TITLE: Is the limit of $f/f'$ necessarily $0$?
QUESTION [8 upvotes]: Let $f:[a,b]\to\mathbb{R}$ be differentiable on $[a,b]$, with $\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}f^\prime(x)=0$, and $f^\prime(x)\ne 0$ in a neighborhood of $a$. Is it necessarily true that 
$$\lim_{x\to a}\frac{f(x)}{f^\prime(x)}=0$$
It doesn't seem necessarily clear to me that this limit should always exist, and that there's not some pathological function for which the limit is nonzero.

REPLY [2 votes]: Other answers have been approaching a construction of a counterexample, and I'm convinced with some more work one should be possible.  What you can say, however, is that
$$\liminf_{x\to a^+} \frac{f(x)}{f'(x)} = 0.$$
First, notice that by Rolle's theorem, $f$ can't have zeros on any interval $(a, a + \delta)$ where $f'$ has no zeros; so $\frac{f}{f'}$ cannot change signs, and from here, it is straightforward to conclude that $\frac{f(x)}{f'(x)} > 0$ for $x \in (a, a + \delta)$.
Now, suppose that $\liminf_{x\to a^+} \frac{f(x)}{f'(x)} = \epsilon > 0$.  Then for some $\delta' > 0$, we have $\frac{f(x)}{f'(x)} > \frac{\epsilon}{2}$ for $x \in (a, a + \delta')$.  It follows that $$\frac{d}{dx}(\log |f(x)|) = \frac{f'(x)}{f(x)} < \frac{2}{\epsilon}$$ for $x \in (a, a + \delta')$.  Therefore, $$\log |f(x)| > \log |f(a + \delta')| - \frac{2}{\epsilon} (a + \delta' - x)$$ for each $x \in (a, a + \delta')$.  This contradicts the assumption that $f(a) = 0$ and $f$ is differentiable and therefore continuous, so $\log |f(x)| \to -\infty$ as $x \to a^+$.