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.

Hint. You may write, as $x$ tends to $+\infty$,
$$
\arctan x=\frac{\pi}{2}-\arctan \frac{1}{x}
$$
and use $$ \arctan u=u-\frac{u^3}{3}+\mathcal{O}(u^4), \quad u \rightarrow 0,$$
to obtain $$
(x^2+1)\arctan x=(x^2+1)\left(\frac{\pi}{2}-\arctan \frac{1}{x}\right)=\frac{\pi  x^2}{2}-x+\frac{\pi }{2}+\mathcal{O}\left(\frac{1}{x}\right)
$$
then
$$
\begin{align}
\frac{x^3}{(x^2+1)\arctan x}&=\frac{x^3}{\frac{\pi  x^2}{2}-x+\frac{\pi }{2}+\mathcal{O}\left(\frac{1}{x}\right)}\\\\&=\frac{2}{\pi}\frac{x}{1-\frac{2}{\pi x}+\mathcal{O}\left(\frac{1}{x^2}\right)}\\\\&=\frac{2 x}{\pi }+\frac{4}{\pi ^2}+\mathcal{O}\left(\frac{1}{x}\right)
\end{align}
$$ Hence $$ \lim\limits_{x\to +\infty} \left({x^3\over(x^2+1)\arctan(x)}-{2x\over \pi}\right)=\frac{4}{\pi ^2}.$$ Hoping this can help you.
