# Can anyone explain to me how f(n)=n^(0.999999) log n=O(n^(0.999999) n^(0.000001))?

pliwraih 2022-07-16 Answered
Can anyone explain to me how
$f\left(n\right)={n}^{0.999999}\mathrm{log}n=O\left({n}^{0.999999}{n}^{0.000001}\right)$?
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## Answers (1)

Alanna Downs
Answered 2022-07-17 Author has 11 answers
It's because $\mathrm{log}n=O\left({n}^{\alpha }\right)$ as $n\to \mathrm{\infty }$ for any fixed $\alpha >0$. They chose $\alpha =0.000001$
To see this, first show that
$\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{log}n}{{n}^{\alpha }}=0$
by using L'Hopital's rule. This implies
$\frac{\mathrm{log}n}{{n}^{\alpha }}=O\left(1\right),$
and multiplying both sides of this by ${n}^{\alpha }$ yields $\mathrm{log}n=O\left({n}^{\alpha }\right)$

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