# How do you find asymptotic discontinuity?

Oct 5, 2014

If a function $f \left(x\right)$ has a vertical asymptote at $a$, then it has a asymptotic (infinite) discontinuity at $a$. In order to find asymptotic discontinuities, you would look for vertical asymptotes. Let us look at the following example.

$f \left(x\right) = \frac{x + 1}{\left(x + 1\right) \left(x - 2\right)}$

In order to have a vertical asymptote, the function has to display "blowing up" or "blowing down" behaviors. In the case of a rational function like $f \left(x\right)$ here, it display such behaviors when the denominator becomes zero.

By setting the denominator equal to zero,

$\left(x + 1\right) \left(x - 2\right) = 0 R i g h t a r r o w x = - 1 , 2$

Now, we have a couple of candidates to consider. Let us make sure that there is a vertical asymptote there.

Is $x = - 1$ a vertical asymptote?

${\lim}_{x \to - 1} \frac{\left(x + 1\right)}{\left(x + 1\right) \left(x - 2\right)}$

by cancelling out $\left(x + 1\right)$'s,

$= {\lim}_{x \to - 1} \frac{1}{x - 2} = \frac{1}{1 - 2} = - 1 \ne \pm \infty$,

which means that $x = - 1$ is NOT a vertical asymptote.

Is $x = 2$ a vertical asymptote?

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

by cancelling out $\left(x + 1\right)$'s,

$= {\lim}_{x \to {2}^{+}} \frac{1}{x - 2} = \frac{1}{0} ^ + = + \infty$,

which means that $x = 2$ IS a vertical asymptote.

Hence, $f$ has an asymptotic discontinuity at $x = 2$.

I hope that this was helpful.