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

You can continue it as the following :
$$\begin{align}\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin^2 x}&=\lim_{x\to 0}\frac{(x\cos x-\sin x)'}{(x\sin^2 x)'}\\&=\lim_{x\to 0}\frac{\color{red}{\cos x}-x\sin x\color{red}{-\cos x}}{\sin^2 x+2x\sin x\cos x}\\&=\lim_{x\to 0}\frac{-x}{\sin x+2x\cos x}\\&=\lim_{x\to 0}\frac{-1}{(\sin x/x)+2\cos x}\\&=\frac{-1}{1+2\cdot 1}.\end{align}$$
