Flaw in proof that a differentiable function has continuous derivative Let f be a function differentiable on $(a,b)$ and continuous on $c\in(a,b)$. If $c+h \in (a,b)$ then by the mean value theorem $\frac{f(c+h)-f(c)}{h}=f'(c+\theta h)$ for $\theta \in [0,1]$. Let $h \xrightarrow{}0$, then $f'(c+\theta h) \xrightarrow{} f'(c)$ by the above.
My reasoning is the following: $\theta$ is a function of $h$, so in fact it is not true that  that $\theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.
 A: It is not true that $\theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $\theta$'s that work for a given $h.$
We only know $f'(y)\to f'(c)$ as $y\to c$ within the set of $y=x+\theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2\sin(1/x)$ shows. 
A: If you look at $f(x)=x^2 \sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2x\sin(1/x)-\cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h \in (0,1)$ there exists $\theta(h) \in (0,1)$ with 
$$h\sin(1/h)=2h\theta(h)\sin(1/(h\theta(h)))-\cos(1/(h\theta(h))).$$
Notice that once $h$ is small enough, $h\theta(h)$ is forced to stay relatively close to the zeros of $\cos(1/x)$, since 
$$|\cos(1/(h\theta(h)))|=|h\sin(1/h)-2h\theta(h)\sin(1/(h\theta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $\cos(1/x)$. If you instead look at something like $f' \left ( \frac{1}{\pi n} \right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
A: You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+\theta h) \to f'(c)$ as $h\to0$.
