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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.

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$\ln^d(x)$ means $(\ln x)^d$, correct? –  MJD May 22 '12 at 20:16
    
Yes. That's how we write it... :-) –  Amihai Zivan May 22 '12 at 20:18
1  
The derivatives of the top get messy. Things may feel simpler if you let $y=\ln x$, then the top is just $y^{\delta}$, the bottom is $e^{\epsilon y}$. –  André Nicolas May 22 '12 at 20:20
    
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 $\varepsilon>0$, $\delta\in\mathbb{R}$ –  Amihai Zivan May 22 '12 at 20:32
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2 Answers 2

up vote 3 down vote accepted

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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very nice. thx! –  Amihai Zivan May 22 '12 at 20:27
    
As per your first comment, I specifically mentioned "disregarding constants" and used ~, as opposed to =. –  Amihai Zivan May 22 '12 at 20:37
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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 $x\geqslant\mathrm e$.
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