I'm trying to show that the following limit is true for all $\delta\in \mathbb{R}$ and $\varepsilon>0$:
$\lim_{x\to\infty}\frac{\ln^{\delta}(x)}{x^{\varepsilon}}=0$
I know that applying L'Hopital's rule $\lceil\delta\rceil$ times so that (disregarding constants):
$\lim_{x\to\infty}\frac{\ln^{\delta}(x)}{x^{\varepsilon}}=\dots\sim\lim_{x\to\infty}\frac{\frac{1}{x^\delta}}{x^{\varepsilon-\delta}}=\lim_{x\to\infty}\frac{1}{x^{\varepsilon}}=0$
Is my proof valid? Is there a more direct proof without using L'Hopital?
Thank you.