Removable Discontinuities (College Board AP® Calculus AB)

Study Guide

Written by: Roger B

Reviewed by: Dan Finlay

A removable discontinuity is a discontinuity in a function that can be 'removed'
- This is done in order to make the function continuous over an interval containing the discontinuity
Recall that a function $f$ is continuous at the point $x = c$ if
- $f (c)$ exists,
- $\lim_{x \to c} f (x)$ exists,
- and $\lim_{x \to c} f (x) = f (c)$
A removable discontinuity is a 'hole' in the function
- It is a point where:
  - $\lim_{x \to c} f (x)$ exists
  - but $f (c)$ doesn't exist

For example, $g (x) = \frac{x}{x}$ is not continuous at $x = 0$ because $g (0)$ doesn't exist
- However the discontinuity at $x = 0$ is removable because $\lim_{x \to c} g (x) = 1$
  - I.e. the limit does exist at the point of discontinuity

To remove a removable discontinuity the function needs to be redefined
- This is done by defining $f (c)$ to be equal to $\lim_{x \to c} f (x)$ at the point of discontinuity
- The function then 'ticks all the boxes' to be continuous at that point
For example, if we instead define the function $g$ by $g (x) = \{\begin{cases} \frac{x}{x}, x \neq 0 \\ 1, x = 0 \end{cases}$
- Then $g$ becomes continuous over all the real numbers, including $x = 0$
- We have removed the removable discontinuity!

Let $f$ be the function defined by $f (x) = \frac{x^{2} + 3 x}{x}$ .

(a) Explain why $f$ is not continuous at $x = 0$ .

Answer:

$f (0)$ is not defined, so the function can't be continuous at $x = 0$

$f (0) = \frac{{(0)}^{2} + 3 (0)}{(0)} = \frac{0}{0}$ which is not defined

$f (0)$ does not exist, therefore $f$ is not continuous at $x = 0$

(b) Explain how the discontinuity at $x = 0$ can be removed.

Answer:

First find the limit at $x = 0$

This can be done by factorising and simplifying

$\frac{x^{2} + 3 x}{x} = \frac{x (x + 3)}{x} = x + 3$

$(0) + 3 = 3$

$\lim_{x \to 0} f (x) = 3$

The limit exists, therefore the discontinuity at $x = 1$ is removable

Redefine the function so that $f (0) = \lim_{x \to 0} f (x)$

The discontinuity can be removed by redefining the function $f$ as $f (x) = \{\begin{cases} \frac{x^{2} + 3 x}{x}, x \neq 0 \\ 3, x = 0 \end{cases}$

Last updated: 6 August 2024

the (exam) results speak for themselves:

Did this page help you?