Proposition 1   Suppose $I$ is an open interval on $\mathbb{R}$ , and $f\colon I\to \sC$ is differentiable at $x\in I$ . Then $f$ is continuous at $x$ . Further, if $f$ is differentiable on $I$ , then $f$ is continuous on $I$ .

Proof. Suppose $x\in I$ . Let us show that $f(y)\to f(x)$ , when $y\to x$ . First, if $y\in I$ is distinct to $x$ , then $$ f(x)-f(y) = \frac{f(x)-f(y)}{x-y} (x-y). $$ Thus, if $f'(x)$ is the derivative of $f$ at $x$ , we have \begin{eqnarray*} \lim_{y\to x} f(x)-f(y) &=& \lim_{y\to x} \frac{f(x)-f(y)}{x-y} (x-y) \\ &=& \lim_{y\to x} \frac{f(x)-f(y)}{x-y}\ \lim_{y\to x} (x-y) \\ &=& f'(x)\ 0 \\ &=& 0, \end{eqnarray*}where the second equality is justified since both limits on the second line exist. The second claim follows since $f$ is continuous on $I$ if and only if $f$ is continuous at $x$ for all $x\in I$ .