# Is $\frac{o(x)}{x^2}=o\left(\frac{1}{x}\right)$ for $x\to0$?

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?

• 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 $0$.
– Syd
Dec 27, 2020 at 22:52
• just note that $o(x)$ is not a function of $x$, but rather a set of functions!
– Syd
Dec 27, 2020 at 22:53

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