1.10 - Exploring Types of Discontinuities

Introduction to Continuity

Continuity is one of the core ideas in calculus. Knowing where functions are continuous and where they break is essential for analyzing how a function behaves.

Informal Definition: The Pencil Test

A function is continuous if you can draw its entire graph without lifting your pencil off the paper.

1.10 - Exploring Types of Discontinuities

This function is continuous. You never need to lift your pencil.

This function is discontinuous at

x = 3

. You have to lift your pencil to keep drawing.

Types of Discontinuities

There are three main types to know:

1. Removable Discontinuity (Hole)

A hole happens when a single point is missing from an otherwise continuous function.

Classification: Removable

Why removable? If you could fill in the hole with a single point, the function would become continuous.

Example: The graph above has a hole at

x = 2

. This discontinuity could be removed by defining the function value at that point.

Same type, different look: sometimes the hole is there but the function is defined somewhere else at that

x

, like a filled dot sitting above or below the hole. That’s still a removable discontinuity. The limit exists, it just doesn’t match the function value, so redefining that one point fixes it. (See Example 4 in 1.11.)

2. Infinite Discontinuity (Vertical Asymptote)

A vertical asymptote happens when the function heads toward infinity (positive or negative) as

x

approaches a specific value.

Classification: Non-removable

Why non-removable? You can’t fix this by filling in a point. The function shoots off to infinity.

Example: This function has a vertical asymptote at

x = -2

. It’s undefined there and can’t be made continuous.

3. Jump Discontinuity

A jump discontinuity happens when the function suddenly jumps from one value to another.

Classification: Non-removable

Why non-removable? The left-hand and right-hand limits are different, so filling in a single point won’t connect the two pieces.

Example: This function has a jump discontinuity at

x = 1

. The two sides don’t connect.

Note: This lesson focuses mostly on holes and vertical asymptotes. Jump discontinuities get covered in the next lesson.

Finding Discontinuities in Rational Functions

For rational functions (fractions), discontinuities happen where the denominator equals zero, because dividing by zero is undefined.

Strategy

Factor both numerator and denominator completely

Set the denominator equal to zero and solve for $x$

Determine the type:

If the factor cancels with the numerator, it’s a hole

If the factor does not cancel, it’s a vertical asymptote

Example 1: Identifying Holes and Vertical Asymptotes

Find all discontinuities of

f(x) = \frac{x^2 - 7x + 10}{x^2 + 4x - 12}

Step 1: Factor the numerator and denominator.

Numerator:

x^2 - 7x + 10 = (x - 2)(x - 5)

Denominator:

x^2 + 4x - 12 = (x - 2)(x + 6)

Step 2: Write the factored form.

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

Step 3: Find where the denominator equals zero.

$x - 2 = 0 \Rightarrow x = 2$

$x + 6 = 0 \Rightarrow x = -6$

Step 4: Determine the type.

x = 2

The factor $(x - 2)$ shows up in both numerator and denominator

It cancels out

Type: hole (removable discontinuity)

x = -6

The factor $(x + 6)$ shows up only in the denominator

It does not cancel

Type: vertical asymptote (non-removable discontinuity)

Answer:

Hole at $x = 2$

Vertical asymptote at $x = -6$

Example 2: Difference of Squares

Find all discontinuities of

g(x) = \frac{x + 3}{x^4 - 1}

Step 1: Factor the denominator.

The denominator is a difference of squares:

x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)

The first factor is also a difference of squares:

x^2 - 1 = (x - 1)(x + 1)

The second factor does not factor further (a sum of squares doesn’t factor over the reals).

Complete factorization:

g(x) = \frac{x + 3}{(x - 1)(x + 1)(x^2 + 1)}

Step 2: Find where the denominator equals zero.

$x - 1 = 0 \Rightarrow x = 1$

$x + 1 = 0 \Rightarrow x = -1$

$x^2 + 1 = 0 \Rightarrow x^2 = -1 \Rightarrow x = \pm i$ (imaginary)

Step 3: Analyze each value.

x = 1

The factor $(x - 1)$ isn’t in the numerator

Does not cancel

Type: vertical asymptote

x = -1

The factor $(x + 1)$ isn’t in the numerator

Does not cancel

Type: vertical asymptote

x = \pm i

These are imaginary numbers

We only care about real discontinuities

Ignore these

Answer:

Vertical asymptote at $x = 1$

Vertical asymptote at $x = -1$

Example 3: Canceling Factors

Find all discontinuities of

h(x) = \frac{(x + 4)(x - 1)}{(x + 4)(x + 2)}

Step 1: It’s already factored.

Step 2: Cancel common factors.

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

Step 3: Identify the discontinuities.

x = -4

The factor $(x + 4)$ canceled

Type: hole

x = -2

The factor $(x + 2)$ did not cancel

Type: vertical asymptote

Answer:

Hole at $x = -4$

Vertical asymptote at $x = -2$

A related case to watch for: if you have

\dfrac{x+4}{(x+4)^2}

, one

(x + 4)

cancels and leaves

\dfrac{1}{x+4}

, which has a vertical asymptote at

x = -4

. The cancellation doesn’t create a separate hole here. It just changes the nature of the discontinuity, and the asymptote is what remains.

Discontinuities in Trigonometric Functions

Trig functions can have discontinuities too, especially when they involve quotients.

Example 4: Tangent Function

Find all discontinuities of

f(x) = \tan(3x)

for

0 \leq x \leq 2\pi

Step 1: Rewrite tangent as a quotient.

\tan(3x) = \frac{\sin(3x)}{\cos(3x)}

Step 2: Find where the denominator equals zero.

We need

\cos(3x) = 0

Step 3: Solve

\cos(\theta) = 0

Cosine equals zero at

\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \ldots

Step 4: Substitute

\theta = 3x

3x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{6}

3x = \frac{3\pi}{2} \Rightarrow x = \frac{\pi}{2}

3x = \frac{5\pi}{2} \Rightarrow x = \frac{5\pi}{6}

3x = \frac{7\pi}{2} \Rightarrow x = \frac{7\pi}{6}

3x = \frac{9\pi}{2} \Rightarrow x = \frac{3\pi}{2}

3x = \frac{11\pi}{2} \Rightarrow x = \frac{11\pi}{6}

3x = \frac{13\pi}{2} \Rightarrow x = \frac{13\pi}{6} > 2\pi \quad \text{(stop here)}

Step 5: Check that all values are in the domain.

All of

\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}

fall between

0

and

2\pi

Answer: Vertical asymptotes at

x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}

Continuous Functions

Not every function has discontinuities. Plenty of functions are continuous everywhere (or across their whole domain).

Example 5: Polynomial Function

Determine whether

f(x) = x^2 - 4

has any discontinuities.

This is a polynomial function. Polynomials are continuous everywhere on the real number line because:

There are no denominators, so no division by zero

There are no even-index radicals, so no domain restrictions

There are no logarithms or other special functions

Answer: This is a continuous function. There are no discontinuities.

Summary Table

Discontinuity Type	Removable?	How to Identify	Graph Feature
Hole	Removable	Factor cancels	Open circle
Vertical asymptote	Non-removable	Factor doesn’t cancel	Vertical line where the function goes to $\pm\infty$
Jump	Non-removable	Different left and right limits	Sudden vertical gap

Classifying with limits. For any function, not just rational ones, the one-sided limits at

x = c

tell you the type:

Removable: $\lim_{x \to c} f(x)$ exists, but $f(c)$ is undefined or doesn’t equal it

Jump: both one-sided limits exist (as real numbers) but aren’t equal

Infinite (vertical asymptote): at least one one-sided limit is $\infty$ or $-\infty$

Practice Problems

Problem 1

Find all discontinuities and classify them:

f(x) = \frac{x^2 - 9}{x^2 - x - 6}

Answer:

Factor:

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

Hole at $x = 3$ (factor cancels)

Vertical asymptote at $x = -2$ (factor doesn’t cancel)

Problem 2

Find all discontinuities and classify them:

g(x) = \frac{x + 5}{x^2 + 10x + 25}

Answer:

Factor:

\dfrac{x+5}{(x+5)^2} = \dfrac{1}{x+5}

after canceling.

One factor of

(x + 5)

cancels and one remains in the denominator, so the remaining factor is what determines the behavior.

Vertical asymptote at $x = -5$

Problem 3

Find all discontinuities:

h(x) = x^3 + 2x - 7

Answer:

This is a polynomial, so it’s continuous everywhere. No discontinuities.

Key Takeaways

A function is continuous if you can draw it without lifting your pencil

Holes (removable) happen when factors cancel in rational functions

Vertical asymptotes (non-removable) happen when denominator factors don’t cancel

To find discontinuities, factor completely and set the denominator equal to zero

Polynomial functions are always continuous

For trig functions, rewrite them as quotients to find where denominators equal zero

Imaginary solutions don’t count as discontinuities. Only real numbers matter

Always say “continuous function” or “no discontinuities” when that’s the case

Back Next