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.
 A: 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 $u > 0.$
$\begin{eqnarray}
lim_{x \to \infty} \frac{x^3}{(1+x^2) \tan x} - \frac{2x}{\pi} = 
 \lim_{u \to {0+}} \frac{1}{u(1+u^2)\tan(1/u)} - \frac{2}{\pi u}  \\
= \frac{2}{u(1+u^2)(\pi-2\tan u)} - \frac{2}{\pi u}  \\
= \frac{2\pi  - 2 (1+u^2)(\pi - 2 \tan u)}{\pi u (1+u^2)(\pi - 2\tan u)}\\
= \frac{2 \pi  - 2 (\pi - 2\tan u + \cdots)}{\pi u(1 + u^2)(\pi - 2 \tan u)}\\
= \frac{4 \tan u + \cdots}{\pi^2 u + \cdots} = \frac{4}{\pi^2}\\
\end{eqnarray}$
A: Let $x=\tan t$, where $t\to\dfrac\pi2$ , and use some basic trigonometric identities, like $\tan t=\dfrac{\sin t}{\cos t}$ and 
$1+\tan^2t=\dfrac1{\cos^2t}$ . Then let $t=\dfrac\pi2-u$, where $u\to0$, and again use some basic trigonometric 
identities, like $\cos\bigg(\dfrac\pi2-u\bigg)=\sin u$ and $\sin\bigg(\dfrac\pi2-u\bigg)=\cos u$. Lastly, use $~\dfrac{\sin u}u\to1~$ when 
$u\to0$, in conjunction with $1-\cos u=2\sin^2\dfrac u2$ and $\sin u=2\sin\dfrac u2\cos\dfrac u2$ . QED. :-) It goes on 
without saying that $\arctan(\tan t)=t$ for $~|t|<\dfrac\pi2$.
A: 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.
