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find: lim (xcosx-log(1+x))/x 2 x→0

find:


lim    (xcosx-log(1+x))/x2


x→0


 


 


 

Grade:12

1 Answers

Vanya Saxena
18 Points
11 years ago

On putting the limits we find the function to be in an indeterminate form.

Thus on applying L hospital's rule we get

x->0lim [cosx-xsinx-(1/1+x)]/2x

Then on differentiating again we get

limx->0[-sinx-sinx-xcosx+(1/1+x)2]/2

on putting the limits we get the answer as  1/2

 

 

 

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