385_James Stewart Calculus 5 Edition Answers

# 385_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 a1 a ax a a(a 1)x 0 x ax+a 1 =lim =lim 35. This limit has the form . lim 2 2 0 x1 x 1 2( x 1) x1 ( x 1) a2 = a(a 1) 2 2x 1e 11 = =0 . L’Hospital’s Rule does not apply. 36. lim 1 x 0 sec x 37. This limit has the form 0 ( ) . We need to write this product as a quotient, but keep in mind 1 that we will have to differentiate both the numerator and the denominator. If we differentiate , ln x we get a complicated expression that results in a more difficult limit. Instead we write the quotient as ln x . 1/2 x lim x ln x x ln x=lim + 0 + x x + 0 2 x 2x 38. lim x e = lim x =lim 1/2 x 0 x x e lim cot 2xsin 6x=lim 0 x 0 2x = lim x e 39. This limit has the form x 1/ x 1 3/2 x 2 x 2x 2x 3/2 3/2 x e 2 x ) =0 + x 2 = lim ( =lim 0 x x = lim 2e =0 x 0 . We’ll change it to the form 0 . 0 sin 6x 6cos 6x 6(1) =lim = =3 2 2 tan 2x x 0 2sec 2x 2(1) 40. lim sin xln x =lim + x 0 x = + 0 ln x 1/ x =lim = lim csc x + csc xcot x + x lim x 0 x sin x x + 0 lim tan x = 1 0=0 0 41. This limit has the form sin x tan x x x + 0 3 2 x 0 . lim x e = lim x x x 3 2 e x 2 = lim x 3x 2 2xe x = lim x 3x 2 2e x = lim x 3 2 4xe =0 x 42. 5 ...
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