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.

Hint. From $\cos (A+B)=\cos A\cos B-\sin A\sin B$ with $B=A$ we have $\cos 2A=\cos^2A-\sin^2A=(1-\sin^2A)-\sin^2 A=1-2\sin^2 A.$ Hence $2\sin^2 A=1-\cos 2A.$  With $A=2x$ therefore $$2\sin^2(x/2)=1-\cos x.$$ So if $\cos x\ne 1$ then $$\frac {x\tan^{-1}x}{1-\cos x}=\frac {x\tan^{-1}x}{2\sin^2(x/2)}=$$ $$=\left(\frac {(x/2)}{\sin (x/2)}\right)^2\cdot 2\cdot \frac {\tan^{-1}x}{x}.$$ Remark: The "half-angle formulae" are usually presented as $$\frac {1-\cos x}{2}=\sin^2(x/2)$$ and $$\frac {1+\cos x}{2}=\cos^2(x/2).$$
