Types of Discontinuities

Not all breaks in a graph are the same. A single missing point is easy to fix; a jump or an asymptote is not. Classifying a discontinuity tells you what’s going on with the limit there, and whether a small change to the function could make it continuous. AP questions often ask you to name the type or to choose a constant that makes a piecewise function continuous.

Key ideas

Three types

Recall that $f$ is continuous at $a$ when $f(a)$ is defined, $\displaystyle\lim_{x \to a} f(x)$ exists, and the two are equal. When this fails, look at the limits:

Type	What the limits do	Graph
Removable	$\displaystyle\lim_{x \to a} f(x)$ exists, but $f(a)$ is undefined or different	a hole (maybe with a dot somewhere else)
Jump	both one-sided limits exist (as 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

The three kinds of discontinuity, each at $x = 2$.

(A function can also fail to have a limit by oscillating, like $\sin\left(\tfrac{1}{x}\right)$ at $0$. That kind is rare on the AP exam.)

Finding discontinuities in rational functions

Factor the numerator and denominator. At a zero of the denominator:

• If the factor cancels, the discontinuity is removable (a hole). The height of the hole is the limit, found from the simplified form.
• If it doesn’t cancel completely (some of that factor is still left in the denominator), the discontinuity is infinite (a vertical asymptote).

Removing a discontinuity

Only a removable discontinuity can be removed. Redefine the function at that one point so that its value equals the limit:

$$f(a) = \lim_{x \to a} f(x)$$

A jump or an infinite discontinuity can’t be fixed by changing a single value, because the limit doesn’t exist.

Making a piecewise function continuous

At a boundary point $a$, the pieces must meet. Set the left-hand limit equal to the right-hand limit (and the function value, which usually comes from one of the pieces):

$$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$$

Each unknown constant needs one equation, so two constants need two boundary points.

Worked examples

Example 1: Classifying for a rational function

Find and classify the discontinuities of $f(x) = \dfrac{x^2 - 4}{x^2 - x - 2}$.

Solution. Factor:

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

The denominator is $0$ at $x = 2$ and $x = -1$.

At $x = 2$, the factor $x - 2$ cancels, leaving $\dfrac{x + 2}{x + 1}$ for $x \ne 2$:

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

The limit exists but $f(2)$ is undefined: a removable discontinuity (a hole at $\left(2, \tfrac{4}{3}\right)$).

At $x = -1$, the factor $x + 1$ doesn’t cancel. The top approaches $-1 + 2 = 1$ while the bottom approaches $0$, so $f$ blows up: an infinite discontinuity (vertical asymptote $x = -1$).

Example 2: A jump

Classify the discontinuity of $p(x) = \begin{cases} 2x - 1, & x \le 1 \\ x + 2, & x \gt 1 \end{cases}$ at $x = 1$.

Solution.

$$\lim_{x \to 1^-} (2x - 1) = 1, \qquad \lim_{x \to 1^+} (x + 2) = 3$$

Both one-sided limits are numbers, but they’re different. This is a jump discontinuity. (The graph jumps up by $2$ at $x = 1$.)

Example 3: Removing a discontinuity

Let $g(x) = \dfrac{x^2 - 5x + 6}{x - 3}$ for $x \ne 3$. What value should $g(3)$ be given so that $g$ is continuous at $x = 3$?

Solution. Find the limit:

$$\lim_{x \to 3} \frac{(x - 2)(x - 3)}{x - 3} = \lim_{x \to 3} (x - 2) = 1$$

Define $g(3) = 1$. Then $\displaystyle\lim_{x \to 3} g(x) = 1 = g(3)$, so $g$ is continuous at $3$.

Example 4: Choosing a constant

Find $k$ so that $f(x) = \begin{cases} kx + 3, & x \lt 2 \\ x^2 - k, & x \ge 2 \end{cases}$ is continuous at $x = 2$.

Solution. The left-hand limit is $2k + 3$. The right-hand limit and the value $f(2)$ are both $4 - k$. Set them equal:

$$2k + 3 = 4 - k \quad\Rightarrow\quad 3k = 1 \quad\Rightarrow\quad k = \frac{1}{3}$$

Check: $2\left(\tfrac{1}{3}\right) + 3 = \tfrac{11}{3}$ and $4 - \tfrac{1}{3} = \tfrac{11}{3}$. The pieces meet at height $\tfrac{11}{3}$.

Common mistakes

Calling every zero of the denominator an asymptote. Factor first. If the factor cancels, it’s a hole (removable), not an asymptote.

Finding the hole’s height from the original formula. Substituting into the original gives $\tfrac{0}{0}$. Use the simplified form to find the limit.

Trying to “remove” a jump. Changing $f(a)$ can’t fix a jump, because the two sides still disagree. Only removable discontinuities can be removed.

Setting the pieces equal as expressions instead of at the point. Solve $2k + 3 = 4 - k$ (the values at $x = 2$), not $kx + 3 = x^2 - k$ for all $x$.

Stopping after one equation with two unknowns. If a piecewise function has two constants, use both boundary points to get two equations.

Practice

1. (Warm-up) Suppose $\displaystyle\lim_{x \to 2} f(x) = 5$ and $f(2) = 1$. What type of discontinuity does $f$ have at $x = 2$?

Solution

The limit exists but doesn’t equal $f(2)$, so it’s removable. Redefining $f(2) = 5$ would remove it.

2. (Warm-up) What type of discontinuity does $f(x) = \dfrac{1}{x + 4}$ have at $x = -4$?

Solution

The top is $1$ and the bottom approaches $0$, so $f$ blows up near $-4$. It’s an infinite discontinuity (vertical asymptote $x = -4$).

3. (Core) Find and classify the discontinuities of $f(x) = \dfrac{x^2 - 1}{x^2 + 2x - 3}$.

Solution

$$f(x) = \frac{(x - 1)(x + 1)}{(x + 3)(x - 1)}$$

At $x = 1$, the factor cancels: $\displaystyle\lim_{x \to 1} \frac{x + 1}{x + 3} = \frac{2}{4} = \frac{1}{2}$. This is removable (hole at $\left(1, \tfrac{1}{2}\right)$).

At $x = -3$, the top approaches $-2$ and the bottom approaches $0$. This is infinite (vertical asymptote $x = -3$).

4. (Core) Let $g(x) = \dfrac{x^2 + 2x - 15}{x - 3}$ for $x \ne 3$. What value of $g(3)$ makes $g$ continuous?

Solution

$$\lim_{x \to 3} \frac{(x + 5)(x - 3)}{x - 3} = \lim_{x \to 3} (x + 5) = 8$$

Define $g(3) = 8$.

5. (Core) Find $k$ so that $f(x) = \begin{cases} x^2 + k, & x \lt 3 \\ kx - 1, & x \ge 3 \end{cases}$ is continuous at $x = 3$.

Solution

$$9 + k = 3k - 1 \quad\Rightarrow\quad 10 = 2k \quad\Rightarrow\quad k = 5$$

Check: both pieces give $14$ at $x = 3$.

6. (Core) Let $h(x) = \begin{cases} \dfrac{\sqrt{x} - 3}{x - 9}, & x \ne 9 \\ c, & x = 9 \end{cases}$ for $x \ge 0$. Find $c$ so that $h$ is continuous at $x = 9$.

Solution

Factor $x - 9 = (\sqrt{x} - 3)(\sqrt{x} + 3)$:

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

So $c = \dfrac{1}{6}$.

7. (Core) Let $f(x) = \begin{cases} \dfrac{\sin x}{x}, & x \ne 0 \\ 2, & x = 0 \end{cases}$ (radians). Classify the discontinuity at $x = 0$, and say how to remove it.

Solution

$\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$ (a special trig limit), but $f(0) = 2$. The limit exists and doesn’t equal $f(0)$, so the discontinuity is removable. Redefine $f(0) = 1$.

8. (Challenge) Find $a$ and $b$ so that $f$ is continuous for all real numbers:

$$f(x) = \begin{cases} x^2, & x \lt 1 \\ ax + b, & 1 \le x \le 3 \\ 12 - x, & x \gt 3 \end{cases}$$

Solution

At $x = 1$: the left side approaches $1$ and the middle piece gives $a + b$. So $a + b = 1$.

At $x = 3$: the middle piece gives $3a + b$ and the right side approaches $12 - 3 = 9$. So $3a + b = 9$.

Subtracting the first equation from the second: $2a = 8$, so $a = 4$ and $b = -3$.

Check: $4(1) - 3 = 1$ and $4(3) - 3 = 9$. Both boundaries match.

9. (Challenge) Find the value of $a$ for which $f(x) = \dfrac{x^2 + ax + 6}{x - 2}$ has a removable discontinuity at $x = 2$. Then find the limit there.

Solution

The discontinuity can only be removable if the factor $x - 2$ cancels, so the top must be $0$ at $x = 2$:

$$4 + 2a + 6 = 0 \quad\Rightarrow\quad a = -5$$

Then

$$\lim_{x \to 2} \frac{x^2 - 5x + 6}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x - 3)}{x - 2} = \lim_{x \to 2} (x - 3) = -1$$

For any other value of $a$, the top is nonzero at $x = 2$ and the discontinuity is infinite.