# how to compute $O(n^{0.000001})$ [duplicate]

this MIT course gives a formula about Big O

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

going through wiki, i cannot find a similar Big O properties or usages.

how to compute $$O(n^{0.000001})$$ to get a quantity like $$\log n$$ ?

## marked as duplicate by Juho, Discrete lizard♦May 17 at 7:04

• I suppose what you are referring to is the claim that for any $c > 0$, we have that $\log n = O(n^c)$. You can prove the claim by just applying the the definition of Big Oh. That is, show that $\lim_{n \to \infty} f(n) / g(n) = \lim_{n \to \infty} \log n / n^c < \infty$ and you are done. – Juho May 17 at 6:58
• @Juho i cannot get your point, would plz give more explanation how do you get the right side from left side of $\lim_{n \to \infty} f(n) / g(n) = \lim_{n \to \infty} \log n / n^c < \infty$, which part of definition let you do this or unlock my question, so that some one else may give more info easy to understand – shi95 May 17 at 7:19
• @Juho thanks for your reply, do you mean you get this formula from the table on Wikipedia? $$O(n^c) = \lim_{n \to \infty} \frac{\log n }{ n^c} < \infty$$ – shi95 May 17 at 7:35