Is there $\epsilon>0$ such that $n\log n=\Omega (n^{1+\epsilon})$

Question

It is obvious that $n\log n=\omega (n)$. But does this necessarily mean that $\exists \epsilon>0$ such that $n\log n=\Omega (n^{1+\epsilon})$? My intuition says yes, but I'm not sure what $\epsilon$ should be.

Answer 1

Answer (Thanks to Antonio Vargas): No such $\epsilon$ exists

Proof: Suppose, for the sake of contradiction, that $\exists \epsilon>0$ with $$n\log n=\Omega (n^{1+\epsilon})$$. This would imply that for all sufficiently large $n$ we'd have $$\frac{\log n}{n^\epsilon} > c$$ for some constant $c>0$. However, $$\lim_{n\rightarrow\infty}\frac{\log n}{n^{\epsilon}}\overbrace{=}^{L'Hopital}\lim_{n\rightarrow\infty}\frac{\frac{1}{n}}{\epsilon n^{\epsilon-1}}=\lim_{n\rightarrow\infty}\frac{1}{\epsilon n^{\epsilon}}\xrightarrow[n\rightarrow\infty]{}0$$ So no such constant exists