Solve $\lim \limits_{x \to 0} (\frac{\sin x}{x})^\frac{1}{1-\cos x}$ without L'Hospital's Rule or Taylor expansion.

Question

I solved it with L'Hospital's Rule, but I had to use it 4 times! Is there any way to avoid it?

Answer 1

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.