How can I solve $\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$? The limit is: $$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$$ 
The limit is the kind of $1^\infty$, since:
$$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x=\left(\frac{2\times\frac{\pi}{2}}{\pi}\right)^\infty=\left(\frac{\pi}{\pi}\right)^\infty=1^\infty$$ 
This is the farthest I went: $$L = \lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$$
$$\ln(L) = \lim_{x\to\infty}\ln\left(\frac{2\arctan(x)}{\pi}\right)^x=\lim_{x\to\infty}x·\ln\left(\frac{2\arctan(x)}{\pi}\right)$$
Which becomes $0·\infty$
and I dont know where to go from here

There is a way to complete your argument, although the work is a little tedious. Observe that
$$
x \cdot \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) =
\frac{\ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right)}{\left( \dfrac{1}{x} \right)}.
$$
As
$$
\lim_{x \to \infty} \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) = \ln \left( \lim_{x \to \infty} \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) = \ln(1) = 0
$$
and
$$
\lim_{x \to \infty} \frac{1}{x} = 0,
$$
we can apply l’Hôpital’s Rule to obtain
\begin{align}
\ln(L)
&= \lim_{x \to \infty} x \cdot \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) \\
&= \lim_{x \to \infty} \frac{\ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right)}{\left( \dfrac{1}{x} \right)} \\
&= \lim_{x \to \infty} \frac{\dfrac{d}{dx} \left[ \ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) \right]}{\dfrac{d}{dx} \left[ \dfrac{1}{x} \right]} \\
&= \lim_{x \to \infty} \frac{\left[ \dfrac{1}{(1 + x^{2}) \cdot {\tan^{-1}}(x)} \right]}{\left( - \dfrac{1}{x^{2}} \right)} \\
&= - \frac{2}{\pi}.
\end{align}
Therefore,
$$
L = e^{\ln(L)} = e^{- 2/\pi}.
$$
