# How do I find the point(s) at which a given piecewise function is discontinuous?

Sep 26, 2015

This is essentially the same problem as determining the points at which any function is discontinuous, except the that points where the definition of the function changes are obvious candidates.

#### Explanation:

A function $f$ is continuous at a point $a$ if both of the one sided limits ${\lim}_{x \to a -} f \left(x\right)$ and ${\lim}_{x \to a +} f \left(x\right)$ exist and are equal to $f \left(a\right)$.

Consider the function (one of my favourites) defined as follows:

$g \left(x\right) = \left\{\left(0 , \text{if " x " is irrational"), (1/q, "if " x = p/q " in lowest terms " p ", } q \in \mathbb{Z} \mathmr{and} q > 0\right)\right.$

This is continuous at every irrational number and discontinuous at every rational number.

Next define the piecewise function $f$ as follows:

$f \left(x\right) = \left\{\begin{matrix}0 & x \le 0 \\ g \left(x\right) & x > 0\end{matrix}\right.$

The problem of finding where $f$ is discontinuous splits into cases, according to intervals.

When $a < 0$, ${\lim}_{x \to a} f \left(x\right) = {\lim}_{x \to a} 0 = 0 = f \left(a\right)$.

When $a > 0$, $f$ is continuous at $a$ if and only if $g$ is continuous at $a$. That is, if and only if $a$ is irrational.

At $a = 0$, we have:

${\lim}_{x \to a -} f \left(x\right) = 0$

We also find:

${\lim}_{x \to a +} f \left(x\right) = {\lim}_{x \to a +} g \left(x\right) = 0$

since if $0 < x < \epsilon$ then $g \left(x\right) < \epsilon$

Since $f \left(0\right) = 0$ too, $f$ is continuous at $0$.