log(n)^k = O(n^epsilon)

Question

Hey I need to prove that for k >= 1 and for $0< \epsilon < 1$: $$\lim_{n \to \infty}\frac{(\log{n})^k}{n^\epsilon} < \infty$$ I should be able to do this without L'Hopitals but couldn't do it without it. can someone give a hint?

Answer 1

Try substituting $n = e^m$, and consider making use of the series expansion of $e^m$ to give a desired bound.