Calculate $\lim\limits_{x \rightarrow \infty}(\frac{2}{\pi}\arctan(x))^{\frac{x^2}{1+2x}}$ I have to calculate next limit: $$\lim\limits_{x \rightarrow \infty}\left(\frac{2}{\pi}\arctan(x)\right)^{\frac{x^2}{1+2x}}$$
So far I got to this point $$e^{\lim\limits{x\rightarrow\infty}\frac{x^2}{1+2x}\frac{2\arctan(x)-\pi}{\pi}}$$
When I start to calculate this limit on $e$ I then came to this $$\frac{1}{\pi} \lim\limits_{x\rightarrow\infty}\frac{x^2(2\arctan(x)-\pi)}{1+2x}$$
And also I must not use L'Hôpital's rule for this one.
Any help?
 A: To avoid direct L'Hopital (but not really, see my explanation below) we note that for positive $x$
$$\arctan(x) = \frac{\pi}{2} - \arctan\left(\frac{1}{x}\right)$$
And we can use this to get asymptotic approximations to the limit
$$= \lim_{x\to\infty} \left(1-\frac{2}{\pi}\arctan\left(\frac{1}{x}\right)\right)^{\frac{x}{2}-\frac{1}{4}+\frac{1}{8x+4}} \sim \lim_{x\to\infty} \left( 1 - \frac{2}{\pi x}\right)^{\frac{x}{2}} = e^{-\frac{1}{\pi}}$$
from the defintion of the limit for $e^x = \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$. This still uses the spirit of L'Hopital, which is asymptotic behaviors of functions.
A: We are to compute
\begin{align*}
\lim_{x\rightarrow\infty}\dfrac{x^{2}}{1+2x}\log\left(\dfrac{2}{\pi}\tan^{-1}x\right).
\end{align*}
With the change of variable $u=1-\dfrac{2}{\pi}\tan^{-1}x$, then 
\begin{align*}
\lim_{x\rightarrow\infty}\dfrac{x^{2}}{1+2x}\log\left(\dfrac{2}{\pi}\tan^{-1}x\right)&=\lim_{u\rightarrow 0^{+}}\dfrac{\cot^{2}\dfrac{\pi}{2}u}{1+2\cot\dfrac{\pi}{2}u}\log(1-u)\\
&=\lim_{u\rightarrow 0^{+}}\dfrac{2}{\pi}\cdot\dfrac{\dfrac{\pi}{2}u}{\sin\dfrac{\pi}{2}u}\cdot\dfrac{\cos^{2}\dfrac{\pi}{2}u}{\sin\dfrac{\pi}{2}u+2\cos\dfrac{\pi}{2}u}\cdot\dfrac{1}{u}\cdot\log(1-u).
\end{align*}
Note that 
\begin{align*}
\lim_{u\rightarrow 0^{+}}\dfrac{1}{u}\cdot\log(1-u)=-\lim_{u\rightarrow 0^{+}}\dfrac{1}{u}\int_{0}^{u}\dfrac{1}{1-t}dt=-\lim_{u\rightarrow 0^{+}}\dfrac{1}{1-\eta_{u}}=-1,
\end{align*}
where $\eta_{u}$ is in between $u$ and $0$, chosen by Mean Value Theorem.
A: Compute the limit of the log in the first place:
$$\frac{x^2}{1+2x}\ln\tfrac\pi2+\frac{x^2}{1+2x}\ln(\arctan x).$$
The first term tends to $\dots$ as $x\to\infty$. For the second term,  set $u=\frac1x$; since $x>0$, we obtain
$$\frac{x^2}{1+2x}\ln(\arctan x)=\frac1{u(u+2)}\Bigl(\frac\pi2-\arctan u\Bigr)=\frac{\pi}{2u(u+2)}-\frac{\arctan u}{u(u+2)}.$$
Can you proceed now?
A: Since $f(x)=\frac{2}{\pi}\arctan(x)-1=-\frac 2 \pi\arctan \frac1x\to 0$ we have
$$\left(\frac{2}{\pi}\arctan(x)\right)^{\frac{x^2}{1+2x}}=\left(\left(1+f(x)\right)^{\frac1{f(x)}}\right)^{f(x)\frac{x^2}{1+2x}}\to e^{-\dfrac 1 \pi}$$
indeed
$$\left(1+f(x)\right)^{\frac1{f(x)}} \to e$$
and by standard limits
$${f(x)\frac{x^2}{1+2x}}=-\frac 2 \pi\frac{\arctan \frac1x}{\frac1x}\frac{x^2}{x+2x^2}\to-\frac2\pi\cdot 1 \cdot \frac12=-\frac1\pi$$
A: 
I thought it would be instructive to present a way forward that relie on only pre-calculus analysis including some elementary inequalities.  To that end we now proceed.

PRIMER:

In THIS ANSWER, I showed using elementary, pre-calculus tools that the arctangent function satisfies the inequalities
$$\frac{x}{\sqrt{1+x^2}}\le \arctan(x)\le x\tag1$$
for $x\ge 0$.  We now use $(1)$ in the development that follows.


Enforcing the substitution $x\mapsto 1/x$ reveals
$$\begin{align}
\lim_{x\to \infty}\left(\frac2\pi \arctan(x)\right)^{\frac{x^2}{1+2x}}&\overbrace{=}^{x\mapsto 1/x}\lim_{x\to 0}\left(\frac2\pi \arctan(1/x)\right)^{\frac1{x(x+2)}}\\\\
&=\lim_{x\to 0}\left(1-\frac2\pi \arctan(x)\right)^{\frac1{x(x+2)}}\tag2
\end{align}$$

Using $(1)$ along with the inequality $\frac1{\sqrt{1+x^2}}\ge 1-\frac12 x^2$ in $(2)$, we find that
$$\left(1-\frac2\pi \left(x-\frac12x^3\right)\right)^{\frac1{x(x+2)}}\le \left(1-\frac2\pi \arctan(x)\right)^{\frac1{x(x+2)}}\le \left(1-\frac2\pi x\right)^{\frac1{x(x+2)}}\tag3$$

Next, recalling that $\lim_{x\to 0}\left(1+ tx\right)^\frac1x=e^t$, it is easy to see that $\lim_{x\to 0}\left(1-\frac2\pi x\right)^{\frac1{x(x+2)}}=e^{-1/\pi}$.
We will now show that the limit of the left-hand side of $(3)$ is also $e^{-1/\pi}$, whence application of the squeeze theorem yields the coveted limit.

Proceeding, we write the left-hand side of $(3)$ as
$$\begin{align}
\left(1-\frac2\pi \left(x-\frac12x^3\right)\right)^{\frac1{x(x+2)}}&=\left(1-\frac2\pi x\right)^{\frac1{x(x+2)}}\times \color{blue}{\left(1+x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}}\tag4
\end{align}$$
Using Bernoulli's Inequality, we have for $0<x<\pi$
$$\begin{align}
1\le \color{blue}{\left(1+x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}}&\le \frac1{\left(1-x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}}\\\\
&\le \frac1{1-\frac{x^2}{(x+2)(\pi -2x)}}\tag5
\end{align}$$
Applying the squeeze theorem to $(5)$, we find that
$$\lim_{x\to 0}\left(1+x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}=1\tag6$$

Using $(6)$ in $(4)$ reveals
$$\lim_{x\to 0}\left(1-\frac2\pi \left(x-\frac12x^3\right)\right)^{\frac1{x(x+2)}}=e^{-1/\pi}\tag7$$

Finally, using $(7)$ in $(3)$ and then equating to $(2)$ yields the coveted limit
$$\lim_{x\to \infty}\left(\frac2\pi \arctan(x)\right)^{\frac{x^2}{1+2x}}=e^{-1/\pi}$$
And we are done!

Tools Used:  Elementary pre-calculus analysis (e.g. Bernoulli's Inequality and other elementary inequalities) only along with the limit definition of the exponential function.

