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

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?

• Try $f(x) = x^2(2+\sin{ 1\over x})$ for $x \neq 0$ and zero otherwise. Nov 22, 2021 at 1:26
• @copper.hat Wow! There's a counterexample. How about more stronger condition that $f^{(n)}(0)$ exists for all $n$? I guess the question is true under this condition. Nov 22, 2021 at 1:42

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 $$\lim_{x\rightarrow 0}\frac{xf'(x)}{f(x)} = \lim_{x\rightarrow 0} x(\ln|f(x)|)' = \lim_{x\rightarrow 0} \frac{2}{x^2}=\infty$$

One condition that works is $$f$$ being analytic at $$x=0$$ with $$f$$ not being constantly zero in any neighborhood of $$x=0$$. Let $$n$$ be the smallest $$k$$ such that $$f^{(k)}(0)\ne 0$$. Then $$f(x)=\frac{f^{(n)}(0)}{n!}x^n+O\left(x^{n+1}\right)\quad(x\to 0)\\ xf'(x)=\frac{f^{(n)}(0)}{(n-1)!}x^n+O\left(x^{n+1}\right)\quad(x\to 0)$$ and $$\lim_{x\to 0}\frac{xf'(x)}{f(x)}=n$$ As @Just a user's example shows, it's not sufficient to assume that $$f$$ is infinitely smooth. What you suggested in the comments (existence of $$f^{(k)}(0)$$ for all $$k$$) is not enough.