Suppose that $h$ is continuous on $[a,b]$, differentiable on $(a,b)$, and that $c \in (a,b)$. Suppose also that $\lim_{x \to c} h'(x)$ exists. Prove that $h'$ is continuous at $c$.
I have found this answer here Prove derivative is continuous but can't follow the process. Any help/detailed explanations would be REALLY appreciated! I'd rather use L'Hospital Rule instead of the Mean Value Theorem, but I can't follow how to go from one step to the other... Thank you!
Well the mean value theorem is the simpler approach from my experience, although my calculus teacher proposed a similar problem in my lectures.
Looking at what the user wrote on the other question, you begin with the definition of a derivative. It is the area under the line for some $h$. Hence, $$f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}$$
When you were first learning calculus this is how you calculated the derivative before you learnt how to derive functions.
Now, we know that if a function is continuous between some $a$ and $b$, it must be continuous or else we would not be able to derive it over $[a,b]$. This means that through using L'Hopital's theorem, we can only derive if it is continuous, and there by giving us a finite value for the answer of what the limit of the function is at c. If we could not differentiate the function over $[a,b]$ then we could not use L'Hopital's to prove that it is continuous at point c. Remember, point c is somewhere between values of $a$ and $b$.
This now brings me to the final equation the user posted. $$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c}f'(x)$$ This is the literal definition of L'Hopitals rule. Because you already know the property of a derivative as I stated above, this is just another manner of writing the proof that $h'$ is continuous at $c$.
