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?

\begin{align}
& \lim_{x\to0} \left( \frac 1 {x \arctan x} - \frac 1 {x^2} \right) = \lim_{x\to0} \frac {x - \arctan x} {x^2 \arctan x} \\[10pt]
= {} & \lim_{x\to0} \frac{1 - \frac 1 {1+x^2}}{\frac{x^2}{1+x^2} + 2x \arctan x  } \qquad \text{(L'Hopital)} \\[10pt]
= {} & \lim_{x\to0} \frac{x^2}{x^2 + 2x(1+x^2) \arctan x} = \lim_{x\to0} \frac{x}{x + 2(1+x^2)\arctan x} \\[10pt]
= {} & \lim_{x\to0} \frac 1 {1 + 4x\arctan x + 2} \qquad \text{(L'Hopital)} \\[10pt]
= {} & \lim_{x\to 0 } \frac 1 {3+4x\arctan x} = \frac 1 3.
\end{align}
