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Lim x->0 (xcosx-log(1+x))/x^2.Limx->0 (xcosx - x )/x^2 (by elimination )Lim x->0 (cosx - x)/xWhich is finally equal to zero but while using expansion and L hospital`s rule i get the answer 1/2.What is wrong in this method ?

Lim x->0 (xcosx-log(1+x))/x^2.Limx->0 (xcosx - x )/x^2 (by elimination )Lim x->0 (cosx - x)/xWhich is finally equal to zero but while using expansion and L hospital`s rule i get the answer 1/2.What is wrong in this method ?

Grade:12

1 Answers

Sourabh Singh IIT Patna
askIITians Faculty 2104 Points
5 years ago
Lim x->0 (xcosx-log(1+x))/x^2

from this to the below written statement

Limx->0 (xcosx - x )/x^2 How you get this ??

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