$$n^{1.001}+nlog (n) = O(nlog(n))$$
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?
Prove or Disprove Big-Oh using ratio and limits
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discrete-mathematics
asymptotics
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1fun fact: $n^\epsilon$ eventually grows faster than $\log n$ for any $\epsilon>0$ – 2017-02-06
1 Answers
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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 $m \rightarrow \infty$.)