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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 by 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 by using expansion and L hospital`s rule i get the answer 1/2.What is wrong in this method ?

Grade:12

1 Answers

Vikas TU
14149 Points
7 years ago
Here if U are using log expansion then write for cos function too.
lim x-> 0 becomes:
 
=> (x(1-x^2/2) – (x – x^2/2))/x^2
=> (1-x^2/2) – 1 + x/2)/x
=> (-x^2/2) + x/2)/x
=> (-x/2 + 0.5)/1
=> 0.5 would be the final ansswer.
 

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