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384_James Stewart Calculus 5 Edition Answers

# 384_James Stewart Calculus 5 Edition Answers - Stewart...

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0 0 lim x ± 0 sin ² 1 x x =lim x ± 0 1/ 1 ² x 2 1 =lim x ± 0 1 1 ² x 2 = 1 1 =1 lim x ± 0 sin x ² x x 3 =lim x ± 0 cos x ² 1 3 x 2 =lim x ± 0 ² sin x 6 x =lim x ± 0 ² cos x 6 = ² 1 6 0 0 lim x ± 0 1 ² cos x x 2 =lim x ± 0 sin x 2 x =lim x ± 0 cos x 2 = 1 2 lim x ± ³ (ln x ) 2 x =lim x ± ³ 2(ln x )(1/ x ) 1 =2lim x ± ³ ln x x =2lim x ± ³ 1/ x 1 =2(0)=0 lim x ± 0 x +sin x x +cos x = 0+0 0+1 = 0 1 =0 lim x ± 0 cos mx ² cos nx x 2 =lim x ± 0 ² m sin mx + n sin nx 2 x =lim x ± 0 ² m 2 cos mx + n 2 cos nx 2 = 1 2 n 2 ² m 2 ( ) ³ ³ lim x ± ³ x ln 1+2 e x ( ) =lim x ± ³ 1 1 1+2 e x ´ 2 e x =lim x ± ³ 1+2 e x 2 e x =lim x ± ³ 2 e x 2 e x =1 lim x ± 0 x tan ² 1 (4 x ) =lim x ± 0 1 1 1+(4 x ) 2 ´ 4 =lim x ± 0 1+16 x 2 4 = 1 4 0 0 lim x ± 1 1 ² x +ln x 1+cos ± x =lim x ± 1 ² 1+1/ x ² ± sin ± x =lim x ± 1 ² 1/ x 2 ² ± 2 cos ± x = ² 1 ² ± 2 ² 1 ( ) = ² 1 ± 2 lim x ± ³ x 2 +2 2 x 2 +1 =lim x ± ³ x 2 +2 2 x 2 +1 = lim x ± ³ x 2 +2 2 x 2 +1 = lim x ± ³ 1+2/ x 2 2+1/ x 2 = 1 2 25. This limit has the form . 26. 27. This limit has the form
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