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While solving a limit as $x\to 0$, I had a doubt: is it true that for $x\to0$ it is $o(x)\cdot\frac{1}{x^2}=o\left(\frac{1}{x}\right)$?

In my opinion yes, because of this: $f(x)=o(g(x))$ for $x\to0$ if $$\lim_{x\to0} \frac{f(x)}{g(x)}=0$$ So if I use this definition with $o(x)\cdot\frac{1}{x}$ in place of $f(x)$ and $\frac{1}{x}$ in place of $g(x)$, I get $$\lim_{x \to 0} \frac{o(x)\cdot\frac{1}{x^2}}{\frac{1}{x}}$$ But it is $$\lim_{x \to 0} \frac{o(x)\cdot\frac{1}{x^2}}{\frac{1}{x}}=\lim_{x \to 0} \frac{o(x)}{x}=0\implies\lim_{x \to 0} \frac{o(x)\cdot\frac{1}{x^2}}{\frac{1}{x}}=0$$ Where $\frac{o(x)}{x} \to 0$ as $x\to0$ because of little-o definition. This shows that for $x\to0$ it is s $o(x)\cdot\frac{1}{x^2}=o\left(\frac{1}{x}\right)$.

Is this correct? I'm not sure if I can use functions that contain themselves little o-notation in the definition of little-o, I mean that I'm not sure if it is correct to use $o(x)\cdot\frac{1}{x^2}$ as $f$. I think that it can be done because $o(x)$ is a function of $x$ itself, so it is reasonable to use $o(x)\cdot\frac{1}{x^2}$ as $f$, can someone confirm me this or tell me if it is wrong?

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    $\begingroup$ you could try the $\epsilon$-$\delta$ definiton instead: $f(x)$ is $o(x)$ as $x$ approaches $0$ iff for every $\epsilon>0$ there exists a $\delta>0$ such that $0<\vert x\vert<\delta$ implies $\vert f(x)\vert < \epsilon\vert x\vert$. So, let $\epsilon>0$. Then, by hypothesis, there exists a $\delta>0$ such that $0<\vert x \vert < \delta$ implies $\vert f(x)\vert < \epsilon \vert x\vert$. Dividing both sides by $\vert x^2\vert$ and calling the lhs $\vert g(x)\vert$ yields $\vert g(x)\vert < \epsilon\vert x^{-1}\vert$, i.e. $g(x)$ is $o(x^{-1})$ as $x$ approaches $0$. $\endgroup$
    – Syd
    Dec 27, 2020 at 22:52
  • $\begingroup$ just note that $o(x)$ is not a function of $x$, but rather a set of functions! $\endgroup$
    – Syd
    Dec 27, 2020 at 22:53

1 Answer 1

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An easy way to think about little $o$'s is to realize that $o(f(x))$ can be written as $o(f(x))=f(x) \varepsilon(x)$ where $\lim \varepsilon(x)=0$ as $x\rightarrow 0$. This is a direct consequence of the definition of $o(x)$.

With that, $$\frac {o(x)}{x^2}=\frac{x\varepsilon(x)}{x^2}=\frac{\varepsilon(x)}{x}=o\left(\frac 1 x\right)$$

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