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
 A: Hint Make change of variables 
$$
1+t=\frac{2\arctan(x)}{\pi}
$$
and recall that
$$
\cot\alpha\sim\frac{1}{\alpha}\quad\mbox{for}\quad\alpha\to 0
$$
A: Series expansion of $\arctan(x)$ about $\infty$ is $$\arctan(x) = \dfrac{\pi}2 - \dfrac1x + \mathcal{O} \left(\dfrac1{x^3}\right)$$
Hence, $$\lim_{x \to \infty}\left(\dfrac2{\pi} \arctan(x)\right)^x = \lim_{x \to \infty} \left(1- \dfrac2{\pi}\dfrac1x + \mathcal{O} \left(\dfrac1{x^3}\right)\right)^x = e^{-2/\pi}$$
A: 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,
$$
L = e^{\ln(L)} = e^{- 2/\pi}.
$$
A: Put $$\cot^{-1}x=y\implies \tan^{-1}x=\frac\pi2-y,y=\cot x$$ and as $x\to\infty,y\to 0$
$$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$$ 
$$=\lim_{y\to 0}\left(\frac{2(\frac\pi2-y)}{\pi}\right)^{\cot y}$$
Let $z=\left(\frac{2(\frac\pi2-y)}{\pi}\right)^{\cot y}$
So, $$\log z=\cot y\log(1-\frac{2y}{\pi})$$
So, $$\lim_{y\to 0}\log z=\lim_{y\to 0}\cot y\log(1-\frac{2y}{\pi})=-\frac2\pi\frac{\left(\lim_{y\to 0}\cos y\right)}{\left(\lim_{y\to 0}\frac{\sin y}y\right)}\left(\lim_{y\to 0}\frac{\log(1-\frac{2y}{\pi})}{(-\frac{2y}{\pi})}\right)$$
$$\lim_{y\to 0}\log z=-\frac2\pi\implies \lim_{y\to 0}z=e^{-\frac2\pi}$$
A: Let's see a basic approach based upon $\lim_{x\to \infty} (1+1/x)^x=e$ and l'Hôpital's rule. Then
$$\lim_{x\to\infty}e^{\left(\displaystyle \frac{\arctan(x)-\frac{\pi}{2}}{\displaystyle\frac{\pi}{2x} }\right)}=\lim_{x\to\infty}e^{\left(\displaystyle \frac{-2x^2}{\pi (x^2+1)}\right)}=e^{\displaystyle\frac{-2}{\pi}}$$
Q.E.D. (Chris)
