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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 $x \rightarrow \infty$

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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 $g(x)=1$.

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

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

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  • $\begingroup$ What is $\epsilon(x)$ but another way to say $\frac{f(x)}{g(x)}$? $\endgroup$ Aug 27 '19 at 16:16
  • $\begingroup$ @MishaLavrov It doesn't require to define a quotient. $\endgroup$ Aug 27 '19 at 16:26
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    $\begingroup$ 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. $\endgroup$ Aug 27 '19 at 17:03

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