Expressing a limit in different way? I know following limit is eventually equal to $0$ but can someone actually show me the steps for general value of k: 
$$\lim\limits_{n \to \infty} \dfrac{(\log n)^k}{n} =0$$
I know that attempting to take the limit of the numerator and the denominator just leads to the $\infty/\infty$ indeterminate form so the L'Hopital's Rule must be used taking derivative of numerator and denominator. I don't know how many times I should take it for the general case.
Edit: Is this accurate for a general form of this limit?
$$[k(k-1)(k-2)...3*2*1]\lim\limits_{n \to \infty}\dfrac{1}{n} = 0$$
 A: Hint.  First use L'Hopital's rule to evaluate
$$\lim_{n\to\infty}\frac{\log n}{n^{1/k}}$$
for $k>0$.
(Note that if you want the limit for $k=0$ or for $k<0$ then it is not an indeterminate form and is much easier.)
A: Maybe You can try in this way:
Let $t=\log n$. Then $n=10^t$. So the $\lim_{t\to \infty}\frac{t^k}{10^t} $.
Is this more clear?
You should take derivative of $\frac{t^k}{10^t}$ until $(k-)(k-2)\dots (k-m)\dots <0$
May it helps.
A: other way:
$\log (n)^k = \log(n^k)$
then   $\lim\limits_{n \to \infty} \dfrac{\log (n^k)}{n} = \lim\limits_{n \to \infty} \dfrac{1}{n} .\log (n^k) =\lim\limits_{n \to \infty} \log((n^k)^ \dfrac{1}{n})= \log(\lim\limits_{n \to \infty}(n^k)^ \dfrac{1}{n})$
then I think we know that $\lim\limits_{n \to \infty}(n^k)^ \dfrac{1}{n} = 1$
then our limit is $\log(1) = 0$
or to avoid use L'Hospital's who know how many time.. do:
$\log(n)^k = k.\log(n)$
then use $ \lim\limits_{n \to \infty}\dfrac{(\log n)^k}{n} = \lim\limits_{n \to \infty} k.\dfrac{(\log n)}{n} = k.\lim\limits_{n \to \infty} \dfrac{(\log n)}{n}$ and then our problem it's $\lim\limits_{n \to \infty} \dfrac{(\log n)}{n}$ and to solve it, use L'Hospital once.
