How can we calculate the following limit? I came across this problem and spent a lot of time on this but couldn't figure it out.
$$\lim_{x\to\infty}\left({\frac{2 \arctan(x)}{\pi}}\right)^x$$

Using a general approach for $1^\infty$ type, I could come to the following expression:
$$ e^\left(\lim_{x\to\infty}{x\frac{2 \arctan(x)}{\pi}-1}\right) $$
But I can't get further from this.
I was able to observe that this is a ($\infty*0$)form so maybe we could transform the whole parenthesis with the $\arctan$ in the denominator so that we could get a $\frac{\infty}{\infty}$ form which could enable us to use L'Hospital's rule here but I that didn't yield a good result.
Is there any better way to do this?
 A: Note that\begin{align}\lim_{x\to\infty}x\left(\frac{2\arctan x}\pi-1\right)&=\lim_{x\to\infty}\frac{\frac{2\arctan x}\pi-1}{\frac1x}\\&=\lim_{x\to\infty}\frac{\frac2{\pi(1+x^2)}}{-\frac1{x^2}}\\&=-\frac2\pi.\end{align}
A: You were going the right way. All that is left is to write the limit with $\frac 1x$ in the denominator, then one application of L'Hospital's rule.
A: If you want to have the limit (and more) and if you enjoy composing Taylor series.
$$y=\Big[\frac{2 }{\pi }\tan ^{-1}(x)\Big]^x \implies \log(y)=x \log\Big[\frac{2 }{\pi }\tan ^{-1}(x)\Big]$$
$$\tan ^{-1}(x)=\frac{\pi }{2}-\frac{1}{x}+\frac{1}{3 x^3}+O\left(\frac{1}{x^5}\right)$$
$$\frac{2 }{\pi }\tan ^{-1}(x)=1-\frac{2}{\pi  x}+\frac{2}{3 \pi  x^3}+O\left(\frac{1}{x^5}\right)$$
$$ \log\Big[\frac{2 }{\pi }\tan ^{-1}(x)\Big]=-\frac{2}{\pi  x}-\frac{2}{\pi ^2 x^2}+O\left(\frac{1}{x^3}\right)$$
$$\log(y)=-\frac{2}{\pi  }-\frac{2}{\pi ^2 x}+O\left(\frac{1}{x^2}\right)$$
$$y=e^{\log(y)}=e^{-2/\pi }\Big[1- \frac{2}{\pi ^2 x}\Big]+O\left(\frac{1}{x^2}\right)$$
