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

Alternative approach using Taylor series:
Using $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + o(x^2)$ we get $\ln(\frac{\sin x}{x}) \sim_0 \ln(1-\frac{x^2}{6})$. This expression has the Taylor expansion $-\frac{x^2}{6}+o(x^2)$.
We can approximate the denominator by $1-\cos x=\frac{x^2}{2} + o(x^2)$.
Then we have:
$\frac{\ln(\frac{\sin x}{x})}{1-\cos x} \sim_0 -\frac{x^2}{6} \frac{2}{x^2} = -\frac{1}{3}$.
This doesn't exactly answer your question, but hopefully it provides a little bit of insight. :)
