$\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x=? $

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.

Let $L$ be the limit, if it exists. Take logs of both sides:

$$\log{L} = \lim_{x \to \infty} x \log{\left ( \frac{2}{\pi} \arctan{x}\right)}$$

Note that

$$\arctan{x} \sim \frac{\pi}{2} - \frac{1}{x} \quad (x \to \infty)$$

This can be seen from the integral form of $\arctan x$:

$$\arctan{x} = \int_0^x \frac{dt}{t^2+1} = \int_0^{\infty} \frac{dt}{t^2+1} - \int_x^{\infty} \frac{dt}{t^2+1}$$

The first integral is $\pi/2$. The second we can approximate because we want to consider $x$ being large, which implies that $t$ is large compared to $1$. Therefore, $t^2+1 \approx t^2$ and the second integral becomes approximately

$$\int_x^{\infty} \frac{dt}{t^2} = \frac{1}{x}$$

So, returning to L, we have

$$\log{L} = \lim_{x \to \infty} x \log{\left ( 1-\frac{2}{\pi x}\right)} = -\frac{2}{\pi}$$

Therefore, the limit is

$$L = e^{-2/\pi}$$