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lim xcosx-log(1+x)/x2
x_0

ashu , 12 Years ago
Grade 11
anser 1 Answers
Rinkoo Gupta
lim (xcosx-log(1+x))/x2
on putting x=0 we get 0/0 which is undefined
so by using L'Hospital rule
lim {-xsinx+cosx-1/(1+x)}/2x again 0/0
again using L'hospital rule
lim {-xcosx-sinx-sinx+1/(1+x)2}/2
=1/2
Thanks & Regards
Rinkoo Gupta
AskIITians faculty

Last Activity: 12 Years ago
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