$\displaystyle \lim_{x\rightarrow1}{\frac{sin3x+sinx-2sin2x}{xsin^2x}}{$

$\displaystyle \lim_{x\rightarrow0}{\frac {sin^{-1}(x)}{x}} $

How Can I solve this?I Tried a lot,and i can't get the answer.

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- Dec 31st 2012, 07:13 AMsrirahulanLiMiTs
$\displaystyle \lim_{x\rightarrow1}{\frac{sin3x+sinx-2sin2x}{xsin^2x}}{$

$\displaystyle \lim_{x\rightarrow0}{\frac {sin^{-1}(x)}{x}} $

How Can I solve this?I Tried a lot,and i can't get the answer. - Dec 31st 2012, 08:03 AMSironRe: LiMiTs
I should split the first limit into three limits. Then use l'Hopital's rule where necessary and for the others try to use some goniometric idenitities to simplify.

For the second one, use l'Hopital's rule. - Dec 31st 2012, 02:45 PMsrirahulanRe: LiMiTs
What is l'Hopital's rule ?

Happy New Year.. - Dec 31st 2012, 03:01 PMPlatoRe: LiMiTs
- Jan 1st 2013, 12:11 AMibduttRe: LiMiTs
For the second limit just try substitution x = sin θ Observe that as x approaches 0, θ also approaches 0