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				   find:
lim    (xcosx-log(1+x))/x2
x→0



6 years ago

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										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


6 years ago

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