Finding the limit of a function with ArcTan I've found difficulties finding this limit ( without using Taylor series approximation, as it's intended for the secondary-school ):
$$
\lim_{x\ \to\ \infty}\left[\,
{x^{3} \over \left(\,x^{2} + 1\,\right)\arctan\left(\,x\,\right)} - {2x \over \pi}
\,\right]
$$
Thanks.

i think the answer is $4/\pi^2.$
i will make a change of variable $u = 1/x$ and will use the fact $\tan 1/u = \pi/2 - \tan u$ for $u > 0.$
$\begin{eqnarray}
lim_{x \to \infty} \frac{x^3}{(1+x^2) \tan x} - \frac{2x}{\pi} = 
 \lim_{u \to {0+}} \frac{1}{u(1+u^2)\tan(1/u)} - \frac{2}{\pi u}  \\
= \frac{2}{u(1+u^2)(\pi-2\tan u)} - \frac{2}{\pi u}  \\
= \frac{2\pi  - 2 (1+u^2)(\pi - 2 \tan u)}{\pi u (1+u^2)(\pi - 2\tan u)}\\
= \frac{2 \pi  - 2 (\pi - 2\tan u + \cdots)}{\pi u(1 + u^2)(\pi - 2 \tan u)}\\
= \frac{4 \tan u + \cdots}{\pi^2 u + \cdots} = \frac{4}{\pi^2}\\
\end{eqnarray}$
