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chapter 2 15 - SECTION 2.3 CALCULATING LIMITS USING THE...

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Unformatted text preview: SECTION 2.3 CALCULATING LIMITS USING THE LLMlT LAWS D 32. (a) 0.5 (b) -- —0.001 0.2886992 —0.0001 0.2886775 20.00001 0.2886754 Al 1 —0.000001 0.2886752 0 0.000001 0.2886751 11m v3 + 3’ — *5 m 0.29 0.00001 0.2886749 .E—>0 ”3 0.0001 0.2886727 0.001 0.2886511 The limit appears to be approximately 0.2887. (C) lim< 3+$_\/§- 3+x~—\/§) 2 lim M 2 lim ; z—>0 w ./3+:c——\/§ $20$(\/3+w+\/§) w—>0\/3+x+\/§ lirral 2 35% Limit Laws 5 and 1 lini)./3+;z:+hmO\/§ [ I — 1 [7 and 11] Nina (3 + as) + \/§ 1 [l 7 d 8] : fi . , an «3 + 0 + x/fi _ 1 2 x/i’: 33. Let f(m) 2 2:112. g(m) 2 332 cos 207m: and h(m) 2 x2. Then ~13 eos207r$ S 1 2 —ac2 S (:2 cos2077m S m2 2 fix) 3 g(a:) S h(:c). So since Tin}J f($) 2 111% Mm) 2 0. by the Squeeze Theorem we have lin}J g(;v) 2 0. 34. Let flat) 2 —\/m. g(a:) 2 W sin(7r/;I:), and h(a:) : M. Then —1 s sin(7r/2:) g 1 :> 2W 3 WsinM/x) S 3:3 + m2 2> f(:c) 3 9(27) 3 h.(a:). So since 311:1?) f(m) 2 £121)ng) 2 0. by the Squeeze Theorem we have linq) 9(17) 2 O. 35.1 5 fix) 3 2:2 +2$+2for all :0. Now limll 2 1 and z—p2 113131 (2:2 + 22: + 2) 2 lim1 m2 + 2 lim1 :0 + lim 2 2 (~1)2 + 2(—1)+ 2 2 1. Therefore. by the Squeeze I—>2 122 m—2—1 Theorem. lim f(:c)2 . 79 ...
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