# Question #c66df

May 21, 2015

A function, $f \left(x\right)$ is discontinuous at a point $x = \hat{x}$
if

$f \left(\hat{x}\right)$ does not exist; but $f \left(\hat{x} \pm \epsilon\right)$ does exist for arbitrarily small values of $\epsilon > 0$

or

${\lim}_{h \rightarrow 0} f \left(\hat{x} \pm h\right) \ne f \left(\hat{x}\right)$

Some simple examples:
$f \left(x\right) = \frac{1}{x}$ is discontinuous at $x = 0$ since the function is not defined at the at point

$f \left(x\right) = \text{Integer} \left(x\right)$ is discontinuous since (for example)
$f \left(3 - h\right) = 2 \text{ for all } h > 0$
but
$f \left(3\right) = 3$