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I understand the general definition of $f(x) \in o(g(x))$ which is usually given in the context when for $x \to x_0$ both $f(x)$ and $g(x)$ grow to $\infty$. In this "standard case" it means that $f$ grows at speed uniformly much slower than that of $g$.

In some cases (e.g. approximation) one wants to use the $o(\cdot)$ notation to quantify the smallness, and one has that both $f(x) \to 0$ and $g(x) \to 0$ whenever $x \to x_0$. In this case, is it correct to say that $f(x) \in o(g(x))$ means that around $x_0$ $f$ approaches $0$ at speed uniformly much faster than that of $g$ ?

PinkCollins
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    In either case it means the fraction $f(x)/g(x)$ goes to zero as $x \to x_0.$ – coffeemath Aug 09 '22 at 08:59
  • Yes, but this should have been clear from the proper definition of little-o. Are you unable to understand that definition? – user21820 Aug 09 '22 at 09:09
  • yes, although I'm an undergrad student, I'm here because I can read and understand some math. However, the pletora of definitions around are quite confusing for a first-time user. Thanks for your interest – PinkCollins Aug 09 '22 at 09:13
  • Words "slower", "faster" and others are only mathematical slang, which may be helpful, but some times may be confusing. Mathematical definition is foundation on which is based any slang. Let me bring one, which, imo, is more better, then mentioned/linked: $o(f),x\to x_0$ as set $$o(f)={g\colon \exists\varepsilon, \exists \delta>0,\varepsilon:U_\delta(x_0)\to \mathbb{R},\lim\limits_{x\to x_0}\varepsilon(x)=0,\forall x \in U_\delta(x_0),g(x)=f(x)\cdot\varepsilon(x) }$$ – zkutch Aug 09 '22 at 11:20

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