$\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x=? $ I`m trying to evaluate this limit and I need some advice how to do that.
 $$\displaystyle\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x $$
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
 A: 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.
A: 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
$$L = e^{-2/\pi}$$
A: Compute the limit of the logarithm and do the substitution $x=1/t$, recalling that, for $t>0$, $\arctan(1/t)=\pi/2-\arctan t$:
\begin{align}
\lim_{x\to\infty}x\log\frac{2\arctan x}{\pi}
&=
\lim_{t\to0^+}\frac{1}{t}\log\Bigl(1-\frac{2}{\pi}\arctan t\Bigr) \\
&=
\lim_{t\to0^+}\frac{-2(\arctan t)/\pi+o(\arctan t)}{t}\\
&=
\lim_{t\to0^+}\frac{-2t/\pi+o(t)}{t}=-\frac{2}{\pi}
\end{align}
so your limit is $e^{-2/\pi}$
