Asymptotic notation for logarithmic function

Question

Can anyone explain to me how

$$ f(n) = n^{0.999999} \log n = O(n^{0.999999} n^{0.000001}) $$

?

Answer 1

It's because $\log n = O(n^\alpha)$ as $n \to \infty$ for any fixed $\alpha > 0$. They chose $\alpha = 0.000001$.

To see this, first show that

$$ \lim_{n \to \infty} \frac{\log n}{n^\alpha} = 0 $$

by using L'Hopital's rule. This implies

$$ \frac{\log n}{n^\alpha} = O(1), $$

and multiplying both sides of this by $n^\alpha$ yields $\log n = O(n^\alpha)$.