Proving that this limit is zero

Question

Let $f$ be a continuous function from $\mathbb{R}$ to $\mathbb{R}$. I want to show that if $f(0)=0$, then there exists an $n\in \mathbb{N}$ such that $$\lim_{x\rightarrow 0}\frac{(f(x))^n}{|x|}=0$$ Here $n$ denotes a power, not a derivative. My first though was to use Taylor series, but I do not know if my fucntion is differentiable. I do not have any clue on what other thing could work. Can someone help me with this?

Answer 1

A counterexample: Using the fact that $$ \lim_{x \to 0} |x| \cdot (\ln(|x|))^n = 0 $$ for all $n > 0$ one can see that the function $f: \Bbb R \to \Bbb R$ defined as $$ f(x) = \begin{cases} 0 & \text{ if } x = 0 \\ \frac{1}{\log(1/|x|)}& \text{ if } 0 < |x| < 1/e \\ 1 & \text{ if } |x| > 1/e \\ \end{cases} $$ is continuous with $$ \lim_{x\to 0}\frac{(f(x))^n}{|x|}=\infty $$ for all $n > 0$.