Define an open set $A \subset \mathbb{R}$ and let $f:A \to \mathbb{R}$, where $A^1$ and $A^*$ denote the set of points at which $f$ is differentiable and strongly differentiable, respectively. Prove that if $a \in A^*$ is a limit point of $A^*$, then $$\lim_{\substack{x \to a \\ x \in A^*}} f^*(x)=\lim_{\substack{x \to a \\ x \in A^1}} f'(x)=f^*(a)=f'(a).$$ I have already proven that if a function is strongly differentiable at a point, it is also classically differentiable at that point. I have also proven that the converse is not always true. Thus, we can conclude that $$A^* \subseteq A^1 \subseteq A.$$ However, beyond this I am unsure how to proceed. Based on the above conclusion, I considered the idea that if we have a sequence converging to a point, then it must have a subsequence that also converges to that same point. Assuming that there is a sequence in $A$ that converges to $a$, it doesn't seem like a stretch to say that there is some sequence in $A^1$ that does as well, and then one in $A^*$ too.
I am not certain if this is the right way to be thinking about this, and I also don't know how to incorporate the idea that $a \in A^*$ is a limit point of $A^*$. That is obviously very important as it is the supposition for the proof. I know that this means every $V_\delta(a)$ contains a point $b \neq a$ where $b \notin A^*$, but not sure how to use this knowledge.
Any help would be greatly appreciated.
Edit: For those unfamiliar with strong differentiability, here is a definition:
A real-valued function defined on an open set $A \subset \mathbb{R}$ is said to be strongly differentiable at $a \in A$ if $$\lim_{\substack{(x_1,x_2) \to (a,a) \\ x_1 \neq x_2}} \frac{f(x_1)-f(x_2)}{x_1-x_2}=f^*(a)$$ assuming that the limit exists and is finite. The number $f^*(a)$ is called the strong derivative of $f$ at $a$.