Two properties of $f(x) = x \ln { (\frac{2}{\pi} \arctan{x}) } $ how do I prove these? Let's look at the function
$$f(x) = x  \ln {  (\frac{2}{\pi}  \arctan{x}) }  $$
This function seems to be strictly increasing after certain point $x=a$.
(1) What is this point $a$ after which it is strictly increasing?
I don't necessarily need to know this value $a$ but I am curious.
I saw the derivative is a sum of two terms, and both go to zero as $x$ goes to infinity. Seems one of them goes with positive, the other one with negative values.
But I don't quite see which term goes to $0$ faster.
(2) Also how do I prove that
$$f(x) \lt \frac{-2}{\pi} $$ for  $x \gt a$ ?
I am sure of all this, it's been checked in WA, I just cannot prove it by hand.
NOTE 1: I already used all these properties of $f(x)$ to find this limit
$\lim_{x \to \infty} (\frac{2}{\pi}  \arctan{x})^x = \lim_{x \to \infty} e^{f(x)} = e^{-2/\pi}  $
but I cannot quite prove these properties so... my calculation of this limit is not very rigorous yet.
Why is it not rigorous? Because I need to make sure the 2nd additional condition here holds true in order to use the composite limit theorem. And that condition will be proved if I have (2) proved.
Here is the WA link for the limit of f(x)
NOTE 2: Please assume I don't know Taylor series yet... Here is the original problem (maybe I should have started my description from there).
$$\lim_{x \to \infty} (\frac{2}{\pi}  \arctan{x})^x =\ ? $$
It is given in a real analysis book right after the chapter about the L'Hopital rules. Using the L'Hopital rules I proved that
$$\lim_{x \to \infty} {f(x)} = -2/\pi \ \ $$
But I cannot prove that $f(x)$ stays below $-2/\pi$ while approaching it... which I think makes me unable to apply the composite limit theorem in order to find $\lim_{x \to \infty} e^{f(x)}$
 A: $$f(x)=x \log \left(\frac{2 \tan ^{-1}(x)}{\pi }\right)\implies f'(x)=\frac{x}{\left(x^2+1\right) \tan ^{-1}(x)}+\log \left(\frac{2 \tan ^{-1}(x)}{\pi}\right)$$ which is impossible to solve analytically. So, a numerical method should be used.
Plotting $f'(x)$, you will notice that its zero is close to $x=1.5$. So, to make things simple, develop $f'(x)$ as a Taylor series around $x=\sqrt 3$. This would give
$$f'(x)=\left(\frac{3 \sqrt{3}}{4 \pi }-\log \left(\frac{3}{2}\right)\right)-\frac{3
   \left(3 \sqrt{3}-2 \pi \right) \left(x-\sqrt{3}\right)}{16 \pi ^2}-\frac{3
   \left(-9 \sqrt{3}-9 \pi +4 \sqrt{3} \pi ^2\right) \left(x-\sqrt{3}\right)^2}{64
   \pi ^3}+O\left(\left(x-\sqrt{3}\right)^3\right)$$ SOlve the quadratic and select the closest root which, numerically, is $1.46814$ while the exact solution is $1.48545$.
This is not very good; so, use the expansion to $O\left(\left(x-\sqrt{3}\right)^2\right)$ which gives
$$x_0=\sqrt 3+\frac{4 \pi  \left(3 \sqrt{3}-4 \pi  \log \left(\frac{3}{2}\right)\right)}{9
   \sqrt{3}-6 \pi }$$ Using Newton method, the iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 1.343134740 \\
 1 & 1.457092451 \\
 2 & 1.484172827 \\
 3 & 1.485448564 \\
 4 & 1.485451248
\end{array}
\right)$$ The second derivative test confirms that this is the minimum.
With regard to the limit when $x\to \infty$, use again Taylor expansion
$$\tan ^{-1}(x)=\frac{\pi }{2}-\frac{1}{x}+\frac{1}{3 x^3}+O\left(\frac{1}{x^5}\right)$$
$$\frac{2 \tan ^{-1}(x)}{\pi }=1-\frac{2}{\pi  x}+\frac{2}{3 \pi  x^3}+O\left(\frac{1}{x^5}\right)$$
$$\log \left(\frac{2 \tan ^{-1}(x)}{\pi }\right)=-\frac{2}{\pi  x}-\frac{2}{\pi ^2 x^2}+O\left(\frac{1}{x^3}\right)$$
$$x\log \left(\frac{2 \tan ^{-1}(x)}{\pi }\right)=-\frac{2}{\pi  }-\frac{2}{\pi ^2 x}+O\left(\frac{1}{x^2}\right)$$
