Evaluating $\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$ I was trying to evaluate
$$\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$$
I have tried taking natural logarithm first:
$\lim_{x\to0}\frac{\ln(\sin x)-\ln x}{1-\cos x}=\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}\quad\quad\text{(L'Hopital Rule)}\\
=\lim_{x\to0}\frac{\cos x}{x\sin x}-\frac{1}{x^2}\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)\\
=\lim_{x\to0}\frac{\cos x-1}{x^2}\quad\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)$
and after this I eventually have the limit equaling $-\frac{1}2$, which means that the original limit should be $\frac{1}{\sqrt{e}}$.
However, I graphed $f(x)=\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$ on Desmos, and it turned out that the limit is approximately $0.7165313$, or $\frac{1}{\sqrt[3]{e}}$.
Therefore I think there's something wrong in my approach, but I couldn't find it. Any suggestions?

$$\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}=\lim_{x\to0}\frac{(x-\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots)-(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots)}{x^3}=\frac{1}{6}-\frac{1}{2}=-\frac{1}{3}$$
