Types of Discontinuity | Math Coach Amy

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Key Ideas Easy Medium Hard

hole: limit exists, value wrong or missing

jump: one-sided limits disagree

VA: function blows up

What to Remember

Removable (hole)

The limit exists, but f is missing or has the wrong y-value. After canceling, the simplified function is defined there.

Jump

The one-sided limits exist and are finite, but they are not equal. Absolute value over a linear factor often jumps.

Vertical asymptote

The denominator is 0 and the numerator is not. The function blows up. After canceling, a factor of the denominator is still 0.

None

Polynomials and rationals whose denominators never zero, such as \(x^2+1\), have no discontinuities.

Level 1

Easy

Name the x-value and the type, with no graphing calculator.

Example

State any discontinuities of \(y = \dfrac{1}{(x - 4)^2}\) and identify the type.

The denominator is 0 at \(x = 4\).

The numerator is 1, which is not 0.

Vertical asymptote at \(x = 4\).

Answer: vertical asymptote at \(x = 4\)

State any discontinuities of \(y = \dfrac{1}{x^2 + 9}\).

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\(x^2 + 9 \geq 9 > 0\) for all real x.

No discontinuities.

Answer: none

State any discontinuities of \(y = \dfrac{|x + 1|}{x + 1}\).

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Undefined at \(x = -1\).

Left-hand limit is \(-1\), right-hand limit is \(1\).

Jump at \(x = -1\).

Answer: jump at \(x = -1\)

State any discontinuities of \(y = \dfrac{x - 2}{x^2 - 5x + 6}\).

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\(x^2 - 5x + 6 = (x - 2)(x - 3)\)

\(y = \dfrac{x - 2}{(x - 2)(x - 3)} = \dfrac{1}{x - 3}\) for \(x \neq 2\).

Hole at \(x = 2\).

Vertical asymptote at \(x = 3\).

Answer: hole at \(x = 2\), vertical asymptote at \(x = 3\)

State any discontinuities of \(y = \dfrac{|x - 5|}{x - 5}\).

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Jump at \(x = 5\).

Answer: jump at \(x = 5\)

State any discontinuities of \(y = \dfrac{1}{(x + 6)^2}\).

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Vertical asymptote at \(x = -6\).

Answer: vertical asymptote at \(x = -6\)

Level 2

Medium

Read removable vs jump vs asymptote from a graph.

Example

The graph of \(f\) is shown. What are all x-values where \(f\) has a removable discontinuity?

Graph of f

A removable discontinuity: the limit exists, but \(f\) is missing or the filled point is in the wrong place.

At \(x = 2\) the open circle is at height 1 and the filled point is at height 3, and the two sides meet at 1.

At \(x = 6\) the two sides meet at 0, but the filled point is at height 2.

At \(x = 4\) the filled point is on the graph, so that is continuous.

\(x = 2\) and \(x = 6\) only.

Answer: \(x = 2\) and \(x = 6\)

Using the graph, is the discontinuity at \(x = 2\) removable or a jump?

Graph of f

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\(\lim_{x \to 2} f(x) = 1\) exists.

\(f(2) = 3 \neq 1\)

Removable.

Answer: removable

Using the graph, find \(\lim_{x \to 4} f(x)\).

Graph of f

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The pieces meet at the filled point \((4, 3)\).

\(3\)

Answer: \(3\)

Using the graph, find \(f(6)\).

Graph of f

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The filled point at \(x = 6\) is at height 2.

\(2\)

Answer: \(2\)

Using the graph of \(f\) with a vertical asymptote, name the type of discontinuity at \(x = 1\).

Graph of f

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The graph blows up at the dashed line.

Vertical asymptote.

Answer: vertical asymptote

Does \(y = \dfrac{x}{x}\) have a hole, a jump, or a vertical asymptote at 0?

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\(\dfrac{x}{x} = 1\) for \(x \neq 0\).

\(\lim_{x \to 0} 1 = 1\)

Hole (removable).

Answer: hole

Level 3

Hard

A rational with two interesting x-values, or a piecewise with mixed types.

Example

Let \(f(x) = \dfrac{2x^3 + 4x^2}{x^2 - x}\). Which statement is true?

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

\(2x^3 + 4x^2 = 2x^2(x + 2)\)

\(f(x) = \dfrac{2x^2(x + 2)}{x(x - 1)} = \dfrac{2x(x + 2)}{x - 1}\) for \(x \neq 0\).

At \(x = 0\) the limit of the simplified form is \(\dfrac{0}{-1} = 0\), so the discontinuity is removable.

At \(x = 1\) the simplified denominator is 0 and the numerator is \(2(3) = 6 eq 0\), so there is a vertical asymptote.

Removable at 0, vertical asymptote at 1.

Answer: removable at \(x = 0\), vertical asymptote at \(x = 1\)

Let \(f(x) = \dfrac{x^2 - 9}{x^2 + x - 12}\). Classify the discontinuities.

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\(x^2 - 9 = (x - 3)(x + 3)\)

\(x^2 + x - 12 = (x + 4)(x - 3)\)

\(f(x) = \dfrac{x + 3}{x + 4}\) for \(x \neq 3\).

Hole at \(x = 3\).

Vertical asymptote at \(x = -4\).

Answer: hole at \(x = 3\), vertical asymptote at \(x = -4\)

Let \(f(x) = \dfrac{x^3 - 4x}{x^2 + x - 6}\). Classify the discontinuities at \(x = 2\) and \(x = -3\).

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\(x^3 - 4x = x(x - 2)(x + 2)\)

\(x^2 + x - 6 = (x + 3)(x - 2)\)

\(f(x) = \dfrac{x(x + 2)}{x + 3}\) for \(x \neq 2\).

Hole at \(x = 2\).

At \(x = -3\) the simplified numerator is \(-3(-1) = 3 \neq 0\), so a vertical asymptote.

Answer: hole at \(x = 2\), vertical asymptote at \(x = -3\)

Let \(f(x) = \begin{cases} x + 2 & x < 1 \\ 0 & x = 1 \\ 4 - x & x > 1 \end{cases}\). What type of discontinuity is at \(x = 1\)?

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Left: \(3\).

Right: \(3\).

\(f(1) = 0 \neq 3\)

Removable.

Answer: removable

Let \(f(x) = \begin{cases} 2x & x < 0 \\ x - 4 & x \geq 0 \end{cases}\). What type of discontinuity is at \(x = 0\)?

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Left: 0. Right: \(-4\).

Jump.

Answer: jump

Let \(f(x) = \dfrac{x^2 + 4}{x^2 + 4}\). State the discontinuities.

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The denominator is never 0.

\(f(x) = 1\) for all x.

None.

Answer: none

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