<s>[INST] Solve $\lim \limits_{x \to 0} (\frac{\sin x}{x})^\frac{1}{1-\cos x}$ without L'Hospital's Rule or Taylor expansion. I solved it with L'Hospital's Rule, but I had to use it 4 times! Is there any way to avoid it?
 [/INST] Can we use some basics: $\lim \limits_{x \to 0} (1+x)^\frac{1}{x} = e$ and $\lim \limits_{x \to 0} \frac{\sin x}{x} = 1$?
So, we separate $1$: $$\large\lim \limits_{x \to 0} \left(\frac{\sin x}{x}\right)^\frac{1}{1-\cos x} = \lim \limits_{x \to 0} \left(1+\frac{\sin x}{x}-1\right)^\frac{1}{1-\cos x}
= \lim \limits_{x \to 0} \left(\left(1+\frac{\sin x}{x}-1\right)^\frac{1}{\frac{\sin x}{x}-1}\right)^\frac{\frac{\sin x}{x}-1}{1-\cos x}=e^{\lim \limits_{x \to 0}\frac{\frac{\sin x}{x}-1}{1-\cos x}},  $$
$$\lim \limits_{x \to 0}\frac{\frac{\sin x}{x}-1}{1-\cos x} = 
\lim \limits_{x \to 0}\frac{\frac{\sin x}{x}-1}{2\left(\frac{x}{2}\right)^2}\cdot \frac{2\left(\frac{x}{2}\right)^2}{2\sin^2\left(\frac{x}{2}\right)} = 2\lim \limits_{x \to 0}\frac{\sin x-x}{x^3},$$
$\lim \limits_{x \to 0}\frac{\sin x-x}{x^3}$ is here.
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