389_James Stewart Calculus 5 Edition Answers

389_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 x x x+1 lim x = lim e x =e 1 x x x x+1 Or: lim ln y x 5/x 60. y=(cos 3x) ln y= 5/x x 1 x+1 x = lim = lim x 5 ln (cos 3x) x lim ln y=5lim x 0 x 0 1 x 1 x 1+ =e 1 ln (cos 3x) 3tan 3x =5lim =0 , x 1 x0 0 so lim (cos 3x) =e =1 . x 0 2 1/x 61. y=(cos x) ln y= 1 2 ln cos x x 2 1/x lim (cos x) x =lim e + 0 62. y= x 2x 3 2 x +5 ln y =e ln cos x lim ln y=lim x + 0 x 2 + 2 =lim x 0 x + 0 tan x sec x 1 =lim = 2x 2 2 + x 0 1/2 =1/ e + 0 2x+1 2x 3 2x+5 ln y=(2x+1) ln 2 8(2x+1) lim ln y = lim ln (2x 3) ln (2x+5) = lim 2/(2x 3) 2/(2x+5) = lim x 2 1/(2x+1) (2x 3)(2x+5) x x x 2/(2x+1) 2 8(2+1/x) = lim =8 (2 3/x)(2+5/x) x lim x 2x 3 2x+5 2x+1 =e 8 63. From the graph, it appears that lim x ln (x+5) ln x =5 . x To prove this, we first note that x+5 5 ln (x+5) ln x=ln =ln 1+ x x ln 1=0 as x . Thus, 9 ...
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