# Classifying Topics of Discontinuity (removable vs. non-removable)

## Key Questions

• 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.

• $f \left(x\right)$ has a removable discontinuity at $x = a$ when ${\lim}_{x \to a} f \left(x\right)$ EXISTS; however, ${\lim}_{x \to a} f \left(a\right) \ne f \left(a\right)$. A removable discontinuity looks like a single point hole in the graph, so it is "removable" by redefining $f \left(a\right)$ equal to the limit value to fill in the hole.

• Recall that a function $f \left(x\right)$ is continuous at $a$ if

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

which can be divided into three conditions:

C1: ${\lim}_{x \to a} f \left(x\right)$ exists.
C2: $f \left(a\right)$ is defined.
C3: C1 = C2

A removable discontinuity occurs when C1 is satisfied, but at least one of C2 or C3 is violated. For example, $f \left(x\right) = \frac{{x}^{2} - 1}{x - 1}$ has a removable discontinuity at $x = 1$ since

lim_{x to 1}{x^2-1}/{x-1} =lim_{x to 1}{(x+1)(x-1)}/{x-1} =lim_{x to 1}(x+1)=2,

but $f \left(1\right)$ is undefined.

• ${\lim}_{x \to {a}^{-}} f \left(x\right) , {\lim}_{x \to {a}^{+}} f \left(x\right)$ are finite and ${\lim}_{x \to {a}^{-}} f \left(x\right) \ne {\lim}_{x \to {a}^{+}} f \left(x\right)$. So it occurs when the left and right limit at $a$ do not match, then we say $f \left(x\right)$ has a jump discontinuity at $a$.

This should not be confused with a point discontinuity where:

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

which means

${\lim}_{x \to a} f \left(x\right)$ exists

and:

${\lim}_{x \to a} f \left(x\right) \ne f \left(a\right)$

It could be the case that $f \left(a\right)$ is finite or simply DNE.