Differentiability at the limit of non-differentiable points

Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function. Suppose there exists a sequence $(x_n)_{n \ge 1} \subseteq \mathbb{R}$ such that:

1. there exists $x_\infty \in \mathbb{R}$ such that $x_n \to x_\infty$ as $n \to \infty$
2. $f$ is not differentiable at each point $x_n$

Do we know that $f$ cannot then be differentiable at $x_\infty$? I suspect it is impossible for $f$ to be differentiable at $x_\infty$ but I have not been able to prove this myself. I have tried approximating the Newton quotients at $x_\infty$ by those at $x_n$, making use of continuity, but to no avail. I have also not found this result anywhere online.

Hints towards the proof of this result or counter-examples would be greatly appreciated.

---

Of course, after posting, I realized that this question may be stated more succinctly. For $f \in C(\mathbb{R})$, is the set $$A := \{x \in \mathbb{R}\ |\ f'(x) \text{ does not exist}\}$$ closed?

Answer 1

Let $g$ be a continuous no-where differentiable function on $[0,1]$ and $f(x)=x[g(x)-g(0)]$. Then $f$ is not differentiable at each of the points $x_n=\frac 1 n$ but it is differentiable at $0$ with $f'(0)=0$.