Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

$$f(x) = \begin{cases}x \ln x &\mbox{if }x>0 ,\\0 &\mbox{if } x=0, \end{cases},$$


$$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x \ln x }{1+ \ln x}=0.$$

Is there a simple proof or counterexample?

share|cite|improve this question
what if $f(x) = 0 \text{ } \forall x \in [0, \delta)$? – Adam Francey Jan 20 at 19:43
@AdamFrancey Then the limit doesn't exist, since it is not well-defined. – Dominik Jan 20 at 19:46
The hypothesis is that the limit exists. How is $f(x)/f'(x)$ defined in that case? – RandyF Jan 20 at 19:47
Right, I missed that the limit already exists. – Adam Francey Jan 20 at 19:53
up vote 8 down vote accepted

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 $\frac{1}{\infty} = 0$.

share|cite|improve this answer
When you say that "such a function does not exist", do you mean a function $f$ satisfying all of the conditions in the OP's first paragraph, for which the stated limit exists and is non-zero? – Brian Tung Jan 20 at 20:44
@BrianTung Yes. – Dominik Jan 20 at 20:48
@Dominik: It would be a good idea to edit your question to make that clearer - at the moment it's quite confusing, because the question asks "is it always the case that L=0?", not "Does a function exist for which L≠0?" :) – psmears Jan 20 at 22:03
@psmears: I've edited it. – Dominik Jan 20 at 22:13

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 $f(x) > 0$.


$$\lim_{x \to 0+}\frac{f(x)}{f'(x)} = L > 0.$$

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 $L =0$.

share|cite|improve this answer
One minor detail: You haven't covered the case $L < 0$. But this can easily be fixed, since then $f > 0$ has to be decreasing, which is a contradiction to $f(0) = 0$. – Dominik Jan 20 at 21:02
@Dominik: Thanks for pointing that out. – RRL Jan 20 at 21:16
If $f(x_n)$ is also zero then $f(x)/f'(x)$ may have a removable discontinuity at $x_n$. – Arin Chaudhuri Jan 20 at 22:59
So you are proposing that both $f(x_n) = 0$ and $f'(x_n) = 0$ at infinitely many points. Since $f(x_n) = f(x_{n+1}) = 0$ and $f$ is continuous and differentiable there are infinitely many intermediate points $x_n < \xi_n < x_{n+1}$ where $f'(\xi_n) = 0$ by Rolle's theorem. Now you need to force $f(\xi_n) = 0$. Now consider the ramifications of that in the context that the limit of the ratio exists. See if you can construct such a function that is not identically zero and you will find a counterexample. – RRL Jan 20 at 23:24
Between any two of infinitely many points where the function is zero there is another point where the function is zero, and the function is continuous and differentiable ... – RRL Jan 20 at 23:36

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.