If a (partial) derivative is continous near x, does that imply that the (partial) derivative at x is either continous or non-existent?

Question

Let $x \in \mathbb{R}^n$, and $A$ be a neighborhood of $x$. Let $f$ be a function with a (partial) derivative that exists and is continous on $A-\{x\}$. Does this imply that the (partial) derivative at $x$ either is continous or doesn´t exist?

Answer 1

No it does not. Well-known example: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined as $$f(x)=\left\{\begin{array}{cc}x^2\sin\left(\frac{1}{x}\right)&,x\neq 0\\ 0&,x=0\end{array}\right\}.$$

Then you can show

$$f'(x)=\left\{\begin{array}{cc}2x\sin\left(\frac{1}{x}\right)-\cos\left( 1 \over x\right)&,x\neq 0\\ 0&,x=0\end{array}\right\}.$$

So $f$ is differentiable on $\mathbb{R}$ and $f'$ is continuous on $\mathbb{R} \setminus \{0\}$ but $f'$ is not continuous at the origin.