Evaluating a limit using L'Hôpital's rule I know that it can be also evaluate using Taylor expansion, but I am intentionally want to solve it using L'Hôpital's rule:  
$$ \lim\limits_{x\to 0} \frac{\sin x}{x}^{\frac{1}{1-\cos x}} = 
\lim\limits_{x\to 0}\exp\left( \frac{\ln(\frac{\sin x}{x})}{1-\cos x} \right)$$
Now, from continuity and L'hopital Rule: 
$$\lim\limits_{x\to 0} \frac{\ln(\frac{\sin x}{x})}{1-\cos x} = 
\lim\limits_{x\to 0} \frac{\frac{x}{\sin x}\cdot\frac{x\cos x - \sin x}{x^2}}{\sin x} = 
\lim\limits_{x\to 0}\frac{\frac{x\cos x - \sin x}{x\sin x}}{\sin x}$$
This is where I got stuck.
If I'm not mistaken the limit is $-\frac{1}{3}$ so the orginial one is $e^{-\frac{1}{3}}$
What should I do different (Or what's is wrong with my calculation?)
Thanks

After rearranging the quotient a bit we can apply L'hopitals rules $2$ more times to get the answer:
$$\lim_{x\to0}\frac{x\cos(x)-\sin(x)}{x\sin^{2}(x)}\underbrace{=}_{\text{l'hopital}}\lim_{x\to0}\frac{-x\sin(x)}{\sin^{2}(x)+2x\sin(x)\cos(x)}=\lim_{x\to0}\frac{-x}{\sin(x)+2x\cos(x)}$$
$$\underbrace{=}_{\text{l'hopital}}\lim_{x\to0}\frac{-1}{3\cos(x)-2x\sin(x)}=\frac{-1}{3}$$
