Until I thought about this, I would have confidently answered yes to my question.

Let $f:\mathbb{R\to\mathbb{R}}$ be differentiable at $x_{0},$ and continuously differentiable on some open neighborhood $\mathcal{N}$ of $x_{o}$. It follows that $f$ is continuous at $x_{o}.$ But, since $\mathcal{N}$ is open, $x_{o}$ is not necessarily in $\mathcal{N}$. So is it possible that

$$\lim_{\delta \to 0} f^{\prime}\left(x_{o}+\delta\right)\ne f^{\prime}\left(x_{o}\right)?$$

For $\Delta x\ne0$ and $x_{o}+\delta+\Delta x\in\mathcal{N},$ the continuity of $f$ gives us,

$$\underset{\delta\to\mathfrak{0}}{\text{Lim}}\frac{\mathit{f}\left(x_{o}+\delta+\Delta x\right)-f\left(x_{o}+\delta\right)}{\Delta x}=\frac{\mathit{f}\left(x_{o}+\Delta x\right)-f\left(x_{o}\right)}{\Delta x},$$

and therefore

$$\underset{\Delta x\to\mathfrak{0}}{\text{Lim}}\left[\underset{\delta\to\mathfrak{0}}{\text{Lim}}\frac{\mathit{f}\left(x_{o}+\delta+\Delta x\right)-f\left(x_{o}+\delta\right)}{\Delta x}\right]=f^{\prime}\left(x_{o}\right).$$

We also have

$$\underset{\Delta x\to\mathfrak{0}}{\text{Lim}}\frac{f\left(x_{o}+\delta+\Delta x\right)-f\left(x_{o}+\delta\right)}{\Delta x}=f^{\prime}\left(x_{o}+\delta\right).$$

But can we conclude $\underset{\delta \to\mathfrak{0}}{\text{Lim}}\mathit{f}^{\prime}\left(x_{o}+\delta\right)=\mathit{f}^{\prime}\left(x_{o}\right)?$ That is, for all $\epsilon>0$ can we find $\delta$ and $\Delta x$ such that $x_{o}+\delta+\Delta x\in\mathcal{N}$ and

$$\epsilon>\left|\frac{f\left(x_{o}+\delta+\Delta x\right)-f\left(x_{o}+\delta\right)}{\Delta x}-\mathit{f}^{\prime}\left(x_{o}\right)\right|?$$

Another way of stating the question is; can we reverse the order of the limit operation so that

$$\underset{\Delta x\to\mathfrak{0}}{\text{Lim}}\left[\underset{\delta\to\mathfrak{0}}{\text{Lim}}\frac{\Delta\mathit{f}_{x_{o}}\left(\delta+\Delta x\right)}{\Delta x}\right]=\underset{\delta\to\mathfrak{0}}{\text{Lim}}\left[\underset{\Delta x\to\mathfrak{0}}{\text{Lim}}\frac{\Delta\mathit{f}_{x_{o}}\left(\delta+\Delta x\right)}{\Delta x}\right]?$$

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    $\begingroup$ I don't understand the question at all. An open neighboghood of $x_0$ surely contains $x_0$. You probably mean a deleted neighborhood of $x_0$. $\endgroup$ Mar 4, 2021 at 5:48
  • $\begingroup$ Based on the title, isn't the "usual" example $f(x)=x^2\sin(1/x)$ exactly giving that? math.stackexchange.com/questions/1391544/… (I am assuming by neighborhood you mean deleted neighborhood, otherwise the question is vacuous) [Edit: "deleted neighborhood", "pointed neighborhood" was a bad French translation) $\endgroup$
    – Clement C.
    Mar 4, 2021 at 5:53
  • $\begingroup$ @ClementC. That's really the issue. What is meant by "neighborhood". See scribd.com/read/282634061/… Theorem III 3.3. I will post a question about that, presently. $\endgroup$ Mar 4, 2021 at 7:05
  • $\begingroup$ @KaviRamaMurthy I do not take "open neighborhood $\mathcal{N}$ of x" to mean $x\in\mathcal{N}$. For the very reason at issue in my question. I took real analysis 26 years ago, and haven't set foot in a classroom in 25 years. So, even though I didn't recall the details, I sensed there was an issue; hence the question. $\endgroup$ Mar 4, 2021 at 7:55
  • $\begingroup$ No problem. The standard terminology is deleted neighorhood but it is not difficult to guess what your question really says. $\endgroup$ Mar 4, 2021 at 7:58

1 Answer 1


If your $\mathcal N$ is deleted neighborhhod of $x_0$ then there is a well known counter-example: $f(x)=x^{2} \sin (\frac 1 x)$ for $x \neq 0$ and $f(0)=0$. In this case $\lim_{x \to 0} f'(x)$ does not even exist.


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