Calculate $\lim\limits_{x \rightarrow \infty}(\frac{2}{\pi}\arctan(x))^{\frac{x^2}{1+2x}}$ I have to calculate next limit: $$\lim\limits_{x \rightarrow \infty}\left(\frac{2}{\pi}\arctan(x)\right)^{\frac{x^2}{1+2x}}$$
So far I got to this point $$e^{\lim\limits{x\rightarrow\infty}\frac{x^2}{1+2x}\frac{2\arctan(x)-\pi}{\pi}}$$
When I start to calculate this limit on $e$ I then came to this $$\frac{1}{\pi} \lim\limits_{x\rightarrow\infty}\frac{x^2(2\arctan(x)-\pi)}{1+2x}$$
And also I must not use L'Hôpital's rule for this one.
Any help?

We are to compute
\begin{align*}
\lim_{x\rightarrow\infty}\dfrac{x^{2}}{1+2x}\log\left(\dfrac{2}{\pi}\tan^{-1}x\right).
\end{align*}
With the change of variable $u=1-\dfrac{2}{\pi}\tan^{-1}x$, then 
\begin{align*}
\lim_{x\rightarrow\infty}\dfrac{x^{2}}{1+2x}\log\left(\dfrac{2}{\pi}\tan^{-1}x\right)&=\lim_{u\rightarrow 0^{+}}\dfrac{\cot^{2}\dfrac{\pi}{2}u}{1+2\cot\dfrac{\pi}{2}u}\log(1-u)\\
&=\lim_{u\rightarrow 0^{+}}\dfrac{2}{\pi}\cdot\dfrac{\dfrac{\pi}{2}u}{\sin\dfrac{\pi}{2}u}\cdot\dfrac{\cos^{2}\dfrac{\pi}{2}u}{\sin\dfrac{\pi}{2}u+2\cos\dfrac{\pi}{2}u}\cdot\dfrac{1}{u}\cdot\log(1-u).
\end{align*}
Note that 
\begin{align*}
\lim_{u\rightarrow 0^{+}}\dfrac{1}{u}\cdot\log(1-u)=-\lim_{u\rightarrow 0^{+}}\dfrac{1}{u}\int_{0}^{u}\dfrac{1}{1-t}dt=-\lim_{u\rightarrow 0^{+}}\dfrac{1}{1-\eta_{u}}=-1,
\end{align*}
where $\eta_{u}$ is in between $u$ and $0$, chosen by Mean Value Theorem.
