# differentiable functions are continuous

###### Proposition 1.

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

 $\displaystyle\lim_{y\to x}f(x)-f(y)$ $\displaystyle=$ $\displaystyle\lim_{y\to x}\frac{f(x)-f(y)}{x-y}(x-y)$ $\displaystyle=$ $\displaystyle\lim_{y\to x}\frac{f(x)-f(y)}{x-y}\ \lim_{y\to x}(x-y)$ $\displaystyle=$ $\displaystyle f^{\prime}(x)\ 0$ $\displaystyle=$ $\displaystyle 0,$

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 $x\in I$. ∎

Title differentiable functions are continuous DifferentiableFunctionsAreContinuous 2013-03-22 14:35:27 2013-03-22 14:35:27 matte (1858) matte (1858) 8 matte (1858) Theorem msc 57R35 msc 26A24 DifferentiableFunctionsAreContinuous2 LimitsOfNaturalLogarithm