lab7_solns_f11

# lab7_solns_f11 - Math 116 Lab 7 Solutions Fall 2011 1 a As...

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Math 116 - Lab 7 - Solutions - Fall 2011. 1. a) As x 1, ln( x 2 ) 0 and ( x - 1) 0. The limit has the indeterminate form 0 0 , so l’Hˆ opital’s rule is applicable. By l’Hˆ opital’s rule, lim x 1 ln( x 2 ) x - 1 = lim x 1 2 x 1 = lim x 1 2 x = 2 . 1. b) lim x 6 5 x +9 8 x +1 = 5 · 6+9 8 · 6+1 = 39 49 . Notice that l’Hˆ opital’s rule is not applicable. 1. c) As t 0+, ( 1 + 3 5 t ) → ∞ and 8 t 0, so it has the indeterminate form 0 . With f ( t ) = ( 1 + 3 5 t ) 8 t , lim t 0+ ln f ( t ) = lim t 0+ 8 t ln ± 1 + 3 5 t ² (0 ·∞ ) = lim t 0+ ln ( 1 + 3 5 t ) 1 8 t ( ) = lim t 0+ - 3 5 t 2 1+ 3 5 t - 1 8 t 2 (l’Hˆ opital) = 0 . Hence, lim t 0+ ( 1 + 3 5 t ) 8 t = lim t 0+ e ln f ( t ) = e lim t 0+ ln f ( t ) = e 0 = 1 . 1. d) As φ 0+, cos(3 φ ) 1 and cot(2 φ ) → ∞ , so it has the indeterminate form 1 . With

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lab7_solns_f11 - Math 116 Lab 7 Solutions Fall 2011 1 a As...

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