Since the limit goes to: . , we can apply L'Hopital.
We have: .
Apply L'Hopital: .
EDIT: Was just a typo in. missed that a sin was missing just noticed the (sin/x)(1/x) bit.
Therefore: .
i dont agree with the last line, lim(ab)=lim (a)lim(b) if both limits exist.
applying l'hopitals once again gives
lim 2sinxcosx/1 as x goes to 0
which is 0
Last edited by jiboom; Feb 28th 2010 at 01:48 PM.
Reason: EDIT: Was just a typo in soroban's post. missed that a sin was missing just noticed the (sin/x)(1/x) bit.