Is the limit of $f/f'$ necessarily $0$? 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.
 A: We can prove that $f'/f$ must be unbounded in any neighborhood of $a$.
First, we impose a fairly weak condition that $f'$ is integrable. By hypothesis $f' \neq 0$ in some interval $(a,b]$. Since derivatives have the intermediate value property (Darboux's theorem) we must have either $f'(x) > 0$ or $f'(x) < 0$ for all $x \in (a,b]$.
Since $f(x) \to 0$ as $x \to a+$, it follows that $f$ is increasing if $f' > 0$ with $f(x) > 0$ and $f'(x)/f(x) > 0$ for $x \in (a,b]$. On the other hand, if $f' < 0$ then $f$ is decreasing with  $f(x) < 0$ and $f'(x)/f(x)  > 0$ for $x \in (a,b].$
Since $f'/f$ is integrable we have
$$\int_{a+\delta}^b \frac{f'(x)}{f(x)} \,dx = \int_{f(a+\delta)}^{f(b)} \frac{dy}{y} = \log f(b) - \log f(a + \delta).$$
Thus, since $\log f(a + \delta) \to -\infty$ as $\delta \to 0+$ we have
$$\int_a^b \frac{f'(x)}{f(x)} \, dx = \lim_{\delta \to 0+} \int_{a+\delta}^b \frac{f'(x)}{f(x)} \,dx = +\infty,$$
and, thus,
$$\limsup_{x \to a+} \frac{f'(x)}{f(x)} = + \infty, \quad \liminf_{x \to a+} \frac{f(x)}{f'(x)} = 0$$
Another approach that does not require $f'$ to be integrable
For any points $y > x >a$ there exists $\xi_{x,y}$ between $x$ and $y$ such that
$$\log f(y) - \log f(x) = \frac{f'(\xi_{x,y})}{f(\xi_{x,y})}(y-x).$$
Thus,
$$\lim_{x \to a+} \frac{f'(\xi_{x,y})}{f(\xi_{x,y})} = [\log f(y) - \lim_{x \to a+} \log f(x)]/(y- a) = + \infty.$$
Hence, on any interval $(a,y]$ no matter how small we can find a sequence of points $(x_n)$ such that $f'(x_n)/f(x_n) \to + \infty.$ The mean value theorem is non-constructive, so we cannot determine that $x_n \to a$ or more generally that $\lim_{x \to a+} \xi_{x,y}= a.$  This shows, at least, that $f'/f$ must be unbounded in any neighborhood of $x=a$, and again
$$\liminf_{x \to a+} \frac{f(x)}{f'(x)} = 0$$
Also, the hypothesis that $f'(x) \to 0$ as $x \to a+$ was never used.
A: 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 $x \to a^+$.
A: I think this is  not necessarily true that the limit is $0$.
Indeed consider a function $f:[0, 1] \rightarrow \mathbb{R}$ such that $f(0)=0$, $f'(0)=0$ and for all $n \in \mathbb{N}$, 
$$f\left( \frac{1}{n} \right) = \frac{1}{n} \quad \text{and} \quad f'\left( \frac{1}{n} \right) = \frac{1}{n^2}$$
I think this is not difficult to show that such a function exists. But such a function would not satisfy your limit property.
A: I think it indeed should always be $0$. Look at the following line of reasoning (where $f(a):=0$):
$$ \lim_{x\to a} \frac{f(x)}{f'(x)} = \lim_{x\to a} \left(\frac{f(x) - f(a)}{x-a}\right)/{f'(x)}\cdot (x-a) = \lim_{x\to a} 1\cdot(x-a) = 0.  $$
It only needs to be verified that
$$ \lim_{x\to a}\left(\frac{f(x) - f(a)}{x-a}\right)/{f'(x)} = 1.$$
I am pretty sure the latter holds but I am unsure about the proof.  
EDIT: Without loss of generality, assume that $x>a$. Using the mean value theorem, we derive that for all $x$ there exists a $a<c_x<x$ such that $(f(x)-f(a))/(x-a) = f'(c_x)$. It follows that
$$ \lim_{x\to a}\left(\frac{f(x) - f(a)}{x-a}\right)/{f'(x)} = \lim_{x\to a} f'(c_x)/f'(x) = 1 $$ 
where the last step is valid if $f'(x)$ is continuous at $a$, since $f'(x)\to 0$ as $x\to a$, this must hold true. Because if it is not continuous, this limit cannot exist.
Edit2 the last claim is not true as seen in the questions. If you want to find a counterexample, it is a good way to start by looking at functions for which the above statement does not hold.
A: The claim is not true.
Let $f'(x) = \frac1{n^2}$ for $x \in [\frac1{2n+1},\frac1{2n}]$, and let $f'(x) = \frac1{2n}$ for $x \in (\frac1{2n},\frac1{2n-1})$. Here $n \ge 1$.
Let $f(x) = \int_0^x f'(u)du$. One verifies that $$f\big(\frac1{2n}\big) \geq \sum_{k=n+1}^{\infty} \int_{\frac1{2k}}^{\frac1{2k-1}}f'(u)du = \sum_{k=n+1}^{\infty}\bigg[ \frac1{2k-1}-\frac1{2k}\bigg] \frac{1}{2k} \geq \sum_{k=n}^{\infty} \frac1{8k^3}\geq \frac{1}{16n^2}.$$ Consequently, $$\frac{f(\frac1{2n})}{f'(\frac1{2n})} \geq \frac1{16}.$$
