TITLE: Proving $n \log n \subseteq O(n^{1+\epsilon})$ without limits
QUESTION [6 upvotes]: How can one show that $n \log n \subseteq O(n^{1+\epsilon})$ where $0 < \epsilon < 1$ without using limits? This question arise from a homework where I used limits to prove the relation. I'd like to know if there is another way.

REPLY [6 votes]: Let $0<\epsilon<1$ be fixed. We need only show that $\log n=O(n^\epsilon)$. Suppose $n\geq 1$ so that $\log n\geq 0$.  Since $$n^\epsilon=e^{\epsilon \log n}=1+\epsilon \log n+\frac{\epsilon^2 \log^2 n}{2}\cdots \geq \epsilon \log n$$ we see that $$\log n \leq \frac{1}{\epsilon}n^\epsilon $$ which proves the desired result.