384_James Stewart Calculus 5 Edition Answers

384_James Stewart - Stewart Calculus ET 5e 0534393217;4 Applications of Differentiation 4.4 Indeterminate Forms and L’Hospital’s Rule 2 1 1 1 x

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Unformatted text preview: Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule 2 1 1/ 1 x 0 sin x 25. This limit has the form . lim =lim =lim 0 x0 x 1 x0 x0 26. lim x 0 sin x x x =lim 3 x cos x 1 2 0 =lim x 3x 27. This limit has the form 0 1 = 2 1x 1 =1 1 sin x cos x 1 =lim = 6x 6 6 x0 0 sin x cos x 1 1 cos x . lim =lim =lim = 2 0 x0 2 2 x 0 2x x0 x 2 (ln x) 2(ln x)(1/x) ln x 1/x 28. lim = lim =2 lim =2 lim =2(0)=0 x 1 x 1 x x x x 29. lim x 0 30. lim x x+sin x 0+0 0 = = =0 . L’Hospital’s Rule does not apply. x+cos x 0+1 1 2 cos mx cos nx =lim 2 0 x x 0 2 msin mx+nsin nx m cos mx+n cos nx 1 2 2 =lim = nm 2x 2 2 x0 x . lim 31. This limit has the form x ( ln 1+2e x ) 1 = lim 1 x 1+2e 32. lim x 0 x 1 tan (4x) =lim x 0 x 2e x ( 1+2e = lim x 2e x x = lim x ) 2e 2e x x =1 2 1 1 2 4 1+16x 1 =lim = 4 4 x0 1+(4x) 1 x+ln x 0 =lim 33. This limit has the form . lim 0 x 1 1+cos x x 1 2 34. lim x x +2 2 2x +1 2 = lim x x +2 2 2x +1 1+1/x =lim sin x x 1 2 = lim x x +2 2 2x +1 2 1/x 2 cos x 2 = lim x = 1+2/x 2 2+1/x = 1 2 ( 1) = 1 2 1 2 4 ...
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This note was uploaded on 08/02/2011 for the course MATH 1B taught by Professor Reshetiken during the Spring '08 term at University of California, Berkeley.

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