Math Core

Knowing that a function is discontinuous at a point is only half the story. How it breaks matters: some breaks are a single misplaced point that can be repaired, while others are genuine gaps or blow-ups. The AP exam expects you to classify discontinuities by their limits and to repair the ones that can be repaired.

Three types of discontinuity

Every discontinuity you will classify in AP Calculus is one of three types, and each is identified by what the limits do at $x = c$ .

Classifying a discontinuity at x = c

Type	Limit behavior	Picture
Removable	$\displaystyle \lim_{x \to c} f(x)$ exists, but $f(c)$ is undefined or $f(c) \ne \lim_{x \to c} f(x)$	a hole, possibly with a dot somewhere else
Jump	both one-sided limits exist (as real numbers) but are different	the graph jumps from one height to another
Infinite	at least one one-sided limit is $\infty$ or $-\infty$	a vertical asymptote

A fourth kind, an oscillating discontinuity like $\sin\left(\dfrac{1}{x}\right)$ at $x = 0$ , also exists, but it is rare on the AP exam.

The graph below shows the first two types.

A removable discontinuity at x = −2 (hole at (−2, 2), point at (−2, −1)) and a jump discontinuity at x = 1.Open in grapher →

At $x = -2$ , both sides approach height $2$ , so $\displaystyle \lim_{x \to -2} f(x) = 2$ . But the filled dot shows $f(-2) = -1$ . The limit exists and does not match the value: a removable discontinuity.
At $x = 1$ , the left side approaches $-1$ and the right side approaches $2$ . Both one-sided limits exist but differ: a jump discontinuity.

And here is the third type:

y = 1/(x − 2) has an infinite discontinuity at x = 2: the left side falls to −∞ and the right side rises to ∞.Open in grapher →

At $x = 2$ , $\displaystyle \lim_{x \to 2^-} \frac{1}{x - 2} = -\infty$ and $\displaystyle \lim_{x \to 2^+} \frac{1}{x - 2} = \infty$ : an infinite discontinuity.

Why "removable"?

A removable discontinuity can be fixed by changing (or supplying) a single function value. If $\displaystyle \lim_{x \to c} f(x) = L$ , define or redefine $f(c) = L$ . The new function is continuous at $c$ , and it agrees with the old one everywhere else.

Jump and infinite discontinuities cannot be removed this way. There is no single value that could make the left and right sides meet, because the limit does not exist.

Classifying rational functions

For a rational function, the trouble spots are the zeros of the denominator. Factor, and look at what cancels.

If a factor $(x - c)$ cancels completely from the denominator, the discontinuity at $c$ is removable (a hole).
If a factor $(x - c)$ remains in the denominator after canceling, the discontinuity at $c$ is infinite (a vertical asymptote).

Worked example: Classifying the discontinuities of a rational function

Find and classify the discontinuities of $f(x) = \dfrac{x^2 - x - 6}{x^2 - 9}$ .

Solution. The denominator is zero at $x = 3$ and $x = -3$ . Factor:

f(x) = \frac{(x - 3)(x + 2)}{(x - 3)(x + 3)} = \frac{x + 2}{x + 3}, \qquad x \ne 3.

At $x = 3$ the factor cancels, and $\displaystyle \lim_{x \to 3} f(x) = \frac{5}{6}$ . Since $f(3)$ is undefined, this is a removable discontinuity (a hole at $\left(3, \dfrac{5}{6}\right)$ ).

At $x = -3$ the factor $(x + 3)$ remains, and the numerator approaches $-1 \ne 0$ . So $f$ is unbounded near $-3$ : $\displaystyle \lim_{x \to -3^+} f(x) = -\infty$ and $\displaystyle \lim_{x \to -3^-} f(x) = \infty$ . This is an infinite discontinuity.

Worked example: Removing a discontinuity

Let $g(x) = \dfrac{\sqrt{x} - 3}{x - 9}$ . What value should $g(9)$ be given so that $g$ is continuous at $x = 9$ ?

Solution. Substituting gives $\dfrac{0}{0}$ . Since $x - 9 = (\sqrt{x} - 3)(\sqrt{x} + 3)$ for $x \ge 0$ ,

\lim_{x \to 9} \frac{\sqrt{x} - 3}{(\sqrt{x} - 3)(\sqrt{x} + 3)} = \lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{6}.

Defining $g(9) = \dfrac{1}{6}$ removes the discontinuity.

Worked example: A jump in a piecewise function

Classify the discontinuity of $p(x) = \begin{cases} x^2, & x < 1 \\ x + 2, & x \ge 1 \end{cases}$ at $x = 1$ .

Solution. $\displaystyle \lim_{x \to 1^-} p(x) = 1$ and $\displaystyle \lim_{x \to 1^+} p(x) = 3$ . Both exist and they differ, so $p$ has a jump discontinuity at $x = 1$ . The size of the jump is $3 - 1 = 2$ .

Common mistake

Do not classify a discontinuity by the formula alone. A zero in the denominator does not always mean a vertical asymptote. You must factor and cancel first (or compute the limit) to tell a hole from an asymptote. Likewise, a piecewise function whose pieces are given by different formulas may still be continuous if the pieces happen to meet.

Tip

A quick test at a zero of the denominator: substitute into the numerator. If the numerator is nonzero, the discontinuity is infinite. If it is zero, you have $\dfrac{0}{0}$ and need to simplify; the discontinuity may be removable.

Practice

Practice 1

What type of discontinuity does $f(x) = \dfrac{x^2 - 4}{x - 2}$ have at $x = 2$ ?

Practice 2

Let $h(x) = \dfrac{x^2 - 3x - 4}{x - 4}$ for $x \ne 4$ . What value should $h(4)$ be given to make $h$ continuous at $x = 4$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Which statement describes the discontinuities of $f(x) = \dfrac{x + 1}{x^2 - 1}$ ?

Practice 4

The graph of $f$ is shown. At what value of $x$ does $f$ have a jump discontinuity?

The graph of f.Open in grapher →

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Let $g(x) = \dfrac{\sqrt{x + 1} - 2}{x - 3}$ for $x \ge -1$ , $x \ne 3$ . What value of $g(3)$ makes $g$ continuous at $x = 3$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Which function has a jump discontinuity at $x = 0$ ?

Practice 7

Let $f(x) = \dfrac{1 - \cos x}{x^2}$ for $x \ne 0$ . What value of $f(0)$ makes $f$ continuous at $x = 0$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Let $f(x) = \begin{cases} 2x + 1, & x < 2 \\ 4, & x = 2 \\ x^2 + 1, & x > 2. \end{cases}$ Which statement is true at $x = 2$ ?