# How do you find discontinuity of a piecewise function?

Oct 3, 2014

In most cases, we should look for a discontinuity at the point where a piecewise defined function changes its formula. You will have to take one-sided limits separately since different formulas will apply depending on from which side you are approaching the point. Here is an example.

Let us examine where $f$ has a discontinuity.

$f \left(x\right) = \left\{\begin{matrix}{x}^{2} \mathmr{if} x < 1 \\ x \mathmr{if} 1 \le x < 2 \\ 2 x - 1 \mathmr{if} 2 \le x\end{matrix}\right.$,

Notice that each piece is a polynomial function, so they are continuous by themselves.

Let us see if $f$ has a discontinuity $x = 1$.

${\lim}_{x \to {1}^{-}} f \left(x\right) = {\lim}_{x \to {1}^{-}} {x}^{2} = {\left(1\right)}^{2} = 1$

${\lim}_{x \to {1}^{+}} f \left(x\right) = {\lim}_{x \to {1}^{+}} x = 1$

Since both limits give 1, ${\lim}_{x \to 1} f \left(x\right) = 1$

$f \left(1\right) = 1$

Since ${\lim}_{x \to 1} f \left(x\right) = f \left(1\right)$, there is no discontinuity at $x = 1$.

Let us see if $f$ has a discontinuity at $x = 2$.

${\lim}_{x \to {2}^{-}} f \left(x\right) = {\lim}_{x \to {2}^{-}} x = 2$

${\lim}_{x \to {2}^{+}} f \left(x\right) = {\lim}_{x \to {2}^{+}} \left(2 x - 1\right) = 2 \left(2\right) - 1 = 3$

Since the limits above are different, ${\lim}_{x \to 2} f \left(x\right)$ does not exist.

Hence, there is a jump discontinuity at $x = 2$.

I hope that this was helpful.