Prove that $\lim\limits_{x\to\infty}\frac{\ln^{\delta}(x)}{x^{\varepsilon}}=0 $ for every $\delta\in\mathbb{R}$ and $\varepsilon >0$ 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.
 A: When you differentiate $\ln^{\delta}(x)$ with respect to $x$, you will get $\delta \dfrac{\ln^{\delta-1}(x)}{x}$ and not as what you have written. It is easier to do it directly, than use L'Hopital's rule.
Replacing $x$ by $\exp(t)$, we get that $$\lim_{x \rightarrow \infty} \dfrac{\ln^{\delta}(x)}{x^{\varepsilon}} = \lim_{t \rightarrow \infty} \dfrac{t^{\delta}}{\exp(\varepsilon t)}$$
Note that $$\exp(\varepsilon t) > \dfrac{(\varepsilon t)^n}{n!}$$ where $n$ is chosen so that it is greater than $\delta$. This is so since $\dfrac{(\varepsilon t)^n}{n!}$ is one term in the taylor expansion of $\exp(\varepsilon t)$ and all the other terms are also positive.
Now you get what you want since
$$\lim_{t \rightarrow \infty} \dfrac{t^{\delta}}{\exp(\varepsilon t)} <  \lim_{t \rightarrow \infty} \dfrac{t^{\delta}}{\dfrac{(\varepsilon t)^n}{n!}} = \lim_{t \rightarrow \infty} \frac{n!}{\varepsilon^n} \dfrac1{t^{n - \delta}} = \frac{n!}{\varepsilon^n} \lim_{t \rightarrow \infty} \dfrac1{t^{n - \delta}} = 0$$
A: Without L'Hopital: 


*

*If $\delta\gt0$, one can use the identity
$$
\frac{\log^\delta x}{x^\varepsilon}=\kappa\cdot\left(\frac{\log z}z\right)^\delta\quad \text{with}\quad z=x^{\varepsilon/\delta}\quad \text{and}\quad \kappa=\left(\frac{\delta}{\varepsilon}\right)^\delta.
$$
Now, $z\to+\infty$ when $x\to+\infty$ because $\varepsilon/\delta\gt0$, and $\dfrac{\log u}u\to0$ when $u\to+\infty$.

*If $\delta\leqslant0$, one can compare the ratio to $\dfrac1{x^\varepsilon}$ for $x\geqslant\mathrm e$.

