Limit with logarithm: $\lim_{n \to \infty} \frac{n^\alpha}{\ln^\beta n}$

Question

What is the limit $\lim_{n \to \infty} \frac{n^\alpha}{\ln^\beta n }$ (ln=natural logarithm) for alfa real and less than zero? I found out it is zero for $\beta\ge0$, since then you can use the arithmetic of limits, but the result should be independent of $\beta$. Thank you.

Answer 1

One may recall that, as $x \to \infty$,

$$ \lim_{x \to \infty} \frac{\ln^b x }{x^a}=0,\quad a>0, \tag1 $$

whatever the value of $b$ is.

Then using $(1)$ with $x:=n$, $a=-\alpha>0$, we obtain

$$ \lim_{n \to \infty} \frac{n^\alpha}{\ln^\beta n }=\lim_{n \to \infty} \frac{\ln^{-\beta} n}{n^{-\alpha} }=0,\tag2 $$ as announced.