What is the difference between a removable and a jump discontinuity?

For a removable discontinuity the two-sided limit exists (both one-sided limits agree) but the function value is missing or wrong, creating a hole. For a jump discontinuity the one-sided limits are finite but unequal, so the two-sided limit does not exist and the graph jumps between two heights.

Is an infinite discontinuity the same as an essential discontinuity?

In AP Calculus, yes. An infinite discontinuity occurs when at least one one-sided limit is +∞ or -∞, producing a vertical asymptote. The College Board framework uses 'infinite (essential)' as the name for this type.

How do I quickly tell which type a rational function has?

Factor the numerator and denominator. If a denominator factor cancels with the numerator, that zero gives a removable hole. If a denominator factor does not cancel, that zero gives an infinite discontinuity with a vertical asymptote.

Which discontinuities can actually be removed?

Only removable discontinuities can be removed, by redefining the function at the single point so its value equals the two-sided limit. Jump and infinite discontinuities cannot be fixed because the two-sided limit does not exist.

AP-CALCBC-1.10

U1.10 Exploring Types of Discontinuities

Learn to classify removable, jump, and infinite discontinuities in AP Calculus BC. Includes definitions, tests, worked examples, and practice questions.

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A voice-first session with the Crimsora tutor on U1.10 Exploring Types of Discontinuities, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

A function can break in more than one way, and AP Calculus wants you to name the break precisely. In topic 1.10 you will learn to look at a point where a function fails to be continuous and decide whether the discontinuity is removable (a hole), a jump, or infinite (also called essential). The tool you use is the one-sided limit: comparing

\lim_{x\to c^-}f(x)

\lim_{x\to c^+}f(x)

, and

f(c)

tells you exactly which case you have. This skill sets up the formal definition of continuity you will meet in 1.11 and the process of removing discontinuities in 1.13, so mastering the classification now pays off across the whole unit.

Why discontinuities have types

Recall from the limit lessons that a two-sided limit

\lim_{x\to c}f(x)

exists only when the left-hand limit and the right-hand limit exist and are equal. A discontinuity at

x=c

means that at least one of three things fails: the limit does not exist,

f(c)

is undefined, or the limit exists but does not equal

f(c)

. The reason we sort discontinuities into categories is that each category behaves differently and can be tested with the same three ingredients.

The three ingredients you always check are the left-hand limit

\lim_{x\to c^-}f(x)

, the right-hand limit

\lim_{x\to c^+}f(x)

, and the function value

f(c)

. Every classification question reduces to comparing these three quantities.

Type	One-sided limits	Two-sided limit	Value $f(c)$
Removable	equal, finite	exists	missing or wrong
Jump	finite but unequal	does not exist	any
Infinite	at least one is $\pm\infty$	does not exist	undefined

A function is continuous at

c

only when all three ingredients agree: both one-sided limits equal each other and equal

f(c)

. Any deviation from this produces one of the three discontinuity types below.

Removable discontinuities (holes)

A discontinuity at

x=c

is removable when

\lim_{x\to c}f(x)

exists as a finite number, but

f

either is undefined at

c

or is defined with a different value. Graphically this looks like a smooth curve with a single point punched out — a hole — sometimes with a stray dot floating above or below.

The classic algebraic signal is a factor that cancels. Consider

f(x)=\frac{x^2-9}{x-3}

. Factoring gives

\frac{(x-3)(x+3)}{x-3}=x+3

for

x\neq 3

, so

\lim_{x\to 3}f(x)=6

. Because the limit exists but

f(3)

is undefined, the discontinuity is removable. You could 'remove' it by defining

f(3)=6

, which is exactly the idea explored in 1.13.

A common misconception is that a hole must come from a

\frac{0}{0}

form. A piecewise function can also have a removable discontinuity if the limit exists but the assigned value disagrees. For example, if

g(x)=x+3

for

x\neq 3

and

g(3)=10

, the limit is still

6

but

g(3)=10

, so there is a removable discontinuity. The defining test is always: does the two-sided limit exist and yet fail to equal

f(c)

Jump and infinite discontinuities

A jump discontinuity occurs when both one-sided limits exist and are finite but are not equal, so the two-sided limit does not exist. The graph literally jumps from one height to another. These appear most often in piecewise functions and in step-like models. For

h(x)=\frac{|x|}{x}

x=0

, the left limit is

-1

and the right limit is

+1

; the mismatch is a jump.

An infinite discontinuity (also called essential) occurs when at least one one-sided limit is

+\infty

-\infty

. This is the vertical-asymptote case studied more in 1.14. For

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

, as

x\to 2^-

the values go to

-\infty

and as

x\to 2^+

they go to

+\infty

, so

x=2

is an infinite discontinuity.

Question	Removable	Jump	Infinite
Do both one-sided limits exist finitely?	yes, equal	yes, unequal	no
Does a vertical asymptote appear?	no	no	yes
Can it be fixed by redefining one point?	yes	no	no

A key distinction: only removable discontinuities can be repaired by changing a single point. Jump and infinite discontinuities are permanent because the limit itself does not exist.

How the exam tests classification

The AP exam probes this skill in three formats. Graph-reading questions show a curve and ask you to identify the type at a marked

x

-value; look for holes (removable), broken levels (jump), and asymptotes (infinite). Analytic questions give a formula, often rational, and ask you to classify each discontinuity after factoring the denominator. Factors that cancel yield removable holes; factors that remain in the denominator yield infinite discontinuities.

For

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

, the

(x-1)

cancels, giving a removable hole at

x=1

, while

(x+2)

survives, giving an infinite discontinuity at

x=-2

. Being able to distinguish these two on one function is a frequently tested pattern.

Piecewise-defined functions test jump versus removable: compute both one-sided limits and compare them to each other and to

f(c)

. A frequent trap is stopping after finding the limit exists without checking

f(c)

, or forgetting that unequal one-sided limits mean the two-sided limit fails to exist. On free-response items you must justify with limit language, not just name the type — state the values of the one-sided limits explicitly to earn full credit.

Key terms

Removable discontinuity.: A point where $\lim_{x\to c}f(x)$ exists finitely but $f(c)$ is undefined or unequal to that limit; appears as a hole.
Jump discontinuity.: A point where both one-sided limits are finite but unequal, so the two-sided limit does not exist.
Infinite (essential) discontinuity.: A point where at least one one-sided limit is $+\infty$ or $-\infty$ , producing a vertical asymptote.
One-sided limit.: The value $f(x)$ approaches as $x$ nears $c$ from only the left ( $c^-$ ) or only the right ( $c^+$ ).
Two-sided limit.: $\lim_{x\to c}f(x)$ ; exists only when the left and right one-sided limits are equal and finite.
Continuity at a point.: The condition that $\lim_{x\to c}f(x)$ exists, $f(c)$ is defined, and the two are equal.
Vertical asymptote.: A vertical line $x=c$ that a graph approaches without bound, signaling an infinite discontinuity.

Worked example

Classify each discontinuity of

f(x)=\dfrac{x^2-x-6}{x^2-9}

Start by factoring numerator and denominator. The numerator factors as

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

and the denominator as

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

.

The function is undefined where the denominator is zero: at

x=3

and

x=-3

. These are the candidate discontinuities.

At

x=3

, the factor

(x-3)

cancels:

f(x)=\frac{(x-3)(x+2)}{(x-3)(x+3)}=\frac{x+2}{x+3}

for

x\neq 3

. Now take the limit:

\lim_{x\to 3}\frac{x+2}{x+3}=\frac{5}{6}

. The limit exists and is finite, but

f(3)

is undefined, so

x=3

is a removable discontinuity (a hole at height

\frac{5}{6}

).

At

x=-3

, the factor

(x+3)

does not cancel; it remains in the denominator. As

x\to -3

, the numerator

x+2\to -1

while the denominator

x+3\to 0

, so the quotient grows without bound. Checking signs,

\lim_{x\to -3^-}f(x)=+\infty

and

\lim_{x\to -3^+}f(x)=-\infty

. Because the values are unbounded,

x=-3

is an infinite (essential) discontinuity with a vertical asymptote.

Conclusion: removable at

x=3

, infinite at

x=-3

Practice questions

The function

f

satisfies

\lim_{x\to 4^-}f(x)=2

\lim_{x\to 4^+}f(x)=7

, and

f(4)=2

. What type of discontinuity does

f

have at

x=4

Removable
Jump
Infinite
No discontinuity

Answer: Jump

The two one-sided limits,

2

and

7

, are both finite but unequal, so the two-sided limit does not exist. Finite but unequal one-sided limits define a jump discontinuity. The value

f(4)=2

is irrelevant to the classification here because the limit itself fails to exist.

Let

g(x)=\dfrac{x+5}{x^2+4x-5}

. Identify and classify every discontinuity.

Answer: Removable at $x=-5$ ; infinite at $x=1$ .

Factor the denominator:

x^2+4x-5=(x+5)(x-1)

. At

x=-5

the

(x+5)

factor cancels with the numerator, leaving

\frac{1}{x-1}

, whose limit as

x\to -5

-\frac{1}{6}

, a finite value — so

x=-5

is a removable hole. At

x=1

the factor

(x-1)

remains in the denominator, and the quotient grows without bound, giving an infinite discontinuity with a vertical asymptote.

A graph shows a curve that rises to an open circle at

(2,5)

, and a separate filled dot sits at

(2,1)

. Classify the discontinuity at

x=2

and explain whether it can be removed.

Answer: Removable discontinuity; it can be removed by redefining $f(2)=5$ .

The open circle at height

5

means both one-sided limits approach

5

, so

\lim_{x\to 2}f(x)=5

exists. The filled dot at height

1

means

f(2)=1\neq 5

. Since the limit exists but disagrees with the function value, the discontinuity is removable and can be fixed by redefining

f(2)

to equal the limit,

5

FAQ

What is the difference between a removable and a jump discontinuity?: For a removable discontinuity the two-sided limit exists (both one-sided limits agree) but the function value is missing or wrong, creating a hole. For a jump discontinuity the one-sided limits are finite but unequal, so the two-sided limit does not exist and the graph jumps between two heights.
Is an infinite discontinuity the same as an essential discontinuity?: In AP Calculus, yes. An infinite discontinuity occurs when at least one one-sided limit is $+\infty$ or $-\infty$ , producing a vertical asymptote. The College Board framework uses 'infinite (essential)' as the name for this type.
How do I quickly tell which type a rational function has?: Factor the numerator and denominator. If a denominator factor cancels with the numerator, that zero gives a removable hole. If a denominator factor does not cancel, that zero gives an infinite discontinuity with a vertical asymptote.
Which discontinuities can actually be removed?: Only removable discontinuities can be removed, by redefining the function at the single point so its value equals the two-sided limit. Jump and infinite discontinuities cannot be fixed because the two-sided limit does not exist.

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