Prove that $\lim_{x\to 0} \frac{f(x)}{f'(x)} = 0$ for $f\in C^1$ and $f(0)=0=f'(0)$ Let $f\in C^1(\mathbb{R})$, with $f(0)=0$ and $f'(0)=0$. Furthermore, assume that in some neighborhood around $0$, $f$ and $f'$ have no additional zeros, so $f^{-1}(\{0\})=\{0\}=(f')^{-1}(\{0\})$.
I want to show that $\lim_{x\to 0} \frac{f(x)}{f'(x)}=0$.
EDIT: According to a comment, this statement might be false. Would it be possible to prove the following, weaker statement: If $\lim_{x\to 0} \frac{f(x)}{f'(x)}=y$, then, $y\in\{0,+\infty,-\infty\}$?
My attempt so far is to write
$$
\lim_{x\to 0} \frac{f(x)}{f'(x)} = \lim_{x\to 0} \lim_{h\to 0} \frac{hf(x)}{f(x+h)-f(x)}
\stackrel{?}{=} \lim_{h\to 0} \lim_{x\to 0} \frac{hf(x)}{f(x+h)-f(x)}
= \lim_{h\to 0} \frac{h\cdot 0}{f(h)-0} = 0.
$$
As indicated by the "?" above the "=", I am not sure how to prove that I am allowed to exchange these limits. I tried to apply the Moore-Osgood theorem. If I understand the theorem correctly, it boils down to showing:

*

*For all $h\neq 0$, the limit $\lim_{x\to 0} \frac{hf(x)}{f(x+h)-f(x)}$ exists. This limit is always equal to $0$, by the same calculation as above.

*For all $x\neq 0$, the limit $\lim_{h\to 0} \frac{hf(x)}{f(x+h)-f(x)}$ exists. This limit is equal to $\frac{f(x)}{f'(x)}$ and thus exists.

*One of the limits converges uniformly, i.e., either the first limit converges uniformly for $h\neq 0$, or the second limit converges uniformly for $x\neq 0$.

Unfortunately, I am stuck at showing uniform convergence of either of the two limits.
I have the following questions:

*

*Is the statement I am trying to prove correct, or do I need further assumptions?

*Is my proof strategy correct so far? Is there a simpler way?

*Is one of the limits actually uniform? If so, can someone give me a hint on how to show it?

 A: One does need further assumptions to be able to deduce
$$\lim_{x \to 0} \frac{f(x)}{f'(x)} = 0\,.$$
A sufficient further assumption would be that $f$ is twice differentiable at $0$ with $f''(0) \neq 0$, for then we would have $f(x) \sim \frac{f''(0)}{2}x^2$ and $f'(x) \sim f''(0)x$. One can construct more complicated weaker assumptions that work.
Without further assumptions, a counterexample is given by
$$f'(x) = \begin{cases} \qquad 0 &\text{if } x = 0, \\ e^{-1/x^2} + \sqrt{\lvert x\rvert}\sin^2 \frac{1}{x} &\text{if } x \neq 0\,. \end{cases}$$
Then $f'(x) \geqslant e^{-1/x^2} > 0$ for all $x \neq 0$ and $f'\bigl(\frac{1}{k\pi}\bigr) = e^{-(\pi k)^2}$ for $k \in \mathbb{N}\setminus \{0\}$. On the other hand, for $0 < x < 1/\pi$ we have
\begin{align}
f(x) &= \int_0^x f'(t)\,dt \\
&> \int_0^x \sqrt{t} \sin^2 \frac{1}{t}\,dt \\
&= \int_{x^{-1}}^{\infty} \frac{\sin^2 u}{u^{5/2}}\,du \tag{$u = 1/t$} \\
&> \int_{x^{-1}}^{x^{-1} + 2\pi} \frac{\sin^2 u}{u^{5/2}}\,du \\
&> \bigl(x^{-1} + 2\pi\bigr)^{-5/2} \int_{x^{-1}}^{x^{-1} + 2\pi} \sin^2 u\,du \\
&= \frac{\pi}{\bigl(x^{-1} + 2\pi\bigr)^{5/2}} \\
&> \frac{\pi}{9\sqrt{3}}x^{5/2}\,.
\end{align}
Thus
$$\frac{f\bigl(\frac{1}{k\pi}\bigr)}{f'\bigl(\frac{1}{k\pi}\bigr)} > c\cdot \frac{e^{(k\pi)^2}}{(k\pi)^{5/2}} \xrightarrow{k \to +\infty} +\infty\,.$$
But since $f'(x)$ can be as large as $\sqrt{x}$ and $f(x) \leqslant b\cdot x^{3/2}$ for $x > 0$, we have
$$\liminf_{x \to 0^+} \frac{f(x)}{f'(x)} = 0\,.$$
By continuity and since $f$ is odd, every real number is a cluster point of $\frac{f(x)}{f'(x)}$ as $x \to 0$.
Concerning your comment

Do you think the following statement is true: If the limit exists, it is either $0$ or $\infty$, but not some other real number.

This is indeed the case. Under the assumptions, $\frac{f(x)}{f'(x)} > 0$ if $x > 0$ is sufficiently small. By perhaps replacing $f$ with $-f$ we can assume that $f'(x) > 0$ for $x > 0$. We always have
$$\liminf_{x \to 0^+} \frac{f(x)}{f'(x)} = 0\,.$$
For $\varepsilon > 0$ small enough that $f'$ has no zero in $(0,\varepsilon]$, let $x_\varepsilon \in (0,\varepsilon]$ be such that
$$f'(x_{\varepsilon}) = \max \{ f'(t) : 0 \leqslant t \leqslant \varepsilon\}\,.$$
Then
$$f(x_{\varepsilon}) = \int_0^{x_{\varepsilon}} f'(t)\,dt \leqslant x_{\varepsilon}\cdot f'(x_{\varepsilon}) \leqslant \varepsilon\cdot f'(x_{\varepsilon})\,.$$
Hence either the limit doesn't exist, or it is $0$.
