calculus - evaluation of limit $limlimits_{xto infty}left(frac{2arctan(x)}{pi}right)^x $

I`m trying to evaluate this limit and I need some advice how to do that.
$$\lim\limits_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x $$
I have a feeling it has to do with a solution of form $1^\infty$ but do not know how to proceed. Any hints/solutions/links will be appreciated

Answer

It could be suitably modified into something involving the limit $(1+\frac1x)^x\rightarrow e$ for $x\to\infty$.

$$
\left(\frac{2\arctan x}{\pi}\right)^x
~=~
\left[1 + \left(\frac{2\arctan x}{\pi}-1\right)\right]^x
$$
Let $f(x)=\left(\frac{2\arctan x}{\pi}-1\right)$; clearly $f(x)\to 0$ for $x\to+\infty$, therefore
$$
\left[1+f(x)\right]^{\frac{1}{f(x)}}\longrightarrow e
$$
Let us focus on the limit of $xf(x)$: using l'Hospital's rule we get

$$
\lim_{x\to+\infty}x\,f(x)
~=~
\lim_{x\to+\infty}
\frac{\frac{2\arctan x-1}{\pi}}{\frac1x}
~\stackrel H=~
\lim_{x\to+\infty}
\frac{\frac{2}{\pi(1+x^2)}}{-\frac1{x^2}}
~=~
-\frac2\pi

$$
Now, putting all together:
$$
\lim_{x\to+\infty}
\left(\frac{2\arctan x}{\pi}\right)^x
~=~
\lim_{x\to+\infty}
\big(1+f(x)\big)^x
~=~
\lim_{x\to+\infty}

\left[\big(1+f(x)\big)^{\frac{1}{f(x)}}\right]^{xf(x)}
~=~
e^{-2/\pi}
$$
Generally, when you run into $1^\infty$ you can work it out in this way.