Why $f \in C^1$ is not necessary for a certain situation in Mean Value Theorem? Consider $f:\mathbb{R} \to \mathbb{R}$ differentiable, where the Mean Value Theorem is applied in the interval $(0,x)$.
Then there is $c_x \in (0,x)$ such that $f'(c_x) = \dfrac{f(x)-f(0)}{x}$
Then I want to take the limit as $ x \to 0$ on both sides:
$\displaystyle\lim_{x \to 0}f'(c_x) = \lim_{x \to 0}\dfrac{f(x)-f(0)}{x}$
Now there is my doubt. This is supposed to be $f'(0)$, but to obtain this am I not using the continuity of the derivative, since I am doing $\displaystyle \lim_{x \to 0}f'(c_x) = f'(\lim_{x\to0}c_x)=f'(0)$?

No, in this case you obtain $f'(0)$ because of
$$
f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x}
$$
by the definition of the derivative, and no arguments for the continuity
of $f'$ are needed.
