Evaluate where $\epsilon>0,k\geqslant 1$ are constants
$$\lim_{n\to \infty}\frac{{\log^k n}}{n^{\epsilon}}$$
L'Hopital can't help here, also I tried to use $\log$ rules but it didn't helped, I know that $\log$ grows slower then polynom, but $n^\epsilon$ is not polynom, how can I evaluate this limit? thank you
Write $$ \frac{(\log n)^k}{n^\varepsilon}= \left(\frac{\log n}{n^{\varepsilon/k}}\right)^{\!k} $$ For $r>0$, we have $$ \lim_{x\to\infty}\frac{\log x}{x^r}= \lim_{x\to\infty}\frac{1/x}{rx^{r-1}}= \lim_{x\to\infty}\frac{1}{rx^r}=0 $$
Write
$$n^{\epsilon} = e^{\epsilon\log(n)}$$
Then you will be able to use Hopital
$$\frac{\log^k(x)}{e^{\epsilon\log(n)}}$$
First derivative will give you for the numerator
$$\frac{k\log^{k-1}(n)}{n}$$
and for the denominator
$$\frac{\epsilon}{n}\ e^{\epsilon\log(n)}$$
Putting together
$$\frac{\frac{k\log^{k-1}(n)}{n}}{\frac{\epsilon}{n}\ e^{\epsilon\log(n)}} = \frac{k\log^{k-1}(n)}{\epsilon\ e^{\epsilon\log(n)}}$$
So if you keep going with derivatives, you will get after other $k-1$ times,
$$\frac{k!}{\epsilon^{k-1}e^{\epsilon\log(n)}}$$
Which for $n\to\infty$ gives
$$\boxed{0}$$
I thought it might be instructive to present a way forward that does not rely on L'Hospital's Rule. Rather, it relies only on an elementary identity and tools.
To begin, note from THIS ANSWER, in which I showed using only the limit definition of the exponential function and Bernoulli's Inequality. that the logarithm function satisfies the inequality
$$\log(x)\le x-1 \tag 1$$
for $x>0$.
Next, for any $\alpha$, the logarithm satisfies $\log(x^\alpha)=\alpha \log(x)$. Using this in $(1)$, we obtain for any $\alpha >0$ and $n>1$
$$0\le \frac{\log^k(n)}{n^{\epsilon}}\le \left(\frac{n^\alpha -1}{\alpha \,n^{\epsilon/k}}\right)^k \tag 2$$
Finally, since $(2)$ is true for all $\alpha>0$, it is true for $0<\alpha<\epsilon/k$. Then, applying the squeeze theorem to $(2)$ for $0<\alpha <\epsilon/k$ reveals
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty} \frac{\log^k(n)}{n^{\epsilon}}=0}$$
as expected!
Yet another answer: define $$ f(x) = \log \bigg( \frac{\log^k x}{x^\epsilon} \bigg) = k\log\log x - \epsilon\log x. $$ Now $$ \lim_{x\to\infty} \frac{f(x)}{\log x} = \lim_{x\to\infty} \frac{\log\log x}{\log x}-\epsilon = 0-\epsilon $$ by a one-step l'Hopital's rule. Therefore $f(x) < -\frac\epsilon2\log x$ when $x$ is large enough, which is enough to show that $$ \lim_{x\to\infty} f(x) = -\infty. $$ Exponentiating both sides, $$ \lim_{x\to\infty} \frac{\log^k x}{x^\epsilon} = \lim_{y\to-\infty} e^y = 0, $$ which of course implies that $$ \lim_{n\to\infty} \frac{\log^k n}{n^\epsilon} = 0. $$
$$n^{\epsilon} = e^{\epsilon\log(n)}$$
Then you will be able to use Hopital (many times, maybe)
$$\frac{\log^k(x)}{e^{\epsilon\log(n)}}$$
– Enrico M. Mar 17 '16 at 23:27