Removable, jump or infinite: telling discontinuities apart

By Marius, private math tutor in Montreal · 2026-10-05

There are three main types of discontinuity, and the two one-sided limits at the break tell them apart. If both one-sided limits are the same finite number, the discontinuity is removable; if both are finite but different, it is a jump; if either one is infinite, it is an infinite discontinuity, with a vertical asymptote. Here is that check on graphs and on formulas, with a self-check at the end.

The three types, on a graph

A discontinuity at $x = a$ is a place where the graph breaks. To name the break, look at the height the graph is heading for just to the left of $a$ and just to the right of $a$ . For now, ignore whatever is plotted at $a$ itself.

All three graphs break at

x = 2

. Left:

y = \frac{x^{3} - 2 x^{2}}{2 x - 4}

, which is

y = \frac{x^{2}}{2}

with the point

(2, 2)

missing. Middle:

y = x + 1

for

x < 2

and

y = 6 - \frac{x}{2}

for

x \geq 2

. Right:

y = \frac{2 x - 3}{x - 2}

Removable. Both sides head for the same height, $2$ . The graph has a hole, and putting one point back would close it.
Jump. The left side heads for $3$ and the right side for $5$ . Both are real numbers; they just disagree, so the graph has to leap from one to the other.
Infinite. Neither side heads for a height at all. The graph runs off the bottom of the picture on the left of $x = 2$ and off the top on the right, hugging the vertical asymptote.

The limit check that sorts them

A sketch can mislead you, so the type is settled by limits, not by the picture. Work in this order:

Find the left-hand limit ${lim}_{x \to a^{-}} f (x)$ and the right-hand limit ${lim}_{x \to a^{+}} f (x)$ .
If either one is $+ \infty$ or $- \infty$ , the discontinuity is infinite. One side is enough.
If both are finite and equal, the two-sided limit exists and the discontinuity is removable.
If both are finite and different, it is a jump. The size of the jump is the right-hand limit minus the left-hand limit.

Do the infinity check first. Two one-sided limits that are both $+ \infty$ look as if they agree, but they do not agree on a number, and only a number can fill a hole.

The type never depends on $f (a)$ . The one-sided limits alone decide it. Whether $f (a)$ is undefined, or defined at some unrelated height, only affects how the break looks on the graph: an empty hole, or a hole with a stray dot above or below it. The trap is to read the type off the value, for instance to call every point where $f$ is undefined “removable”. The function $\frac{1}{x}$ is undefined at $0$ , and its discontinuity there is infinite.

Reading the type off a formula

Before computing anything, a formula tells you where to look.

A factor of the denominator that cancels completely with the numerator leaves a hole: after cancelling, the simplified expression has a finite limit there. Removable.
A factor of the denominator that is still there after cancelling makes the values grow without bound. Infinite.
The junction of a piecewise rule is where jumps come from, because two different formulas meet there. Take each side's limit from its own formula.
Anywhere else, a formula built from the usual functions is continuous as long as it is defined. The only other places to check are points where some part of it is undefined, such as $x = \frac{π}{2}$ for $\tan x$ .

Worked example

Find and classify every discontinuity of

f (x) = {\begin{matrix} \frac{x^{2} + x}{x^{2} - 1}, & x < 2 \\ x + 1, & x \geq 2 \end{matrix}

Step 1: the candidates. The denominator $x^{2} - 1$ is zero at $x = - 1$ and $x = 1$ , and both lie in the region $x < 2$ where that formula is used. The other candidate is the junction, $x = 2$ . Everywhere else $f$ is a rational expression with a non-zero denominator or a polynomial, so it is continuous.

Step 2: factor.

\frac{x^{2} + x}{x^{2} - 1} = \frac{x (x + 1)}{(x - 1) (x + 1)} = \frac{x}{x - 1} for x \neq - 1

At $x = - 1$ : the factor $x + 1$ cancelled, so the limit is the simplified expression's value:

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

Removable: a hole at $(- 1, \frac{1}{2})$ .

At $x = 1$ : in $\frac{x}{x - 1}$ the numerator tends to $1$ while the denominator tends to $0$ . Just left of $1$ , $x - 1$ is a small negative number; just right of it, a small positive one:

lim_{x \to 1^{-}} f (x) = - \infty lim_{x \to 1^{+}} f (x) = + \infty

Infinite: the line $x = 1$ is a vertical asymptote.

At $x = 2$ : each side uses its own formula.

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

Both finite, and different: a jump of $3 - 2 = 1$ . The value $f (2) = 3$ is defined, and it changes nothing.

So $f$ has one discontinuity of each type: removable at $- 1$ , infinite at $1$ , jump at $2$ .

When none of the three names fits

The College Board's AP course framework names these three types: removable discontinuities, jump discontinuities, and discontinuities due to vertical asymptotes (the infinite type on this page). In the same unit it lists three ways a limit can fail to exist: the function is unbounded, it oscillates, or the limits from the left and from the right differ. The infinite type is unbounded, and unequal sides is the jump. Oscillation is the odd one out: it is a way for a limit to fail, not one of the named types, and question 7 below has an example.

Check your understanding

No calculator needed.

Name the type of discontinuity at $x = a$ in each case. (a) Both one-sided limits are $4$ , and $f (a) = 1$ . (b) The left-hand limit is $4$ and the right-hand limit is $- 4$ . (c) The left-hand limit is $4$ and the right-hand limit is $+ \infty$ .
Find and classify every discontinuity of $g (x) = \frac{x^{2} - 4 x}{x^{3} - 16 x}$ .
Let $h (x) = 3 - x$ for $x < 2$ and $h (x) = \sqrt{x + 2}$ for $x \geq 2$ . What type of discontinuity does $h$ have at $x = 2$ , and how big is it?
At $x = 3$ , a function has ${lim}_{x \to 3^{-}} f (x) = 7$ and $f (3) = 7$ , and its discontinuity there is a jump. What can you say about ${lim}_{x \to 3^{+}} f (x)$ ?
Find the error. Let $p (x) = \frac{1}{(x - 1)^{2}}$ for $x \neq 1$ and $p (1) = 0$ . A student writes: “Both one-sided limits at $1$ are $+ \infty$ , so they agree, the limit exists, and the discontinuity is removable. Redefine $p (1)$ to fix it.”
Classify the discontinuity of $e^{1 / x}$ at $x = 0$ .
Does the discontinuity of $\sin (\frac{1}{x})$ at $x = 0$ fit any of the three types? Explain.
True or false: a rational function (a polynomial divided by a polynomial) can have a jump discontinuity. Give a reason.

Show the answers

(a) Removable: the limit exists and equals $4$ , and $f (a) = 1$ is just the wrong height. Redefining $f (a) = 4$ removes it. (b) Jump: both limits are finite and different; the graph drops by $8$ . (c) Infinite: one infinite side is enough.
$x^{3} - 16 x = x (x - 4) (x + 4)$ and $x^{2} - 4 x = x (x - 4)$ , so $g (x) = \frac{1}{x + 4}$ for $x \neq 0$ and $x \neq 4$ . Removable at $x = 0$ (the limit is $\frac{1}{4}$ ) and at $x = 4$ (the limit is $\frac{1}{8}$ ), because those factors cancel. Infinite at $x = - 4$ , where the factor survives: the limit is $- \infty$ from the left and $+ \infty$ from the right.
A jump of $1$ , upward. From the left, $3 - 2 = 1$ ; from the right, $\sqrt{4} = 2$ . Both are finite and they differ by $2 - 1 = 1$ .
It is a real number, and it is not $7$ . A jump needs both one-sided limits to be finite and different, and the left one is $7$ . The value $f (3)$ tells you nothing more.
$+ \infty$ is not a number the values approach; it records that they grow without bound. A removable discontinuity needs a finite common limit, so this one is infinite, with a vertical asymptote at $x = 1$ . No value of $p (1)$ can repair it.
Infinite. From the left, $\frac{1}{x} \to - \infty$ , so $e^{1 / x} \to 0$ . From the right, $\frac{1}{x} \to + \infty$ , so $e^{1 / x} \to + \infty$ . One infinite side is enough to make the line $x = 0$ a vertical asymptote, so the discontinuity is infinite; it is not a jump, because a jump needs both sides finite.
No. As $x \to 0$ , $\frac{1}{x}$ runs through infinitely many full turns, so $\sin (\frac{1}{x})$ keeps returning to $1$ and to $- 1$ on both sides, and neither one-sided limit exists. That rules out removable and jump. The values never leave the interval from $- 1$ to $1$ , so it is not infinite either. The AP framework lists this very limit among its examples of limits that do not exist, beside the statement that oscillation is one way that happens. Some teachers call the break an “oscillating discontinuity”, but that is a description, not a fourth official type.
False. Cancel every common factor first. At a zero of the original denominator, either the simplified denominator is not zero there, and then both one-sided limits equal the simplified expression's value (removable), or it is zero while the simplified numerator is not, and then the values grow without bound (infinite). Two finite, different one-sided limits never happen.

Practise it

The free AP Calculus AB Unit 1 worksheet has a rational function whose discontinuities you classify, one-sided limits to read off a graph, and a discontinuity to remove, alongside asymptotes and the Intermediate Value Theorem. If you are in CEGEP, the Calculus I Limits and Continuity worksheet covers the same three types. Classify each break with the one-sided limits first, then check your answer against the graph.

Want one-on-one help with this? I offer AP Calculus AB tutoring and exam preparation in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.

Free worksheets for this

Sources

AP Calculus AB and BC Course and Exam Description (Unit 1, Topics 1.3 and 1.10) — College Board. Checked 2026-10-05.