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?

  • $\begingroup$ what if $f(x) = 0 \text{ } \forall x \in [0, \delta)$? $\endgroup$ Commented Jan 20, 2016 at 19:43
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    $\begingroup$ @AdamFrancey Then the limit doesn't exist, since it is not well-defined. $\endgroup$
    – Dominik
    Commented Jan 20, 2016 at 19:46
  • $\begingroup$ The hypothesis is that the limit exists. How is $f(x)/f'(x)$ defined in that case? $\endgroup$
    – RandyF
    Commented Jan 20, 2016 at 19:47
  • $\begingroup$ Right, I missed that the limit already exists. $\endgroup$ Commented Jan 20, 2016 at 19:53

3 Answers 3


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

  • 2
    $\begingroup$ 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? $\endgroup$
    – Brian Tung
    Commented Jan 20, 2016 at 20:44
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    $\begingroup$ @BrianTung Yes. $\endgroup$
    – Dominik
    Commented Jan 20, 2016 at 20:48
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    $\begingroup$ @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?" :) $\endgroup$
    – psmears
    Commented Jan 20, 2016 at 22:03
  • $\begingroup$ @psmears: I've edited it. $\endgroup$
    – Dominik
    Commented Jan 20, 2016 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 $L =0$.

  • $\begingroup$ 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 $f(0) = 0$. $\endgroup$
    – Dominik
    Commented Jan 20, 2016 at 21:02
  • $\begingroup$ @Dominik: Thanks for pointing that out. $\endgroup$
    – RRL
    Commented Jan 20, 2016 at 21:16
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    $\begingroup$ If $f(x_n)$ is also zero then $f(x)/f'(x)$ may have a removable discontinuity at $x_n$. $\endgroup$ Commented Jan 20, 2016 at 22:59
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    $\begingroup$ 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. $\endgroup$
    – RRL
    Commented Jan 20, 2016 at 23:24
  • 1
    $\begingroup$ 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 ... $\endgroup$
    – RRL
    Commented Jan 20, 2016 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.


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