Evaluating $\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}$ How do I evaluate
$$\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{\dfrac{1}{1-\cos x}}\ ?$$
I tried using the fact that $\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}} = \exp\left(\ln\bigg(\frac{\sin x}{x}\right)\frac{1}{1-\cos x}\bigg)$ and then I am already stuck.

Let us start with $$A=\left(\frac{\sin (x)}{x}\right)^{\frac{1}{1-\cos (x)}}$$ Take logarithms; so $$\log(A)=\frac{1}{1-\cos (x)}\log\left(\frac{\sin (x)}{x}\right)$$ Now, use Taylor series $$\sin(x)\approx x-\frac{x^3}{6}$$ $$\frac{\sin (x)}{x}\approx 1-\frac{x^2}{6}$$ Now consider that,for small $y$, $\log(1+y)\approx y-\frac{y^2}{2}$ So $$\log\left(\frac{\sin (x)}{x}\right)\approx -\frac{x^2}{6}-\frac{x^4}{180}$$ Also $$\cos(x)\approx 1-\frac{x^2}{2}+\frac{x^4}{24}$$ Using all of the above, we then have $$\log(A)\approx \frac{-\frac{x^2}{6}-\frac{x^4}{180}}{\frac{x^2}{2}-\frac{x^4}{24}}= \frac{-\frac{1}{6}-\frac{x^2}{180}}{\frac{1}{2}-\frac{x^2}{24}} \approx -\frac{1}{3}-\frac{7 x^2}{180}$$ which shows the limit of $\log(A)$ and how it is approached.
I am sure that you can take from here.
