Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?

You may use the Taylor expansion, as $u \to 0$,
$$
\frac1{1-u}=1+u+O(u^2)
$$ and the Taylor expansion, as $x \to 0$,
$$
\arctan x=x-\frac{x^3}3+O(x^5),
$$ giving
$$
\begin{align}
\frac1{x\arctan x}&=\frac1{x(x-\frac{x^3}3+O(x^5))}\\\\
&=\frac1{x^2}\frac1{(1-\frac{x^2}3+O(x^4))}\\\\
&=\frac1{x^2}(1+\frac{x^2}3+O(x^4))\\\\
&=\frac1{x^2}+\frac13+O(x^2)
\end{align}
$$and

$$
\lim_{x \to 0}\left(\frac1{x\arctan x}- \frac1{x^2}\right)=\frac13.
$$

