1.13 - Removing Discontinuities

Introduction

We’ve covered the different types of discontinuities. Now we’ll focus on removable discontinuities (holes) and how to remove them so the function becomes continuous.

Key idea: removing a discontinuity means filling in the hole with the right function value, which makes the function continuous at that point.

What is a Removable Discontinuity?

A removable discontinuity (hole) happens when

\lim_{x \to c} f(x) \text{ exists, but } f(c) \neq \lim_{x \to c} f(x)

or when

f(c)

is undefined entirely.

In plain terms: the function approaches a specific value from both sides, but either there’s an open circle (hole) at that point, or the function is defined at a different value than where the limit is pointing.

This discontinuity can be removed by defining

f(2) = 3

, which fills in the hole.

Finding the Hole Coordinates

To remove a discontinuity, you need the $(x, y)$ coordinates of the hole.

Step 1: Find the $x$ -value of the hole.

Factor the function completely

Identify which factor cancels (that’s what creates the hole)

Set that factor equal to zero and solve for $x$

Step 2: Find the $y$ -value of the hole.

Take the limit as $x$ approaches the hole’s $x$ -value

That limit is the $y$ -coordinate of the hole

Step 3: State how to remove the discontinuity.

Define the function at that point: $f(c) =$ the limit value

Example 1: Basic Hole

Find the coordinates of the removable discontinuity and state how to remove it:

f(x) = \frac{x^2 - 16}{x - 4}

Step 1: Find the $x$ -value of the hole

Factor the numerator:

f(x) = \frac{(x - 4)(x + 4)}{x - 4}

The factor

(x - 4)

cancels, which creates a hole. Set the canceled factor equal to zero:

x - 4 = 0 \Rightarrow x = 4

Step 2: Find the $y$ -value of the hole

After canceling:

f(x) = x + 4 \quad \text{(for } x \neq 4\text{)}

Take the limit:

\lim_{x \to 4} f(x) = \lim_{x \to 4} (x + 4) = 4 + 4 = 8

Step 3: State the hole coordinates

The hole is at the point $(4, 8)$ .

How to remove it: define

f(4) = 8

. That fills in the hole and makes the function continuous.

Example 2: Difference of Squares

Find the removable discontinuity and how to remove it:

g(x) = \frac{x^2 - 25}{x + 5}

Factor the numerator (difference of squares):

g(x) = \frac{(x - 5)(x + 5)}{x + 5}

Find the $x$ -value: the factor

(x + 5)

cancels, so

x + 5 = 0 \Rightarrow x = -5

Find the $y$ -value: after canceling,

g(x) = x - 5

, so

\lim_{x \to -5} g(x) = -5 - 5 = -10

Answer: hole at $(-5, -10)$ . To remove the discontinuity, define

g(-5) = -10

Example 3: Given That the Function is Continuous

The function

h

is continuous for all real numbers. For

x \neq 3

h(x) = \frac{x^2 - x - 6}{x - 3}

Find

h(3)

Since we’re told

h

is continuous, there has to be a value at

x = 3

that fills in what would otherwise be a hole.

Factor the numerator:

h(x) = \frac{(x - 3)(x + 2)}{x - 3}

Cancel the common factor:

h(x) = x + 2 \quad \text{(for } x \neq 3\text{)}

Find the limit:

\lim_{x \to 3} h(x) = 3 + 2 = 5

Since

h

is continuous,

h(3) = 5

Answer:

h(3) = 5

Explanation: for the function to be continuous, the value at

x = 3

has to equal the limit as

x

approaches 3. That’s what removes the discontinuity.

Example 4: Piecewise Function with an Unknown Constant

The function

f

is continuous for all real numbers, where

f(x) = \begin{cases} \frac{x^2 + 2x - 8}{x - 2} & \text{if } x \neq 2 \\ k & \text{if } x = 2 \end{cases}

Find the value of

k

For

f

to be continuous at

x = 2

, the value

k

has to equal the limit as

x

approaches 2.

Factor:

\frac{x^2 + 2x - 8}{x - 2} = \frac{(x - 2)(x + 4)}{x - 2}

Cancel:

f(x) = x + 4 \quad \text{(for } x \neq 2\text{)}

Find the limit:

\lim_{x \to 2} f(x) = 2 + 4 = 6

For continuity,

k = 6

Answer:

k = 6

Verification: when

k = 6

, the function value at

x = 2

matches the limit, so

f

is continuous.

Example 5: Piecewise with Two Pieces

The function

g

is continuous everywhere, where

g(x) = \begin{cases} \frac{x^2 - 9}{x + 3} & \text{if } x \neq -3 \\ a & \text{if } x = -3 \end{cases}

Find the value of

a

that makes

g

continuous.

Step 1: Simplify the first piece.

\frac{x^2 - 9}{x + 3} = \frac{(x - 3)(x + 3)}{x + 3} = x - 3

Step 2: For continuity at

x = -3

, the two pieces have to meet:

a = \lim_{x \to -3} (x - 3)

Step 3: Evaluate.

a = -3 - 3 = -6

Answer:

a = -6

This makes

g(-3) = -6

, which matches the limit, so

g

is continuous.

Example 6: Multiple Factors

Find all removable discontinuities and state their coordinates:

f(x) = \frac{(x - 1)(x + 2)}{(x - 1)(x - 5)}

Identify the canceling factor:

(x - 1)

shows up in both the numerator and denominator, so it cancels.

$x$ -value of the hole:

x - 1 = 0 \Rightarrow x = 1

Simplified function:

f(x) = \frac{x + 2}{x - 5} \quad \text{(for } x \neq 1\text{)}

$y$ -value of the hole:

\lim_{x \to 1} f(x) = \frac{1 + 2}{1 - 5} = \frac{3}{-4} = -\frac{3}{4}

Answer: removable discontinuity (hole) at

\left(1, -\frac{3}{4}\right)

Note: there’s also a vertical asymptote at

x = 5

(non-removable), but that isn’t a hole.

Example 7: Different Expressions on Each Side

When a piecewise function uses a different expression on each side of a boundary, three values have to match for continuity there: the left expression’s value at the boundary, the right expression’s value at the boundary, and the function’s actual value at the boundary.

Find the value of

k

that makes

f

continuous at

x = 2

f(x) = \begin{cases} kx + 1 & \text{if } x < 2 \\ kx^2 - 3 & \text{if } x \geq 2 \end{cases}

Step 1: Left-hand limit (first piece):

\lim_{x \to 2^-} f(x) = k(2) + 1 = 2k + 1

Step 2: Right-hand limit and function value (second piece, since

x = 2

satisfies

x \geq 2

\lim_{x \to 2^+} f(x) = f(2) = k(2)^2 - 3 = 4k - 3

Step 3: Set the two sides equal and solve:

2k + 1 = 4k - 3 \Rightarrow 4 = 2k \Rightarrow k = 2

Answer:

k = 2

Verification: with

k = 2

, the left piece gives

2(2) + 1 = 5

, the right piece gives

2(2)^2 - 3 = 5

, and

f(2) = 5

. All three match.

When a Discontinuity Can’t Be Removed

Redefining one point only works when the limit exists. If the limit doesn’t exist, no choice of

f(c)

can fix it:

Jump: the one-sided limits are different, so no single value matches both. For example, $g(x) = x + 1$ for $x < 1$ and $g(x) = x^2 + 3$ for $x \geq 1$ approaches $2$ from the left and $4$ from the right.

Vertical asymptote: the function is unbounded, so there’s no finite value to fill in. For example, $\frac{1}{x - 2}$ at $x = 2$ .

These are the non-removable discontinuities.

Practice Problems

Problem 1

Find the coordinates of the hole:

f(x) = \frac{x^2 + 5x + 6}{x + 2}

Answer:

Factor:

\dfrac{(x+2)(x+3)}{x+2} = x + 3

Hole at

x = -2

. Limit:

-2 + 3 = 1

Hole at $(-2, 1)$

Problem 2

The function

g

is continuous. For

x \neq 4

g(x) = \frac{x^2 - 16}{x - 4}

Find

g(4)

Answer:

Factor:

\dfrac{(x-4)(x+4)}{x-4} = x + 4

g(4) = 4 + 4 = 8

Problem 3

Find

k

so that

h

is continuous:

h(x) = \begin{cases} \frac{x^2 - 7x + 12}{x - 3} & \text{if } x \neq 3 \\ k & \text{if } x = 3 \end{cases}

Answer:

Factor:

\dfrac{(x-3)(x-4)}{x-3} = x - 4

k = 3 - 4 = -1

Key Takeaways

A removable discontinuity is a hole you can fill in to make the function continuous

To find the $x$ -coordinate of a hole, identify and solve the canceling factor

To find the $y$ -coordinate of a hole, evaluate the limit after canceling

To remove the discontinuity, define the function value to equal the limit

When you’re told a function “is continuous,” the function value has to equal the limit at any potential hole

For piecewise functions, the value at the boundary has to equal the limit from the continuous piece

When the pieces use different expressions, set the left expression equal to the right expression at the boundary (and make sure $f(c)$ matches too)

Canceling factors create holes. Factors that don’t cancel create vertical asymptotes (non-removable)

Jumps and vertical asymptotes can’t be removed, because the limit doesn’t exist

The notation “ $f(x)$ for $x \neq c$ ” is a signal that there’s probably a hole at $x = c$

Always factor completely before identifying holes

After finding the hole coordinates, state “define $f(c) = L$ ,” where $L$ is the limit value

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