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 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
Canonical name	DifferentiableFunctionsAreContinuous
Date of creation	2013-03-22 14:35:27
Last modified on	2013-03-22 14:35:27
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	8
Author	matte (1858)
Entry type	Theorem
Classification	msc 57R35
Classification	msc 26A24
Related topic	DifferentiableFunctionsAreContinuous2
Related topic	LimitsOfNaturalLogarithm