Find supremum and infimum of $f(x)$ such that $\lim_{x\to\pm\infty}\frac{f(x)}{|x|}=\infty$

Question

If $\lim_{x\to\pm\infty}\frac{f(x)}{|x|}=\infty$, with $f:\mathbb{R}\to\mathbb{R}$ differentiable, then can I say by applying de l'Hopital that $\lim_{x\to\infty}f'(x)=+\infty$ and $\lim_{x\to-\infty}f'(x)=-\infty$, and so $\sup{f'(x)}=+\infty$, $\inf{f'(x)}=-\infty$? My doubt: how can I be sure that the limits of first derivative exist?

Answer 1

L'Hopital's Rule cannot be applied, but you can use a very elementary argument to show that the supremum if $f'$ is $+\infty$ and infimum is $-\infty$. If $\sup f' <\infty$, say $f'(x) <M$ for all $x$ then $\frac {f(x)-f(0)} {|x|}< M $ by Mean Value Theorem so $\lim \sup_{x \to \infty} \frac {f(x)} {|x|}\le M$ contradicting the hypothesis. Similarly, $\inf f'=-\infty$.