How do we argue that in math-terms?

Question

If we have $\,\lim\limits_{n\to \infty}\dfrac{\log n}{n}\,,\,$ I know that this is converging to $0$.

I know that $\,n\,$ is a "stronger function" than $\,\log n\,,\,$ but is there a "mathematical way" of saying that ?

In little-o notation this is written $\log n = o(n)$. – Qiaochu Yuan Feb 09 '23 at 07:48 — Qiaochu Yuan, Feb 09 '23 at 07:48

score 2 · Answer 1 · edited Feb 09 '23 at 08:18

2

Yes, by using the little-$\mathrm o$ notation.
We say that $\,f(x) = \mathrm o(g(x))\,$ iff $\,\lim\limits_{x\rightarrow \infty} \dfrac{f(x)}{g(x)} =0\,.$
In your case, $\,f(x)=\ln x\,$ and $\,g(x)=x\,$.

edited Feb 09 '23 at 08:18

Angelo

12,328

answered Feb 09 '23 at 07:52

Mhd Bhsn

71

Thank you so much – kasper kaspersen Feb 09 '23 at 07:53

How do we argue that in math-terms?

1 Answers1