<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] Surprised no one explicitly used L'Hopital's rule.
$\displaystyle \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^2}.$
Derivative of the numerator is therefore
$\displaystyle \tan^{-1}(x) + \frac{x}{1 + x^2}.$
Derivative of the denominator is $\sin(x)$
Since both derivatives evaluate as $(0)$ at $(x = 0)$, L'Hopital's rule must be re-applied.
Next derivative of numerator is
$\displaystyle \frac{1}{1 + x^2} + \frac{1 + x^2 - 2x^2}{\left(1 + x^2\right)^2}.$
At $(x=0)$, this evaluates to $(2)$.
Next derivative of denominator is $\cos(x)$.
At $(x = 0)$, this evaluates to $(1)$.
Therefore, the limit is $\displaystyle \frac{2}{1} = 2.$
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