<s>[INST] 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.
 [/INST] Without using L'Hopital's rule or Taylor series:
$\begin{align} \lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}&=\lim_{x \to 0}\frac{ \frac{\tan^{-1} x}{x}}{\frac{1-\cos x}{x^2}} \; \text{(Divided Numerator and Denominator by $x^2$)}\\ 
&=\frac{\lim\limits_{x \to 0} \frac{\tan^{-1} x}{x}}{\lim\limits_{x \to 0}\frac{1-\cos x}{x^2}}\\
&=\frac{1}{1 \over 2}=2 \end{align}$
$\begin{align}\because \lim_{x \to 0} \frac{\tan^{-1} x}{x}&=\lim_{y \to 0} \frac{y}{\tan y} \;\;(\text{substituted}\; y=\tan^{-1} x, \;y\to 0\text{ as}\; x \to 0)\\
&=\frac1{\lim\limits_{y \to 0} \frac{\tan y}{y}}\\
&= \frac{1}{1}=1 \;\quad \left(\because \lim\limits_{y \to 0} \frac{\tan y}{y} =1\right)\end{align} $
And,
$\begin{align}\lim\limits_{x \to 0}\frac{1-\cos x}{x^2}&=\lim\limits_{x \to 0}\frac {2\sin^2\left(\frac{x}{2}\right)}{4\left(\frac{x}{2}\right)^2}\\&={1\over 2}\left(\lim\limits_{\frac{x}{2}\to 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2\\&={1\over 2}(1)^2 \quad \; \left(\because \lim\limits_{y\to 0}\frac{\sin y}{y}=1\right)\\&={1\over 2}\end{align}$
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