Prove that $\lim_{x\to 0} \frac{f(x)}{f'(x)} = 0$ for $f\in C^1$ and $f(0)=0=f'(0)$ https://ift.tt/eA8V8J

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:

1. 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.
2. 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.
3. 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:

1. Is the statement I am trying to prove correct, or do I need further assumptions?
2. Is my proof strategy correct so far? Is there a simpler way?
3. Is one of the limits actually uniform? If so, can someone give me a hint on how to show it?

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