If a function $f$ is defined and differentiable in an open interval A, then it's also continuous in A. Since it's differentiable, for a random real number a that belongs to A:
$f'(a) = \lim_{x\to a}{f(x) - f(a)\over x - a} $
Since $f$ is continuous, $\lim_{x\to a}f(x) = f(a)$ so the limit $\lim_{x\to a}{f(x) - f(a)\over x - a}$ is ${0\over 0}$, so we can use L'Hospital's rule and say that $f'(a) = \lim_{x\to a}{f(x) - f(a)\over x - a} = \lim_{x\to a}{f'(x)}$
(pretty much the same thing if it's a closed interval)
So that means that $f'$ is continuous, which is wrong since it's just a random function and it's possible to make a derivative of a function non continuous. So where's the mistake?
You can not use l'Hospital here because you need to know about the existence of $$\lim_{x\to a} f'(x)$$ first. You are messing the order of implication of l'Hospital. To show that a differentiable function is continuous have a look at $$f(x)-f(a)= \frac{f(x) -f(a)}{x-a} \cdot (x-a)$$ and use the known limit results.