Introduction

Welcome to the FiveHive article for AP Calculus Unit 1.10: Exploring Types of Discontinuities!

Now is the part of the unit where we shift gears. The first part of the unit was talking about limits and how to evaluate them. Now we will talk more about continuity, but we will still use limits. 

In this article, we will talk about the different types of discontinuities at a point. If a function is not discontinuous at that point, it is continuous at that point. A more formal definition of continuity will be given in the next article, as we will only discuss the different types of discontinuities. 

Discontinuities

There are three types of discontinuities that will be discussed in this article: removable, jump, and infinite.

A removable discontinuity (also called a point discontinuity) is when the limit of the function at a value exists, but the function is not defined at that value. 

For example, $f(x)=\frac{x^2-36}{x-6}$ has a removable discontinuity at $x=6$ because the limit as $x$ approaches $6$ of $f(x)$ is $12$, but $f(6)$ is undefined. 

Since the limit still exists and function is defined everywhere else $(6,12)$ is called a hole in the graph.

A jump discontinuity is when the function “jumps” $y$-values (or whatever the dependent variable of the function is). 

This happens when the left side limit of the function does not equal the right side limit of the function.  

The piecewise defined function, $h(x)=\{\begin{array}{ll}{x}^2-5x& \text{if }x\le 1\\ -3x+4& \text{if }x>1\end{array}$ has a jump discontinuity at $x=1$. 

The left side limit is $-4$, which does not equal the right side limit of $1$. 

An infinite discontinuity is when the function is discontinuous due to a vertical asymptote. 

The most simple example is the function $g(x)=x^{-1}$ which has an infinite discontinuity at $x=0$ due to its vertical asymptote.

Trig functions like $\csc x$ are discontinuous at every integer multiple of $\pi$ due to the repeated vertical asymptotes.

It is possible for a function to be discontinuous on more than just a few points or a pattern on its domain. The Dirichlet function $f(x)=\{\begin{array}{ll}1& \text{if }x\text{ is rational}\\ 0& \text{if }x\text{ is irrational}\end{array}$ is discontinuous everywhere because it is impossible to take a limit of it as all the ones and zeroes are surrounded by the other making every limit nonexistent.

Practice

Congratulations, you have made it through 10 topics of AP Calculus! Now, it is time for some practice!