= Solution
{c}
Choose $h_n=\min\{1/n,(b-c)/2\}>0$. Then $h_n\to0$ and $[c,c+h_n]\subset(a,b)$. The <mean value theorem> states that a <continuous function> on a <closed interval>, <differentiable> in its interior, has an interior <derivative> equal to its endpoint <difference quotient>. It gives $x_n\in(c,c+h_n)$ with
$$
f'(x_n)=\frac{f(c+h_n)-f(c)}{h_n}.
$$
Here $x_n\ne c$, $x_n\to c$, and the right-hand side tends to $f'(c)$ by <differentiability> at $c$. If $L=\lim_{x\to c}f'(x)$ exists, the same values $f'(x_n)$ must tend to $L$. Uniqueness of the <limit of a sequence> gives
$$
\boxed{\lim_{x\to c}f'(x)=f'(c).}
$$
Thus a <derivative> cannot have a <removable discontinuity> at a point where the original <function> is <differentiable>. The proof also excludes an infinite <limit of a function> there, since the displayed <difference quotients> have a finite <limit of a sequence>.
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