Does $\lim_{x\to0}\frac{xf'(x)}{f(x)}$ exist when $f(0)=0$, $f'(0)=0$

Question

Suppose $f(0)=0$, $f$ is differentiable and $\frac{f'(x)}{\frac{f(x)}{x}}=\frac{xf'(x)}{f(x)}$ is well defined in some interval $(-\epsilon,\epsilon)$ without $0$. By definition,
$$\lim_{x\to0}\frac{f(x)}{x}=f'(0)=\lim_{x\to0}f'(x)$$
When $f'(0)\neq0$,
$$\lim_{x\to0}\frac{xf'(x)}{f(x)}=\frac{f'(0)}{f'(0)}=1$$
so I guessed that $\lim_{x\to0}\frac{xf'(x)}{f(x)}=1$ when $f'(0)=0$ too, but it was not since
$$\lim_{x\to0}\frac{xf'(x)}{f(x)}=\lim_{x\to0}\frac{nx^n}{x^n}=n$$
where $f(x)=x^n$. So now my question is that does $\lim_{x\to0}\frac{xf'(x)}{f(x)}$ always exist when $f'(0)=0$. I tried to L'Hospital and squeeze but I can't prove it because the limit does not converges to unique value. Can we prove this or any counterexamples?

Answer 1

Unfortunately there is no way to save this. The well-known $C^\infty$-function $$f(x)=\begin{cases} e^{-1/x^2} & x\not=0 \\ 0 &x= 0 \end{cases}$$

satifies $f^{(n)}(0)=0$ for all $n$, but $$\lim_{x\rightarrow 0}\frac{xf'(x)}{f(x)} = \lim_{x\rightarrow 0} x(\ln|f(x)|)' = \lim_{x\rightarrow 0} \frac{2}{x^2}=\infty$$