Behavior of derivative near the zero of a function. Suppose the function $f:[0,\delta) \to \mathbb{R}$ is continuous, differentiable in $(0,\delta)$ and $f(0)=0$.
If the limit $\displaystyle \lim_{x \to 0+}\frac{f(x)}{f'(x)}= L$ exists, then is it always the case that $L = 0$.
This seems to be true both for functions with well-behaved derivatives such as $f(x) = x,$
$$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x}{1}=0,$$
as well as functions with "bad" derivatives such as 
$$f(x) = \begin{cases}x \ln x &\mbox{if }x>0 ,\\0  &\mbox{if } x=0,  \end{cases},$$
where
$$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x \ln x }{1+ \ln x}=0.$$
Is there a simple proof or counterexample?
 A: There is no function $f$ that satisfies your assumptions and also satisfies $\lim \limits_{x \to 0+} \frac{f(x)}{f'(x)} \ne 0$.
Let's assume otherwise. If every interval $(0, \epsilon)$ contains a root of $f$, then the limit can only be $0$ [note that we assume that the limit exists]; Therefore we can assume that $f$ has no roots. Wlog $f > 0$ holds on $(0, \delta)$. Now we can consider $g(x) := \ln(f(x))$. Then $g$ is differentiable on $(0, \delta)$ and satisfies $\lim \limits_{x \to 0+} g(x) = -\infty$ and $\lim \limits_{x \to 0+} g'(x) = \frac{1}{L} < \infty$. 
This means that on some interval $(0, \epsilon)$ the inequality $|g'(x)| \le \frac{1}{|L|} + 1$ holds. Now the mean-value theorem implies that for every $x \in (0, \epsilon)$ the inequaliy $\left|\frac{g(x) - g(\epsilon)}{x - \epsilon}\right| \le \frac{1}{|L|} + 1$ holds, i.e. $g(x) \ge -(\frac{1}{L} + 1)(\epsilon - x) + g(\epsilon)$, which is a contradiction to $\lim \limits_{x \to 0+} g(x) = -\infty$.
Note that this argumentation also works for $L = \infty$ if we set $\frac{1}{\infty} = 0$.
A: There is an interval $(0,\delta')$ where $f'(x) \neq 0$. Otherwise $0$ is a limit point of zeros of $f'(x)$ and the limit could not exist. By the mean value theorem, we have for each $x \in (0,\delta')$ a number $\theta$ between $0$ and $x$ such that 
$$f(x) = f'(\theta)x \neq 0.$$
Hence, $f(x) > 0$ or $f(x) < 0.$  Assume WLOG that $f(x) > 0$.
Suppose 
$$\lim_{x \to 0+}\frac{f(x)}{f'(x)} = L > 0.$$
There exists $0 < \delta'' < \delta'$ such that for $0 < x < \delta''$ we have
$$\frac{L}{2} < \frac{f(x)}{f'(x)} < \frac{3L}{2}.$$
Also $L,f(x) > 0 \implies f'(x) > 0$ implies $f$ is increasing.
Hence, for all $x \in (0,\delta'')$
$$\frac{f(x)}{f'(x)} = \frac{x f'(\theta)}{f'(x)} = x \frac{f(\theta)}{f(x)} \frac{f'(\theta)}{f(\theta)} \frac{f(x)}{f'(x)} < x (1)\frac{2}{L}\frac{3L}{2} = 3x \to 0.$$
Therefore, we cannot have $L>0$ and can have only $L =0$.
A: Since $\dfrac{f(x)}{f'(x)} \to L$ as $x \to 0^{+}$ it follows that $f'(x) \neq 0$ as $x \to 0^{+}$ and by Darboux Theorem (or simply intermediate value property of derivatives) it implies that $f'(x)$ is of constant sign as $x \to 0^{+}$. Let's assume that $f'(x) > 0$ so that $f(x) > f(0) = 0$ as $x \to 0$. It follows that $L \geq 0$.
Let us suppose that $L > 0$ then we know that $$\frac{f(x)}{f'(x)} > \frac{L}{2}\tag{1}$$ as $x \to 0^{+}$. We now consider the function $g(x) = e^{-2x/L}f(x)$. Then $$g'(x) = e^{-2x/L}f'(x)\left\{1 - \frac{2}{L}\cdot\frac{f(x)}{f'(x)}\right\} < 0\tag{2}$$ as $x \to 0^{+}$. Thus we see that $g(x)$ is a strictly decreasing function of $x$ in some interval of the form $[0, \delta)$ and $g(0) = 0$ implies that $g(x) < 0$ in the interval $(0, \delta)$. But clearly since $f(x) > 0$ in this interval, it implies that $g(x) > 0$. This contradiction proves that $L = 0$.
The case when $f'(x) < 0$ as $x \to 0^{+}$ can also be covered in exactly the same manner.

Note: The equation $(1)$ above reminds me of the famous question dealing with $f'(x) \leq cf(x)$ (see problem 2, also see this question on MSE) and I used the same technique here.
