Exploring Types of Discontinuities (AP Calculus AB 1.10)

A function is continuous at $x = c$ when the value it heads toward near $c$ is the value it actually takes at $c$ : in symbols, $\lim\limits_{x \to c} f(x) = f(c)$ , with both sides existing. A discontinuity is a point where that fails, and the graph breaks there. The breaks come in three kinds, and you tell them apart by asking what the one-sided limits do. Topic 1.11 takes the definition of continuity apart condition by condition; this topic is about recognizing and naming the ways it fails.

Three ways a graph can break

Three graphs, each with a discontinuity at x = 2. Removable: a smooth curve heading to the height 1 from both sides, with an open circle at (2, 1) and the function's value drawn as a separate filled dot at (2, 3). Jump: a segment rising to an open circle at (2, 2) from the left, and a separate segment starting at the filled point (2, 3) on the right. Infinite: a curve with a dashed vertical asymptote at x = 2, falling without bound as x approaches 2 from the left and rising without bound as x approaches 2 from the right.

Removable. The two one-sided limits exist and are equal, so $\lim\limits_{x \to c} f(x)$ exists and is a finite number $L$ . The trouble is only at the single point: either $f(c)$ is undefined (a hole in the graph) or $f(c) \neq L$ (a hole with the point displaced above or below it). It is called removable because changing or supplying one value, $f(c) = L$ , repairs it; topic 1.13 does exactly that.

Jump. Both one-sided limits exist as finite numbers but are different. The graph arrives at one height from the left and leaves from another height on the right, so no choice of $f(c)$ can make it continuous. The value $f(c)$ , if it exists, may match one side, the other, or neither.

Infinite. At least one one-sided limit is $\infty$ or $-\infty$ , and the line $x = c$ is a vertical asymptote. One side is enough: $e^{1/x}$ approaches 0 as $x \to 0^-$ but grows without bound as $x \to 0^+$ , and that single infinite side makes $x = 0$ an infinite discontinuity. Topic 1.14 covers infinite limits and asymptotes in more detail.

Whether $f(c)$ exists does not decide the type. The graph of $\frac{x^2-1}{x-1}$ has a removable discontinuity at 1, $\frac{|x|}{x}$ has a jump at 0, and $\frac{1}{x^2}$ has an infinite discontinuity at 0, and none of the three is defined at its point. The limits decide the type; the value only matters for deciding whether there is a discontinuity at all.

Not every discontinuity fits these three. $\sin\left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$ infinitely often near 0, so neither one-sided limit exists, finite or infinite. AP questions classify only the three named types, but this example is why the classification starts by computing the one-sided limits rather than guessing from a formula.

Where each type comes from

You can predict where to look from the way a function is built.

Rational functions (a polynomial over a polynomial) break only where the denominator is 0, and the numerator decides how. If the numerator is also 0 there, $x - c$ is a factor of both; cancel it and look again. If the cancelled form has a nonzero denominator at $c$ , the limit is finite and the discontinuity is removable. If the numerator is not 0 at $c$ , the fraction is a nonzero number over something approaching 0, and its size grows without bound, so the discontinuity is infinite. A rational function never has a jump: near $c$ the same formula applies on both sides, so if both one-sided limits are finite they are the same number.

Piecewise functions can break at the points where the rule changes, and that is where jumps come from. Each one-sided limit at a boundary is computed with the piece that applies on that side. A piece's own formula can also break inside its interval, for instance a fraction whose denominator is 0 there.

Absolute values produce jumps when they appear in a quotient: $\frac{|x - c|}{x - c}$ equals $-1$ for $x < c$ and $1$ for $x > c$ .

Worked examples

Worked exampleExample 1: a rational function with two kinds of break

Let $r(x) = \dfrac{x^2 + x - 12}{x^2 - 9}$ . Find every value of $x$ where $r$ is discontinuous and classify each discontinuity, with justification.

Show solution

The denominator $x^2 - 9$ is 0 at $x = 3$ and $x = -3$ , so those are the only candidates; elsewhere $r$ is a quotient of polynomials with a nonzero denominator. Factor both parts:

r(x) = \frac{(x + 4)(x - 3)}{(x - 3)(x + 3)} = \frac{x + 4}{x + 3} \quad \text{for } x \neq 3,\ x \neq -3.

At $x = 3$ : the numerator is also 0, and after cancelling,

\lim_{x \to 3} r(x) = \frac{3 + 4}{3 + 3} = \frac{7}{6}.

The limit exists but $r(3)$ is undefined, so $r$ has a removable discontinuity at $x = 3$ .

At $x = -3$ : the factor $x + 3$ did not cancel, and the numerator $x + 4$ approaches $1$ . As $x \to -3^+$ , $x + 3$ is a small positive number, so $\frac{x+4}{x+3} \to \infty$ ; as $x \to -3^-$ , it is a small negative number, so the quotient $\to -\infty$ . Since a one-sided limit is infinite, $r$ has an infinite discontinuity (a vertical asymptote) at $x = -3$ .

Worked exampleExample 2: all three types in one piecewise function

Let

g(x) = \begin{cases} 3 - x & x < 1 \\ 5 & x = 1 \\ x^2 + 1 & 1 < x \le 3 \\ \dfrac{7}{x - 4} & x > 3 \end{cases}

Classify each discontinuity of $g$ .

Show solution

Each formula is a polynomial or a quotient of polynomials, so the only places $g$ can break are the boundaries $x = 1$ and $x = 3$ and the point $x = 4$ , where the last piece's denominator is 0 (and $4 > 3$ , so that piece is in use there).

At $x = 1$ : from the left use $3 - x$ , giving $2$ ; from the right use $x^2 + 1$ , giving $2$ . The limit exists and equals 2, but $g(1) = 5$ . That is a removable discontinuity.

At $x = 3$ : from the left use $x^2 + 1$ , giving $10$ ; from the right use $\frac{7}{x - 4}$ , giving $\frac{7}{-1} = -7$ . Both are finite and they differ, so this is a jump discontinuity. (Here $g(3) = 10$ , matching the left side, but that doesn't change the type.)

At $x = 4$ : the numerator is 7 while $x - 4 \to 0$ . As $x \to 4^+$ the quotient $\to \infty$ , and as $x \to 4^-$ it $\to -\infty$ . That is an infinite discontinuity.

Worked exampleExample 3: an exam-style question with a parameter (no calculator)

Let $f(x) = \dfrac{x^2 + ax - 10}{x - 2}$ , where $a$ is a constant.

(a) For what value of $a$ does $f$ have a removable discontinuity at $x = 2$ ? For that value, find $\lim\limits_{x \to 2} f(x)$ .

(b) For every other value of $a$ , what type of discontinuity does $f$ have at $x = 2$ ? Explain.

Show solution

(a) The denominator approaches 0, so the limit can be finite only if the numerator also approaches 0; otherwise the quotient grows without bound. Set the numerator equal to 0 at $x = 2$ :

4 + 2a - 10 = 0 \quad\Longrightarrow\quad a = 3.

With $a = 3$ the numerator factors as $(x + 5)(x - 2)$ , so $f(x) = x + 5$ for $x \neq 2$ , and $\lim\limits_{x \to 2} f(x) = 7$ . Since $f(2)$ is undefined, the discontinuity is removable.

(b) If $a \neq 3$ , the numerator approaches $2a - 6 \neq 0$ while the denominator approaches 0, so the quotient's absolute value grows without bound from each side. The discontinuity is infinite, with a vertical asymptote at $x = 2$ . For example, with $a = 4$ the numerator approaches 2, and $f(x) \to \infty$ as $x \to 2^+$ and $f(x) \to -\infty$ as $x \to 2^-$ .

(c) No. A jump needs two different finite one-sided limits. Near $x = 2$ the same formula applies on both sides, and parts (a) and (b) cover every $a$ : either both one-sided limits equal 7, or at least one is infinite.

Where students lose points

On the exam

Classification questions come as multiple choice (a formula, a piecewise rule or a graph, and a choice of types or of the points where one type occurs) and as free-response parts asking whether a function is continuous at a point, with justification. The justification the exam rewards names the limits: for a jump, the two different one-sided limits; for a removable discontinuity, the existing limit and the missing or different value; for an infinite one, the infinite one-sided limit. "The limit does not exist" on its own is not a justification; give the one-sided behavior that shows it.

Check yourself

Question 1
Let $f(x) = \dfrac{x - 2}{x^3 - 4x}$ . At which value of $x$ does $f$ have a removable discontinuity?
1. A
  $x = -2$
2. B
  $x = 0$
3. C
  $x = 2$
4. D
  $f$ has no removable discontinuity
Show answer
C
The denominator factors as $x(x - 2)(x + 2)$ , so $f(x) = \frac{1}{x(x + 2)}$ for $x \neq 2$ . At $x = 2$ the limit is $\frac{1}{8}$ , finite, so that discontinuity is removable. At $x = 0$ and $x = -2$ the numerator is not 0 while the denominator is, so those are infinite discontinuities; choices A and B treat every zero of the denominator the same way.
Question 2
Let $g(x) = \dfrac{|x - 3|}{x - 3} + x$ . Find $\lim\limits_{x \to 3^-} g(x)$ and $\lim\limits_{x \to 3^+} g(x)$ , and classify the discontinuity at $x = 3$ .
Show answer
The left-hand limit is $2$ and the right-hand limit is $4$ , so $g$ has a jump discontinuity at $x = 3$ .
For $x < 3$ , $|x - 3| = -(x - 3)$ , so $g(x) = -1 + x \to 2$ . For $x > 3$ , $g(x) = 1 + x \to 4$ . Both one-sided limits are finite and they differ, which is a jump.
Question 3
A function $f$ is defined for all $x$ near $c$ except at $x = c$ itself. Which of the following must be true?
1. A
  $f$ has a vertical asymptote at $x = c$
2. B
  $\lim\limits_{x \to c} f(x)$ does not exist
3. C
  $f$ is not continuous at $x = c$
4. D
  $f$ has a removable discontinuity at $x = c$
Show answer
C
Continuity at $c$ requires $f(c)$ to exist, so a missing value always means a discontinuity. Which type it is depends on the one-sided limits, so A, B and D can each be false: $\frac{x^2 - 1}{x - 1}$ at 1 has a limit (so B and A fail), and $\frac{1}{x}$ at 0 is not removable (so D fails).
Question 4
Classify the discontinuity of $h(x) = e^{1/x}$ at $x = 0$ , giving both one-sided limits.
Show answer
$\lim\limits_{x \to 0^-} e^{1/x} = 0$ and $\lim\limits_{x \to 0^+} e^{1/x} = \infty$ , so $x = 0$ is an infinite discontinuity.
As $x \to 0^-$ , $\frac{1}{x} \to -\infty$ and $e^{1/x} \to 0$ ; as $x \to 0^+$ , $\frac{1}{x} \to \infty$ and $e^{1/x} \to \infty$ . One infinite one-sided limit is enough for an infinite discontinuity, even though the other side is finite. It is not a jump, because a jump needs both sides finite.

Exploring Types of Discontinuities

Three ways a graph can break

Where each type comes from

Worked examples

Where students lose points

On the exam

Check yourself

Go deeper