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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.

  • 0
    By the way, I forgot to add, I found it quite interesting (and fascinating, perhaps) that $\sum\frac{1}{n\ln^{1+\delta}(n)} $converges for all \delta>0, while $\sum\frac{1}{n^{1-\varepsilon}\ln^{\delta}(n)}$ diverges for all \varepsilon>0, $\delta\in\mathbb{R}$2012-05-22

2 Answers 2

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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 $\lim_{t \rightarrow \infty} \dfrac{t^{\delta}}{\exp(\varepsilon t)} < \lim_{t \rightarrow \infty} \dfrac{t^{\delta}}{\dfrac{(\varepsilon t)^n}{n!}} = \lim_{t \rightarrow \infty} \frac{n!}{\varepsilon^n} \dfrac1{t^{n - \delta}} = \frac{n!}{\varepsilon^n} \lim_{t \rightarrow \infty} \dfrac1{t^{n - \delta}} = 0$

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    As per your first comment, I specifically mentioned "disregarding constants" and used ~, as opposed to =.2012-05-22
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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 $x\geqslant\mathrm e$.