I'm working on a question that states:

Consider a differentiable function $f:\mathbb{R}\to\mathbb{R}$. Furthermore consider the sequences $(x_n)_n$ and $(y_n)_n$ which both have limit $0$ and $x_n < y_n$ for all $n\in\mathbb{N}$. Is the following true?

$$\lim_{n\to\infty} \frac{f(y_n)-f(x_n)}{y_n-x_n} = f'(0). $$

If yes give a proof, if no give a counter example and find the extra requirements for it to be true.

I thought it was true and as a proof I did the following:

Given that $f$ is differentiable and that $x_n < y_n$, by the mean value theorem there exists a $c_n \in (x_n,y_n)$ such that

$$ \frac{f(y_n)-f(x_n)}{y_n-x_n} = f'(c_n). $$

As $x_n$ and $y_n$ both converge to $0$ we have that $c_n$ will also converge to $0$. Therefore, by taking the limit for $n$ to infinity we find that

$$\lim_{n\to\infty} \frac{f(y_n)-f(x_n)}{y_n-x_n} = \lim_{n\to\infty}f'(c_n) = f'(0). $$

At this point I realised that I had used the continuity of $f'$ which was not given in the question. I am now wondering if this is a necessary condition for the problem or if it can be solved without using the continuity of $f'$. If it is a necessary condition what is a suitable counter example where $f'$ isn't continuous? Any help is appreciated.

  • $\begingroup$ You are correct that continuity of $f'$ is not needed. For counterexamples, search for "differentiable at one point" (here is one example); $f'$ may not even exist outside of zero. $\endgroup$
    – angryavian
    Aug 3, 2021 at 18:59
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    $\begingroup$ Sorry, I missed that condition in your post, so you are right that my counterexample does not apply. Still, I think one can come up with an example where $f'$ is not continuous at $0$; you can probably find one with some effort. In any case, the claim only requires differentiability at $0$; see the duplicate question marked by José Carlos Santos. $\endgroup$
    – angryavian
    Aug 3, 2021 at 19:21
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    $\begingroup$ There is a few typos: it has to be $[f(x_n) - f(y_n)] / (x_n-y_n)$. You have $[f(y_n) - f(x_n)]$ in the numerators so your limit is off by a minus sign compared to what you want. $\endgroup$
    – Winther
    Aug 5, 2021 at 9:39
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    $\begingroup$ You could try to find a counterexample with $f(x) = x^2\sin(1/x)$, which has a discontinuous derivative at $0$. I think $x_n = 1/n$, $y_n=e^{1/n}-1$ might work. $\endgroup$
    – Milten
    Aug 5, 2021 at 9:58
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    $\begingroup$ I though of using the suggested function $f(x)=x^2\sin(1/x)$ at $x \neq 0$ and $f(x)=0$ at $0$ with the sequences $x_n = 1/(2\pi n)$ and $y_n = 1/(2\pi (n+1/4)$. That way the limit tends to $2/\pi$ which is not equal to $f'(0)=0$. Is this correct? $\endgroup$
    – MathsBBB
    Aug 6, 2021 at 8:22

1 Answer 1


A counterexample has been given in the comment: $f(x) = x^2 \sin (1/x)$. Actually, any differentiable functions so that $f'(x)$ does not converge to $f'(0)$ as $x\to 0$ suffices: in this case, there is $\epsilon_0>0$ and a sequence $x_n \to 0$ so that $|f'(x_n) - f'(0)| >\epsilon_0$ for all $n$. For each $n$, one can find by the definition of $f'(x_n)$, $y_n\in (x_n, x_n+1/n)$ so that

$$\left| \frac{f(y_n)-f(x_n)}{y_n-x_n} - f'(x_n)\right| < \frac 12 \epsilon_0.$$ Then $y_n\to 0$, $x_n < y_n$. Also,

\begin{align} \epsilon_0 &< |f'(x_n) - f'(0)|\\ & <\left| \frac{f(y_n)-f(x_n)}{y_n-x_n} - f'(x_n)\right| +\left|\frac{f(y_n)-f(x_n)}{y_n-x_n} - f'(0)\right| \\ &< \frac 12 \epsilon_0 +\left|\frac{f(y_n)-f(x_n)}{y_n-x_n} - f'(0)\right| \end{align} and this implies $$\left| \frac{f(y_n)-f(x_n)}{y_n-x_n} - f'(0)\right|> \frac 12 \epsilon_0.$$

Thus $ \frac{f(y_n)-f(x_n)}{y_n-x_n}$ does not converge to $f'(0)$.

Together with what you did, the limit holds for all such sequences $(x_n), (y_n)$ if and only if $f'$ is continuous at $0$.


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