Is it possible for a function to be continuously derivable over its entire open domain except for a removable discontinuity?

Question

For example, is there a function $f \in \mathcal{C}(\mathbb{R}) \cap \mathcal{C}^1(\mathbb{R} \setminus \{ 0 \})$ such that

$$ \exists \lim_{x \to 0} f'(x) = \lim_{x \to 0} \lim_{y \to x} \frac{f(y)-f(x)}{y-x} $$

but

$$ \nexists f'(0)=\lim _{x \to 0} \frac{f(x)-f(0)}{x} $$

Answer 1

As someone commented, it is impossible, since we can use L'Hôpital:

$$ \lim_{x \to 0} \frac{f(x)-f(0)}{x} = \lim_{x \to 0} f'(x) = f'(0) $$