Does every continuous function have an order of vanishing? I am trying to understand in what ways we can characterise growth of continuous functions around a point. One particular way to do so is by comparing a functions growth to a power of $x$ as follows.
For a continuous function $f:[0,\infty) \to \mathbb{R}$ define its order of vanishing (at $x=0$) to be the number $0\leq \alpha \leq \infty$ such that
$$\lim_{x \to 0} \left(\frac{f(x)}{x^\alpha}\right) =\gamma \neq \cases{0 \\ \pm \infty} $$
i.e. the $\alpha$ so the above limit is defined, finite and non-zero.
In the case for a function $f(x)$ where $\frac{f(x)}{x^n} \to 0 \quad \forall n \in[0,\infty)$ define the order of vanishing of $f(x)$, namely $\alpha$, to be $\infty$.
For example with these definitions $\alpha=\infty$ when $f(x)=\exp \left(-\frac{1}{x}\right)$ as then $f(x)$ would grow slower than any power of $x$ around $x=0$.
My question is: Does the order of vanishing always exist for each continuous function $f$ on $[0,\infty)$? i.e. is this quantity well defined?
If not then are there extra conditions should I stipulate on $f$ to guarantee existence?
My thoughts on the matter as are follows,

*

*If such an $\alpha$ exists it is necessarily unique, for if $\beta<\alpha$ then $$\frac{f(x)}{x^\beta}=\frac{f(x)}{x^\alpha}\cdot x^{\alpha-\beta} \to 0$$ since by assumption the limit of $\frac{f(x)}{x^\alpha}$ is finite and non-zero. Hence $\beta$ cannot be an order of vanishing for $f$. Similarly if $\beta>\alpha$ then $$\frac{f(x)}{x^\beta}=\frac{f(x)}{x^\alpha}\cdot \frac{1}{x^{\beta-\alpha}} \to \pm \infty$$ again, $\beta$ cannot be an order of vanishing of $f$. Therefore if $\alpha$ exists it is unique.


*We cannot have $\frac{f(x)}{x^n}\to \pm \infty \quad \forall n \in [0,\infty)$ Since for $n=0$ this would imply $f(0)$ non finite.


*I suspect the existence of such an $\alpha$ may have a topological proof, maybe we can partition the domain of $\alpha \in [0,\infty)$ up into parts $S$ and $L$, i.e. $[0,\infty)=S \cup L$ where
$$S:=\bigg\{ \alpha \in [0,\infty) \bigg | \frac{f(x)}{x^\alpha} \to \pm \infty \text{ as } x \to 0 \bigg \}$$
$$L:=\bigg\{ \alpha \in [0,\infty) \bigg | \frac{f(x)}{x^\alpha} \to 0 \text{ as } x \to 0 \bigg \}$$
Maybe this violates the connectivity of $[0,\infty)$?
 A: $g(x)=\begin{cases}\frac1{\log \lvert x\rvert}&\text{if }x\ne 0\\ 0&\text{if }x =0\end{cases}$ satisfies $$\lim_{x\to 0}\left\lvert\lvert x\rvert^{-\alpha}g(x)\right\rvert=\begin{cases}\infty&\text{if }\alpha>0\\ 0&\text{if }\alpha=0\end{cases}$$
A: As I have commented, $f(x) = x\sin\frac{1}{x}$ with extended defintion $f(0)=0$ is a counter example. Intuitively, we may want $f(x)$ to have $1$ as the order of vanishing.
We can do this by defining the order of vanishing to be $$\sup\{\alpha:\lim_{x\rightarrow 0^+}\frac{f(x)}{x^{\alpha}}=0\}$$
Note that the set includes $\alpha=0$ (if we assume $f(0)=0$, i.e. $f(x)$ indeed vanishes at $0$), hence nonempty and $\sup$ exists (could be $\infty$). This defintion is clearly the same as yours if exists.
This definition is equivalent to the following one:
$$\inf \{\alpha: \limsup_{x\rightarrow 0^+} \frac{f(x)}{x^\alpha}=\infty\}$$
If the set is empty, then $\inf$ is understood to be $\infty$.
Using these definitions, the $g(x)$ constructed by @Saucy O'Path has $0$ as its order of vanishing.
