chapter 2 16 - 80 E CHAPTERZ LlMITS AND DERIVATIVES 36...

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Unformatted text preview: 80 E CHAPTERZ LlMITS AND DERIVATIVES 36. 3m§f(m)£$3+2for0§:c§2. Now lim3m=3and lim ($34.2) = limx3+lim2213+2:3. :E—rl z—>1 x—>1 x—rl Therefore. by the Squeeze Theorem. lim flag) 2 3. z:—>1 37. #1 g cos(2/:c) S 1 => —m4 S m4 cos(2/:z:) 3 x4. Since lim (-3104) 2 0 and lim x4 = 0, we have :c—>0 1—»0 lim [:34 cos(2/a:)] = 0 by the Squeeze Theorem. (It—’0 38. —1 S sin(7r/;v) g 1 :> e”1 g e‘mM/Il g 61 => fi/e S fiesmh/x) S fie. Since hm (fi/e) : 0 and lim (fie) : 0. we have lim [fl 65“](T/m)] : O by the Squeeze Theorem. IHO+ $404” mao‘l’ 39.Ifm>74.then|w+4|:x+4,so lim |$+4|: lim (m+4):i4+4:0. $474+ za—4 Ifzc<—4.thenla:+4|:—(m+4).so lim lav—Fill: lim 7(w+4):—(74+4):0. m—>~4— zH—4’ Since the right and left limits are equal. lim4 lm + 4| 2 0. . +4| . ,($+4) . 40.1fm<—4.th +4:— 4.. 1 la” :1 ——= _ =7 enlw l (”7+ “0.3317 $+4 .7137 m+4 .33—( 1) 1 . la: 1 2| . m — 2 , 41. Ifa: > 2. then |m - 2] = :c — 2. so 11m : 11m : 11m 1 = 1. Ifm < 2. then 1—>2+ $ — 2 x—>2+ I} — 2 m—>2+ . —2 . — ~2 . , , . la: — 2| : — (ac 7 2). so 11m I20 I 2 11111 i; : 11m —1 : —1. The r1ght and left limits are (E—>2‘ m 7 2 z—>2‘ IE — 2 1:4»2’ . _ luv # 2| . d1fferent. so 11m does not CXISt. $—>2 J: i 2 3 42. Ifrc > 2 then |2cc —3| : 2w — 3. so 2% 7 3:1: 21:2 — 3a: m (2116 — 3) 3 1‘ ___—: 1' ’: l‘ —: l' =1.5.If' <—.th $7111} l2$ * 3| $713+ 2m # 3 kiln“ 2m # 3 3—31r15+ :2: ‘r 2 en 21:2 ~ 356 2m2 — 3m . 20(2317 # 3) 27 : —2. l' ———-= 1' ————:l ———-= l' —:—1.5. l m 3' 3 w .0 .Jm7 l2. 7 31 .3117 7 (2. 7 3) .3“ 7 (2. 7 3) 1111.735 . . . . . . 2:02 —32: . The rlght and lett lim1ts are d1fferent, so hm ———- does not ex1st. m—v1.5 |2$ —‘ 3| . , 1 1 , 1 1 . 2 . . 43. Smce |ccl : —:c fora: < 0. we have 11m — — — : 11m — — — : 11m —.wh1ch does not ex1st x—)0- CE lm| 240* at —$ 25—.0- :1: since the denominator approaches 0 and the numerator does not. 1 1 1 1 . 44. Since lml = asform > 0. we have lim <— — ——> 2 lim <— — —> : 11m 0 : 0. I_,0+ :3 Ian] 3070+ :1: m z—>0+ 45. (a) (b) (i) Since sgnw : 1 for :1: > 0. lim sgna: : lim 1 : 1. mHO+ $—>0+ (ii) Since sgna: : —1 fora: < 0. lim sgn :c : lim —1 : —1. z—>0— arr->0- (iii) Since lim sgn :1: ¢ lim sgn as. lim sgnm does not exist. m—>O’ ma0+ z—>O (iv) Since lsgnml : 1 form 75 0.11rq)|sgn:z:l= lim 1 : 1. z—>0 ...
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