Limit of a derivative of inverse

Question

Prove for a function $f: \mathbb{R}\rightarrow\mathbb{R}$, $f(x) = \pi x -\arctan(x)$ that it is bijective. Moreover, prove that its inverse function is also differential bijection. Also, verify if they exist and calculate the following limits: $\lim_{x\to +\infty}(f^{-1})'(x)$ and $\lim_{x\to +\infty}\frac{(f^{-1})(x)}{x}.$

I proved it is bijection, namely it is injective because its derivative is strictly greater than zero for all $x \in \mathbb{R}$ which implies it is strictly increasing which then implies it is injective.
It is continuous on $\mathbb{R}$ so by Intermediate Value Theorem it is surjective.

However, I am not sure how to verify these two limits or their existence at all. I suppose the latter limit tells something about asymptote of the inverse, but I really do not manage to connect the dots.

Answer 1

You're right for the first part: $$ f'(x)=\pi-\frac{1}{1+x^2}>0 $$ because $1+x^2\ge1$. Moreover $$ \lim_{x\to-\infty}f(x)=-\infty \qquad \lim_{x\to\infty}f(x)=\infty $$ and the IVT allows you to finish.

The inverse function theorem says that, denoting $g=f^{-1}$ for simplicity, $$ g'(f(x))=\frac{1}{f'(x)}=\frac{1+x^2}{\pi(1+x^2)-1} $$ or $$ g'(y)=\frac{1+g(y)^2}{\pi(1+g(y)^2)-1} $$ What you need is $$ \lim_{y\to\infty}g'(y)= \lim_{x\to\infty}g'(f(x))= \lim_{x\to\infty}\frac{1+x^2}{\pi(1+x^2)-1}=\frac{1}{\pi} $$ For the other limit, use l'Hôpital.

Answer 2

For the first limit observe that $$ (f^{-1})'(x)=\frac1{f'(f^{-1}(x))}=\frac1{\pi-\dfrac{1}{1+(f^{-1}(x))^2}}. $$ For the second limit, from the inequality $$ \pi\,x-\frac\pi2\le f(x)\le\pi\,x,\quad x\ge0, $$ we obtain (setting $y=f(x)$) $$ \pi\,f^{-1}(y)-\frac\pi2\le y\le\pi\,f^{-1}(y),\quad y\ge0. $$