389_James Stewart Calculus 5 Edition Answers

# 389_James Stewart Calculus 5 Edition Answers - Stewart...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online