# Little-o notation: limit in the definition

In Apostol, little-o $$f(x) = o(g(x))$$ is defined as $$\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}$$. However, in Wiki it is defined as $$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)}$$. How do we determine the limit point from the notation? Is it possible that $$f(x) = o(g(x))$$ if $$x \rightarrow 0$$, while $$f(x) \neq o(g(x))$$, if $$x \rightarrow \infty$$

We usually point out the limit of $$x$$ in the text, such as "$$f=o(g)$$ as $$x\to0$$" for the first case and "$$f=o(g)$$ as $$x\to\infty$$" for the second case.

Yes. For example, $$f(x)=x$$ and $$g(x)=1$$.

I'm gonna cite another way to define it without limit of $$\frac{f(x)}{g(x)}$$, maybe this will result easier:

Let be $$f : U \longmapsto \mathbb{R},x_{0}$$ accumulation point of $$f$$. We say that $$f = o(g(x)) \hspace{0.2cm} x \to x_{0}$$

If exists $$\epsilon : U \longmapsto \mathbb{R} :$$

$$f(x) = \epsilon(x) \cdot g(x), \hspace{0.3cm} \lim\limits_{x \to x_{0}} \epsilon(x) = 0$$

• What is $\epsilon(x)$ but another way to say $\frac{f(x)}{g(x)}$? Aug 27, 2019 at 16:16
• @MishaLavrov It doesn't require to define a quotient. Aug 27, 2019 at 16:26
• While in my opinion, this is a better way of defining little-o, all the explanation is about things the OP did not ask, while the only thing that in any way addresses the OP's question is tacked on without discussion. Namely, including $x \to x_0$ in the notation. Aug 27, 2019 at 17:03