if $\frac{o(x)}{x} \rightarrow 0$, then $o(x) \rightarrow 0$? Peano rest

Question

I have an question about peano rest in taylor series. Let consider that in use of example:

$$ f(x) = \cos(x) $$

Let consider $f$ as series in use of taylor with peano rest (for example to $x^4$) $$ f(x) = \cos(x) = 1 - \frac{x^2}{2!}+\frac{x^4}{4!}+o(x^4)$$ where $o(x^4)$ is "small o notation"

What I know

From theorem I know that $$\frac{o(x^4)}{x^4} \rightarrow 0 \text{ when } x\rightarrow 0$$

My doubts

Does it mean that $$o(x^4) \rightarrow 0 \text{ when } x\rightarrow 0 \text{?}$$ I think yes because $x$ is getting smaller and smaller and we divide by them but it is only idea, not formal substantiation

Answer 1

You're correct. For, suppose $f(x)=o(x^4)$ as $x \to 0$. Then, $$\lim_{x \to 0} f(x) = \lim_{x \to 0}\left( x^4 \cdot \frac{f(x)}{x^4}\right) = 0.$$