A function is continuously differentiable in an open neighborhood of $x_o$ and differentiable at $x_o$, is it continuously differentiable at $x_o$?

Question

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

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

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.