differentiable functions are continuous

Proposition 1.

Suppose $I$ is an open interval on $\mathrm{R}$,and $f\mathrm{:}I\mathrm{\to }\mathrm{C}$ is differentiable at $x\mathrm{\in }I$. 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 $f^{\prime }(x)$ is the derivative of $f$ at $x$, we have

$\underset{y\to x}{lim}f(x)-f(y)$$=$$\underset{y\to x}{lim}{\displaystyle \frac{f(x)-f(y)}{x-y}}(x-y)$

$=$$\underset{y\to x}{lim}{\displaystyle \frac{f(x)-f(y)}{x-y}}\underset{y\to x}{lim}(x-y)$

$=$$f^{\prime }(x)\mathrm{\hspace{0.25em}0}$

$=$$0,$

where the second equality is justified since both limits on the second lineexist. The second claim follows since $f$ is continuous on $I$ if and onlyif $f$ is continuous at $x$ for all $x\in I$.∎