Q:

Prove or Disprove Big-Oh using ratio and limits

$$n^{1.001}+nlog (n) = O(nlog(n))$$
So from there we can break it down into a ratio where we are using limits and approaching infinity:
$$\frac{n^{1.001}+nlog (n)}{nlog(n)}$$
A bit rusty on the arithmetic, should I find a derivative next? 

A:

You are right to look at the ratio.  Discarding the $+1$, we get:
$$
{n^{1.001} \over n \log(n)} = {n^{0.001} \over \log(n)}. 
$$
Taking the $\log$ of the latter ratio, we get:
$$
0.001 \log(n) - \log (\log(n)),
$$
which grows without bound as $n \rightarrow \infty$.  (To see this, let $m = \log(n)$ and examine the expression as $m \rightarrow \infty$.)

