382_James Stewart Calculus 5 Edition Answers

# 382_James Stewart Calculus 5 Edition Answers - Stewart...

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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 . Therefore, lim ln y x q ( x) (f) lim x p(x) =lim a x p(x) q ( x) =lim y=lim e a x p(x) 1/q ( x ) a x ln y = . a 0 is an indeterminate form of type . a 5. This limit has the form 0 . We can simply factor the numerator to evaluate this limit. 0 2 x1 (x+1)(x 1) lim = lim = lim (x 1)= 2 x+1 x 1 x+1 x 1 x 1 x+2 6. lim x 2 = lim 2 x x +3x+2 x +2 1 = lim =1 2 ( x+1)( x+2) x 2 x+1 9 8 x1 9 9 0 49 9x = lim x = (1)= 7. This limit has the form . lim 5 =lim 4 5x 1 5 5 0 x1 x 1 x 1 5x a x1 8. lim x b 1 =lim x x1 1 ax bx a1 b1 9. This limit has the form = a b 0 . lim 0 x 2 cos x = lim 1 sin x + ( /2) x + ( /2) sin x = lim cos x x tan x= . + ( /2) 2 x+tan x 1+sec x 1+1 10. lim =lim = =2 1 x 0 sin x x 0 cos x t t 0 e1 e =lim = 11. This limit has the form . lim 3 0 t0 t 0 3t 2 t 12. lim t 0 e 3t t 1 t since e 1 and 3t 2 + 0 as t 0. 3t 3e =lim =3 t01 2 2 tan px psec px p(1) p 0 13. This limit has the form . lim =lim = = 2 2 q 0 x 0 tan qx x 0 qsec qx q(1) 2 ...
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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 Berkeley.

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