What is the limit of $ \lim_{x\to 0}( \sin(x)/x )^{1/(1-\cos x)} $ What is the limit of 
$$ \lim\limits_{x\to 0}\left( \frac{\sin x}{x} \right)^{\frac{1}{1-\cos x}} $$
By applying L'Hôpital's rule many times, I find the limit is $ e^{-\frac{1}{3}} $. Are there other methods more straightforward than mine?
 A: Let $$f(x) = \left(\frac{\sin(x)}{x}\right)^{1/(1-\cos(x))}$$
For $|x| < \pi$, $\sin(x)/x > 0$ and
$$\log(f(x)) = \frac{\log(\sin(x)/x)}{1 - \cos(x)}$$
Now use Maclaurin series: $$\frac{\sin(x)}{x} = 1 - \frac{x^2}{6} + \ldots $$
so $$ \log \left(\frac{\sin(x)}{x}\right) = - \frac{x^2}{6} + \ldots $$
while
$$1 - \cos(x) = \frac{x^2}{2} + \ldots $$
so
$$ \lim_{x \to 0} \log(f(x)) = -\frac{1}{3}$$
and thus
$$ \lim_{x \to 0} f(x) = \exp(-1/3)$$
A: We have
$$\left( \frac{\sin x}{x} \right)^{\frac{1}{1-\cos x}}
=\left[\left( 1-\left(1-\frac{\sin x}{x}\right) \right)^{\frac1{\left(1-\frac{\sin x}{x}\right)}}\right]^{\frac{1-\frac{\sin x}{x}}{1-\cos x}} \to (e^{-1})^\frac13=e^{-\frac13}$$
indeed
$$\frac{1-\frac{\sin x}{x}}{1-\cos x}=\frac{\frac16 x^2+o(x^2)}{\frac12x^2+o(x^2)}\to \frac13$$
A: The easiest way is to use Taylor expansion:
\begin{align}\left( \frac{\sin x}{x} \right)^{\frac{1}{1-\cos x}}&=\exp\left(\frac{1}{1-\cos x}\ln\left(\frac{\sin x}{x}\right)\right)\\&=\exp\left(\frac{1}{\frac{x^2}{2}+o(x^2)}\ln\left(1-\frac{x^2}{6}+o(x^2)\right)\right)\\&\sim \exp\left(\frac{-x^2/6}{x^2/2}\right)=e^{-1/3}\end{align}
