How to find the limit of $\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$ How to find the limit of the following function?
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$$
What I tried is as follows.
$$\tan^{-1} x = y \implies \tan y = x $$
$$\frac{x \tan^{-1} x}{1-\cos x}=\frac{\tan y \cdot y}{1-\cos(\tan y) }$$
But it didn't work.
Please Help me.

$\tan y \approx y + \frac{y^3}{3}$ from the Taylor series. We can discard the $y^3/3$ term to get $\arctan x \approx x$, as we only need the $x^2$ term and below.
Then:
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x} = \lim_{x \to 0} \frac{x^2}{1-\cos x} \frac{1 + \cos x}{1 + \cos x} = 2.$$
