 Eample CAS commands (values of del may vary for a specified eps): Maple: f := -> (^-8)/(-);0 := ; plot( f(), =0-0+, color=black, (a) title=section, 6(a) ); L := limit( f(), =0 ); epsilon := 0; (c) plot( [f(),l-epsilon,l+epsilon], =0-000+00, color=black, linestyle=[,,], title=section, 6(c) ); q := fsolve( abs( f()-l ) = epsilon, =0-0+ ); (d) delta := abs(0-q); plot( [f(),l-epsilon,l+epsilon], =0-delta0+delta, color=black, title=section, 6(d) ); for eps in [0, 0005, 000 ] do (e) q := fsolve( abs( f()-l ) = eps, =0-0+ );

76 Chapter Limits and Continuity delta := abs(0-q); head := sprintf(section, 6(e)\n epsilon = %5f, delta = %5f\n, eps, delta ); print(plot( [f(),l-eps,l+eps], =0-delta0+delta, color=black, linestyle=[,,], title=head )); end do: Mathematica (assigned function and values for 0, eps and del may vary): Clear[f, ] y: L eps; y: L eps; 0 ; f[_]: ( (7 )Sqrt[] 5)/( ) Plot[f[], {, 0 0, 0 0}] L: Limit[f[], Ä 0] eps 0; del 0; Plot[{f[], y, y},{, 0 del, 0 del}, PlotRange Ä {L eps, L eps}] ONE-SIDED LIMITS AND LIMITS AT INFINITY (a) True True (c) False (d) True (e) True (f) True (g) False (h) False (i) False (j) False (k) True (l) False (a) True False (c) False (d) True (e) True (f) True (g) True (h) True (i) True (j) False (k) True (a) lim f(), lim f() Ä b Ä c $ No, lim f() does not eist because lim f() lim f() Ä Ä b Á Ä c (c) lim f(), lim f() Ä% c Ä% b $ (d) Yes, lim f() because lim f() lim f() Ä% Ä% c Ä% b (a) lim f(), lim f(), f() Ä b Ä c $ Yes, lim f() because lim f() lim f() Ä Ä b Ä c (c) lim f() ( ), lim f() ( ) Ä c Ä b (d) Yes, lim f() because lim f() lim f() Ä Ä c Ä b 5 (a) No, lim f() does not eist since sin ˆ does not approach any single value as approaches 0 Ä! b lim f() lim 0 0 Ä! c Ä! c (c) lim f() does not eist because lim f() does not eist Ä! Ä! b 6 (a) Yes, lim g() 0 by the sandwich theorem since g() when 0 Ä! b È Ÿ Ÿ È No, lim g() does not eist since È is not defined for 0 Ä! c (c) No, lim g() does not eist since lim g() does not eist Ä! Ä! c

Section One-Sided Limits and Limits at Infinity 77 7 (a Ñ lim f() lim f() Ä c Ä b (c) Yes, lim f() since the right-hand and left-hand Ä limits eist and equal 8 (a) lim f() 0 lim f() Ä b Ä c (c) Yes, lim f() 0 since the right-hand and left-hand Ä limits eist and equal 0 9 (a) domain: 0 Ÿ Ÿ range: 0 y Ÿ and y lim f() eists for c belonging to Ä c (0ß ) (ß ) (c) (d) 0 0 (a) domain: range: Ÿ y Ÿ lim f() eists for c belonging to Ä c ( _ß ) ( ß ) (ß _) (c) none (d) none 05 / lim lim 0 Ä!Þ& c É É É È É É È 05 / Ä b! lim ˆ ˆ 5 ˆ () ˆ Ä b Š ( ) 5 ( ) ( ) lim ˆ ˆ 6 ˆ ˆ ˆ 6 ˆ ˆ ˆ 7 ˆ Ä c 7 7 7 Èh h5è5 Èh h5è5 Èh h5è5 5 lim lim h Ä! b h Š h Ä! b h Š Èh h5è 5 ah h5b5 h(h ) 0 È È È5 È5 È5 lim lim h Ä! b hš Èh h5 5 h Ä! b hš Èh h5 5

78 Chapter Limits and Continuity È6 È5h h 6 È6 È5h h 6 È6 È5h h 6 6 lim lim h Ä! c h h Ä! c Š h Š È6 È 5h h 6 6 a5h h 6b h(5h ) (0 ) È È È6 È6 È6 lim lim h Ä! c h Š È6 5h h 6 h Ä! c h Š È6 5h h 6 kk () 7 (a) lim ( ) lim ( ) for Ä b Ä b ( ) ak k b lim ( ) ( ) Ä b k k ( ) lim ( ) lim ( ) ( ) for Ä c Ä c ( ) ak k b lim ( )( ) ( ) Ä c È ( ) È ( ) 8 (a) lim lim for Ä b k k Ä b () ak k b lim È È Ä b È ( ) È ( ) lim lim ( ) for Ä c k k Ä c () ak k b lim È È Ä c ÚÛ ) ÚÛ ) 9 (a) lim lim ) Ä$ b ) ) Ä$ c ) 0 (a) lim at ÚtÛb 0 lim t t t Ä% b t Ä% c a Ú Ûb È sin ) sin lim lim (where ) ) Ä! È ) Ä! È ) sin kt k sin kt k sin ) sin ) lim lim lim k lim k k (where kt) t Ä! t t Ä! kt ) Ä! ) ) Ä! ) ) sin y sin y sin y sin ) lim lim lim lim (where y) y Ä! y y Ä! y y Ä! y Ä! ) ) ) h h lim lim lim (where h) h Ä! c sin h ˆ h Ä! c sin h h Ä! c ˆ sin h Œ sin ) lim ) h ) Ä! c ) tan ˆ sin sin sin cos 5 lim lim lim lim lim Ä! Ä! Ä! cos Š Ä! cos Š Ä! t t t cos t 6 lim lim lim lim cos t t Ä! tan t t Ä! ˆ sin t t Ä! sin t Š Œ sin t t Ä! lim cos t tä! t csc 7 lim lim lim lim () Ä! cos 5 ˆ Ä! sin cos 5 Š Ä! sin Š Ä! cos 5 ˆ 6 cos 8 lim 6 (cot )(csc ) lim lim ˆ cos Ä! Ä! sin sin Ä! sin sin 9 lim lim ˆ lim ˆ lim Ä! Ä! Ä! Ä! lim Š lim ˆ sin lim Š ()() Ä! Ä! cos sin Ä! cos cos sin cos sin cos sin cos sin cos sin sin sin 0 lim lim 0 () 0 Ä! ˆ Ä! ˆ

sin( cos t) sin ) lim lim since cos t 0 as t 0 t Ä! cos t Ä Ä ) Ä! ) ) sin (sin h) sin ) lim lim since sin h 0 as h 0 h Ä! sin h Ä Ä ) Ä! ) ) sin ) sin ) ) sin ) ) lim lim lim ) Ä! sin ) ˆ ) Ä! sin ) ) ˆ ) Ä! ) sin ) lim lim ˆ lim ˆ Ä! Ä! Ä! sin 5 sin 5 5 5 sin 5 5 5 sin sin 5 5 sin 5 lim tan lim sin lim sin 8 Ä! sin 8 ˆ Ä! cos sin 8 ˆ Ä! cos sin 8 8 lim ˆ ˆ sin ˆ 8 8 Ä! cos sin 8 8 8 sin y cot 5y sin y sin y cos 5y sin y sin y cos 5y 5y 6 lim lim lim y Ä! y cot y y Ä! y cos y sin 5y y Ä! Š y Š cos y Š sin 5y Š 5y lim y Ä! Š sin y Š sin y Š 5y Š cos 5y ˆ y y sin 5y cos y 5 5 5 Section One-Sided Limits and Limits at Infinity 79 Note: In these eercises we use the result lim 0 whenever 0 This result follows immediately from Ä _ mn Î n Eample 6 and the power rule in Theorem 8: lim ˆ lim ˆ lim 0 0 Ä _ Ä _ Š Ä _ mn Î mn Î mn Î mn Î 7 (a) m 8 (a) 9 (a) 0 (a) 8 8 5 5 (a) (a) sin sin Ÿ Ÿ Ê lim 0 by the Sandwich Theorem Ä_ cos ) cos ) ) Ÿ ) Ÿ ) Ê lim ) 0 by the Sandwich Theorem ) Ä_ tsin t ˆ sin t 00 t t 5 lim lim t Ä_ tcos t t Ä_ ˆ cos t 0 6 lim lim lim r Ä_ r 7 5 sin r r r Ä_ 5ˆ r Ä_ 0 0 7 sin r t rsin r ˆ sin r 0 r r sin 7 e Ÿ e Ÿ e Ê lim e sin 0 by the Sandwich Theorem Ä_ 8 lim e cos 0 cos 0 0 0 Ä_ a bˆ ˆ e e e e c c e 0 e e 9 lim lim lim lim Ä_ eec Ä_ e Ä_ e Ä_ e b 0 e e

80 Chapter Limits and Continuity e c 0 sin / 0 e 50 lim lim Ä_ a b sina/b Ä_ 5 (a) lim lim (same process as part (a)) Ä_ 5 7 7 Ä_ 5 5 5 7 7 Š $ $ 5 (a) lim lim Ä_ $ 7 7 Ä_ (same process as part (a)) $ 5 (a) lim lim 0 0 (same process as part (a)) Ä_ Ä_ 7 7 5 (a) lim lim 0 0 (same process as part (a)) Ä_ Ä_ 7 7 $ 55 (a) lim lim 7 (same process as part (a)) Ä_ $ 6 6 ( Ä_ $ 56 (a) lim lim 0 (same process as part (a)) Ä_ $! Ä_ $ 0 0 & % ' 57 (a) lim lim 0 Ä_ ' Ä_ 0 (same process as part (a)) 9 9 $ 9 5 6 $ % % 58 (a) lim lim Ä_ % 5 6 Ä_ 9 (same process as part (a)) $ 5 5 $ 59 (a) lim $ lim Ä_ Ä_ (same process as part (a)) % 60 (a) lim lim Ä_ % 7$ 7 9 7 7 9 Ä_ (same process as part (a)) % È c Š Š È Š 7 Š Î Î Î 6 lim lim 0 6 lim lim Ä_ 7 Ä_ Ä_ È Ä_ $ & È È Š 6 lim Ä_ ÐÎ&Ñ cðî$ñ Î& $ & È È lim ÐÎ&Ñ cðî$ñ lim Ä_ Ä_ Š Î& c c% 6 lim c c$ lim Ä_ Ä 7 &Î$ Î$ *Î& )Î& 65 lim lim )Î& Ä_ È Ä Î& $Î& Î! 7 $ È5 66 lim Î$ lim Ä_ Ä_ 5 5 Î$ Î$

Section One-Sided Limits and Limits at Infinity 8 67 Yes If lim f() L lim f(), then lim f() L If lim f() lim f(), then lim f() does not ei Ä ab Ä ac Á Ä a Ä ab Ä ac st Ä a 68 Since lim f() L if and only if lim f() L and lim f() L, then lim f() can be found by calculating Ä c Ä cb Ä cc Ä c lim f() Ä c b 69 If f is an odd function of, then f( ) f() Given lim f(), then lim f() $ Ä! b Ä! c 70 If f is an even function of, then f( ) f() Given lim f() 7 then lim f() 7 However, nothing Ä c Ä b can be said about lim f() because we don't know lim f() Ä c Ä b f() 7 Yes If lim then the ratio of the polynomials' leading coefficients is, so lim as well Ä_ g() Ä_ g() 7 Yes, it can have a horizontal or oblique asymptote f() f() 7 At most horizontal asymptote: If lim L, then the ratio of the polynomials' leading coefficients is L, so Ä_ g() f() lim L as well Ä_ g() È a ba b È È 7 lim È È lim È È È lim Ä_ Ä_ È È Ä_ É É lim lim Ä_ È È Ä_ 75 For any % 0, take N Then for all N we have that kf() kk kk kk 0 % 76 For any % 0, take N Then for all y N we have that kf() kk kk kk 0 % È 77 I (5ß5 $ ) Ê 5 &$ Also, 5 % Ê 5 % Ê &% Choose $ % Ê lim È 5 0 Ä& b È 78 I (%$ ß% ) Ê %$ Also, % % Ê % % Ê %% Choose $ % Ê lim È % 0 Ä% c 79 As Ä 0 the number is always Thus, ¹ ( ) ¹ Ê kk % % Ê 0 % which is always true independent of the value of Hence we can choose any $ 0 with $! Ê lim Ä! c k k 80 Since Ä we have and k k Then, ¹ ¹ k k % Ê 0 % which is always true so long as Hence we can choose any $!, and thus $ Ê ¹ kk ¹ % Thus, lim kk Ä b 8 (a) lim ÚÛ 00 Just observe that if 00 0, then ÚÛ 00 Thus if we choose, we have for any Ä %!! b $ number %! that 00 00 $ ÊlÚÛ00ll00 00 l!% lim 99 Just observe that if 99 00 then 99 Thus if we choose, we have for any Ä %!! c ÚÛ ÚÛ $ number %! that 00 $ 00 ÊlÚÛ99ll99 99 l!%

8 Chapter Limits and Continuity (c) Since lim ÚÛÁ lim we conclude that lim does not eist Ä %!! b Ä %!! c ÚÛ ÚÛ Ä %!! 8 (a) lim f() lim È È0 0; È 0 % Ê % È % Ê! % for positive Choose $ % Ä! b Ä! b Ê lim f() 0 Ä! b lim f() lim sin 0 by the sandwich theorem since sin for all 0 Ä! c ˆ Ä! c Ÿ ˆ Ÿ Á Since k 0k k 0k whenever k k È, we choose È and obtain sin ˆ % % $ % 0 % if $ 0 (c) The function f has limit 0 at! 0 since both the right-hand and left-hand limits eist and equal 0 cos ) 8 lim sin lim sin, 8 lim lim, Ä _ Ä 0 Ä_ Ä! ) ) ) ˆ ) ˆ ) ) ) c t 85 lim lim lim, t Ä _ 5 Ä _ t Ä 0 5t ˆ 5 Î 86 lim ˆ z lim z, ˆ z Ä_ z Ä! b 87 lim ˆ ˆ cos lim ( )(cos ) ()(), ˆ Ä _ ) ) ) ) Ä 0 88 lim ˆ cos ˆ sin lim a cos b( sin ) (0 )( 0), ˆ Ä_ ) ) ) ) ) Ä! b 5 INFINITE LIMITS AND VERTICAL ASYMPTOTES cos positive positive lim lim Ä! b _ Š positive Ä! c _ Š positive lim lim Ä c _ Š Ä$ b _ Š positive 5 positive 5 lim 6 lim Ä) b 8 _ Š positive Ä& c 0 _ Š positive 7 lim 8 lim Ä( (7) _ Š positive Ä! () _ Š positive positive c 9 (a) lim lim Ä! b Î$ _ Ä! c 0 (a) lim lim Ä! b Î& _ Ä! Î$ Î& lim lim lim lim Ä! Î& Ä! Î& _ Ä! Î$ Ä! Î$ _ a b a b lim tan _ lim sec _ Ä ˆ c Ä ˆ c b 5 lim ( csc )) _ ) Ä! c 6 lim ( cot )) _ and lim ( cot ), so the limit does not eist ) Ä! b ) Ä! c ) _

Section 5 Infinite Limits and Vertical Asymptotes 8 7 (a) lim lim Ä b Ä b ()() _ Š positive positive lim lim Ä c Ä c ()() _ Š positive (c) lim lim Ä b Ä b ()() _ Š positive (d) lim lim Ä c Ä c ()() _ Š positive 8 (a) lim lim Ä b Ä b ()() _ Š positive positive positive lim lim Ä c Ä c ()() _ Š positive (c) lim lim Ä b Ä b ()() _ Š positive (d) lim lim Ä c Ä c ()() _ Š 9 (a) lim 0 lim Ä! b Ä! b _ Š lim 0 lim Ä! c Ä! c _ Š positive Î$ Î$ Î$ (c) lim 0 Ä $ È Î$ (d) lim Ä ˆ positive 0 (a) lim lim Ä b _ Š positive Ä c _ Š ( )( ) 0 (c) lim lim 0 Ä b Ä b (d) lim Ä! c positive ( )( ) (a) lim lim Ä! b $ Ä! b ( ) _ Š positive ( )( ) lim lim lim, Ä b $ Ä b () Ä b Á ( )( ) (c) lim lim lim, Ä c $ Ä c () Ä c Á ( )( ) (d) lim lim lim, Ä $ Ä () Ä Á ( )( ) (e) lim lim Ä! $ Ä! ( ) _ Š positive ( )( ) ( ) ( )( ) ( ) () 8 (a) lim $ lim lim Ä b Ä b Ä b ( )( ) ( ) lim lim lim Ä b $ Ä b ( )( ) Ä b ( ) _ Š positive ( )( ) ( ) (c) lim lim lim Ä 0c $ Ä! c ( )( ) Ä! c ( ) _ Š positive ( )( ) ( ) 0 (d) lim lim lim 0 Ä b $ Ä b ( )( ) Ä b ( ) ()() (e) lim Ä! b ( ) _ Š positive positive and lim Ä! c ( ) _ Š positive so the function has no limit as Ä 0 (a) lim lim _ t Ä! b t Ä! c _ tî$ tî$

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