What is required in order to say a continuous function is differentiable at a point? If I know a function is continuous on $[0, 1]$, and that $\lim_{h \in \mathbb Q \to 0} {{f(c+h)-f(c)}\over h }$ exists, can I say $f'(c)$ exists?
I feel like this is an obvious yes, since by definition of derivative, $f'(c) = \lim_{h \to 0} {{f(c+h)-f(c)}\over h }$.  Hence if that limit exists and is finite, the function at $c$ is differentiable, and so $f(c)$ exists.
Does the fact that $h \in \mathbb Q$ change anything? 
And if so, where does $f(x)$ being continuous on $[0,1]$ come in?
 A: To answer your last question, $f$ being continuous at a point is merely a consequence of it being differentiable.  The reverse is not true.
Edit: Oh, I see.  
It does not hold that just because $\lim_{h \in \mathbb{Q}: h \to 0} \frac{f(c+h) - f(c)}{h}$ exists that $f$ is differentiable at a point $c$.  Consider the indicator function of the rationals defined by $\chi_{\mathbb{Q}}(x) = 1$ if $x \in \mathbb{Q}$ and 0 otherwise.  Then we have that $\chi_{\mathbb{Q}}(h) - \chi_{\mathbb{Q}}(0) = 0$ for every rational $h$, but the function is clearly not differentiable at $0$, since it's not even continuous.  
We'll show however that if $f$ is continuous and the above limit exists, then $f$ is differentiable at $c$.  Let $A = \lim_{h \in \mathbb{Q}: h \to 0} \frac{f(c +h) - f(c)}{h}$.  Fix an $\epsilon > 0 $.  Find a $\delta > 0$ so that if $h$ is a nonzero rational which satisfies $|h| < \delta$ then $|\frac{f(c+h) - f(c) - hA}{h}| < \epsilon$.  Let $s$ be any nonzero real satisfying $|s| < \delta$.  Find a sequence $\{h_n\}_{n=1}^{\infty}$ of nonzero rationals so that $h_n \to s$ and $|h_n| < \delta$ for every $n$.  By continuity of $f$, we have $|\frac{f(c+s) - f(c) - sA}{s}| = \lim_{n\to \infty} |\frac{f(c+h_n)-f(c) -h_nA}{h_n}| \le \epsilon$ .
