<s>[INST] 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?
 [/INST] Let me try an approach without Taylor series, starting with your step
$$\lim_{x \to 0} \frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{\sin x}=\lim_{x \to 0}\frac{x\cos x - \sin x}{x \sin^2 x}$$
We continue from here:
$$
\begin{align}
\lim_{x \to 0}\frac{x\cos x - \sin x}{x \sin^2 x} &\overset{\mathrm{H}}{=} \lim_{x \to 0} \  \frac{-x \sin x}{\sin^2x + 2x\sin x \cos x} \\
&= \lim_{x \to 0} \ \frac{-x}{\sin x + 2x \cos x}
\overset{\mathrm{H}}{=} \lim_{x \to 0} \ \frac{-1}{3\cos x-2\sin x}=\frac{-1}{3}
\end{align}
$$
and the result follows from there.
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