# What's the exact definition of little o notation

I have seen the following definitions of little-o notation,

$$(1)$$ Let $$f,g:\mathbb{R}\rightarrow \mathbb{R}$$ . Then I say $$f(x)=o(g(x))$$ iff, $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=0$$.

$$(2)$$ On the other hand I have also seen this definition on some websites, If $$f(x)=o(g(x))$$. Then there exists a $$k\in\mathbb{N}$$ such that $$f(x)\leq k g(x)$$ for all $$x\geq x_{0}$$ for some $$x_{0}\in\mathbb{R}$$

My question:-

I was just wondering how are they both equivalent. I am unable to get the intuition behind. How to understand this equivalence intuitively?

• It's clear that these definitions are not equivalent. Take $f=g=1$ to see that (2) doesn't imply (1). Jul 12, 2022 at 19:49
• I got the fault in my question as mentioned by @John White. Jul 12, 2022 at 20:06
• As an English sentence, the one you've written in (2) looks more like a theorem than a definition, the theorem being the true statement that $f\in o(g)\Rightarrow f\in O(g)$. Jul 13, 2022 at 6:56

• It should be $f(x) = o(g(x))$ if $\textbf{for every}$ $k \in \mathbb{R}^+$, there exists $x_0$ s.t. $f(x) \leq k g(x)$ Jul 12, 2022 at 20:03
• @JohnWhite You mean $f(x)=O(g(x))$