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everyone, I'm trying to show that: $$\forall\, c\in\mathbb{R},\, d>0\,:\,\lim_{n\to\infty}\frac{\left(\log\left(n\right)\right)^{c}}{n^{d}}=0$$ Intuitively it's completely clear but formalizing it is proving to be a bit difficult. I'd appreciate some help evaluating $\left|\frac{\left(\log\left(n\right)\right)^{c}}{n^{d}}\right|$ in a way that would lead to the result.

Thanks a lot.

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marked as duplicate by Micah, Davide Giraudo, Hagen von Eitzen, Tara B, Ben Millwood Mar 9 '13 at 16:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Which methods and results are you allowed to use? – Norbert Mar 9 '13 at 15:39
up vote 2 down vote accepted

Substitute $x=e^t$ , then $$\lim_{x\to\infty}\frac{\left(\log\left(x\right)\right)^{c}}{x^{d}}=\lim_{t\to\infty}\frac{t^{c}}{e^{td}}$$

Now, Use L'Hopital's rule.

EDIT: On @Peter's suggestion,

Since functional limit $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{\left(\log\left(x\right)\right)^{c}}{x^{d}}=0$$

Thus, for any sequence $\{a_n\}\to\infty $ $$\lim_{n\to\infty}f(a_n)=0$$

Choose $\{a_n\}=n$

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You could make a comment on sequential vs. functional limits, for example, but I don't see anything wrong here. – Pedro Tamaroff Mar 9 '13 at 15:45
This is quite simple indeed, thanks :) – Serpahimz Mar 9 '13 at 16:10

If $c<0$ then $(\log(n))^c\to0$ and $n^d\to\infty$ as $n\to\infty$,whence the claim.

If $c=0$, also $n^d\to\infty$ gives us the desired result.

Assume $c>0$. First show that $$\tag1 \lim_{x\to\infty}\frac{\ln x}{x}=0.$$ Then, substituting $x\leftarrow x^a$ with $a>0$ (so that $x\to+\infty\iff x^a\to+\infty$), this turns into $$ \lim_{x\to+\infty}\frac{\ln x^a}{x^a}=\lim_{x\to\infty}\frac{a\ln x}{x^a}=0,$$ hence $$ \lim_{x\to+\infty}\frac{\ln x}{x^a}=0.$$ By taking $c$th powers, $c>0$, and letting $a=\frac dc$, we find $$ \lim_{x\to+\infty}\frac{(\ln x)^c}{x^d}=0.$$

How to show $(1)$? The inequality $e^y\ge 1+y$ for $y\in\mathbb R$ should be well-known. Then $e^y=(e^{y/2})^2\ge (1+\frac y2)^2=1+y+\frac 14y^2$ for $y\ge -2$ so that substituting $x\leftarrow e^y$, the claim in $(1)$ follows from $$0\le \lim_{y\to+\infty}\frac y{e^y}\le \lim_{y\to+\infty}\frac y{1+y+\frac14y^2}\le \lim_{y\to+\infty}\frac y{\frac14y^2}=0.$$

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That's a nice solution also, thanks! – Serpahimz Mar 9 '13 at 16:12

One way is to use L'hopital's rule on:

\begin{equation} f(x)=\frac{(\log x)^c}{x^d} \end{equation}

and conclude the result for $f(n)$.

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Using L'hopital's rule is quite messy, for starters c,d aren't necessarily integers and even if you bound the original function from above by $\left|\frac{\left(\log\left(n\right)\right)^{\left\lceil c\right\rceil }}{n^{\left\lfloor d\right\rfloor }}\right|$ it's still messy to derivative $\left(\log\left(n\right)\right)^{\left\lceil c\right\rceil }$ over and over. – Serpahimz Mar 9 '13 at 15:42
I meant in the sense as @Avatar. I'm pretty sure that you don't need c and d as integers. – Nirav Mar 9 '13 at 15:43
@Serpahimz There is no need for $c,d$ to be integers. In fact, after $c$ derivations you should get to a denominator $1$, and a very large numerator. – Pedro Tamaroff Mar 9 '13 at 15:46
You indeed don't need c,d to be integers, with the substitution suggested above and derivating $\left\lfloor c\right\rfloor$ it works out nicely. Derivating the original form is of course a complete mess :) – Serpahimz Mar 9 '13 at 16:11

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