Unit 1Types of Discontinuities

Concept 4.275Built for May 2027 AP Exam

Types of Discontinuities

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

A discontinuity is not one condition but four, and the type is determined entirely by the two one-sided limits. If both exist and agree while the value is missing or displaced, the break is removable. If both exist but disagree, it is a jump. If at least one side is unbounded, the break is infinite and the graph has a vertical asymptote. If neither side settles at all, the break comes from oscillation. The name matters because it determines what you are allowed to do next: a removable break can be repaired by redefining a single value, a jump cannot be repaired at all, and an infinite break is not a defect in the definition but a statement about the function's behavior.

On the AP Exam

AP Calculus BC tests this skill through the representations and reasoning described below.

What you’ll see: A graph with a visible break to classify, a rational or piecewise formula whose behavior at a point must be diagnosed, or a table of values that shows one column heading in two directions.
What you’ll do: Compute the two one-sided limits, compare them, and let the comparison name the break. Only then decide whether the discontinuity is removable.
Memory hook: Two sides, one verdict: same, different, or infinite.
Use it on the exam: For MCQs: Classify from the graph in seconds. Agreeing sides with a hole is removable; disagreeing sides is a jump; an arrow along an asymptote is infinite. For FRQs: State the one-sided limits explicitly before naming the type. A classification without the supporting limits is an assertion, and it is regularly denied credit. On released BC papers: Classifying a discontinuity has not appeared as a scored free-response part across the 2017–2026 BC papers. The topic lives in the multiple-choice sections, where a graph or a formula is presented and four labels are offered, and it also does the diagnostic work inside free-response parts that ask about a vertical asymptote without naming the classification. The count is zero, and the payoff is indirect: knowing that a jump cannot be repaired prevents the wrong repair in every part that touches a piecewise function.

Concept Review

Four Signatures

Figure 1 — The three classic breaks, each with its diagnostic signature.

Caption. The shapes are different because the one-sided limits are different. A hole has agreeing sides and no value; a step has disagreeing sides; an asymptote has at least one side that never settles on a number.

Type	Left-Hand Limit	Right-Hand Limit	Value at $a$	Repairable?
Removable	$L$	$L$	Missing, or $\neq L$	Yes, by defining $f(a) = L$
Jump	$L_1$	$L_2 \neq L_1$	Any value or none	No
Infinite	$\pm\infty$ or DNE	$\pm\infty$ or DNE	Not finite	No
Oscillatory	Does not exist	Does not exist	Any value or none	No

The right-hand column is the part students skip. "Removable" is not a description of how the graph looks; it is a claim that a single redefinition restores continuity, and only the first row supports that claim.

Three small panels showing a removable hole, a jump between two one-sided levels, and a vertical asymptote, each labelled with its diagnostic signature

Reading the Type Off a Formula

For a rational expression, the first diagnostic is whether the numerator also vanishes at the point:

Denominator zero, numerator nonzero gives an infinite discontinuity, with the sign determined by the sign of each side.
Both zero does not yet decide anything. Factor or rationalize; if the obstruction cancels, the break is removable; if it does not, the break is infinite.
Neither zero gives continuity at that point, not a discontinuity.

For a piecewise definition, the diagnostic is the comparison of the two branch limits at the join. Equal limits with a mismatched value is removable; unequal limits is a jump.

Oscillation, the Fourth Type

The function $\sin(1/x)$ near $0$ has no one-sided limit on either side: it oscillates between $-1$ and $1$ without settling. Neither side is unbounded and neither side converges, so the break belongs to no category that involves a limit value. The exam mentions this type rarely, but recognising it prevents the error of calling an oscillation "infinite" simply because the graph looks crowded.

Why the Classification Matters

The name of the break determines the correct response:

Removable: report the limit, and if asked, note that redefining the function at $a$ produces a continuous extension.
Jump: report that the two-sided limit does not exist, and give both one-sided values. There is no repair.
Infinite: report that the limit does not exist because the function is unbounded, and state the behavior on each side using $\infty$ or $-\infty$ notation.
Oscillatory: report that the limit does not exist because the function fails to settle, and say so in words rather than with an infinity symbol.

Common Misconception Check

Myth 1: "Every discontinuity can be removed by defining the missing value."
- Truth: Only a removable one. A jump has two competing limit values, and no single definition satisfies both.
Myth 2: "A zero in the denominator always means a vertical asymptote."
- Truth: It means the function is undefined there. Whether the graph runs to infinity depends on the numerator, and a common factor produces a hole instead.
Myth 3: "If the limit does not exist, the discontinuity is infinite."
- Truth: Oscillation also produces a failed limit with no unbounded behavior. The distinction is whether either side runs away or merely refuses to settle.

Topic-Specific Study Method

Classify Before You Compute Anything Else

Borrowed idea: Name the break and the remedy follows. A hole wants a definition. A jump wants honesty: the limit genuinely does not exist, and no redefinition repairs it. An asymptote wants a different question, because the function is not small there at all. Classify first. Every repair you will attempt depends on the classification being right, and the classification is a ten-second reading of two one-sided limits, so the order of operations matters more than the algebra that follows.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
Graph with a hole	Two branches meeting an open circle	Do the sides agree, and is the value missing?	Both one-sided limits, then compare to the plotted point
Graph with a step	Two levels joined by a vertical gap	Do the sides disagree, and by how much?	Report $L_1$ and $L_2$ separately, then DNE
Graph with an asymptote	A branch leaving the frame	Which side diverges, and with what sign?	Describe each side with $\pm\infty$
Rational formula	Denominator zero at $a$	Does the numerator vanish too?	Check the numerator before naming the type
Piecewise formula	Two expressions joined at $c$	Are the branch limits equal?	Compare the two one-sided limits at $c$

Executable Action Steps: ① Write both one-sided limits before naming anything. The type is a function of those two numbers, and naming first invites the wrong category. ② Check the numerator whenever the denominator vanishes. A shared factor converts an apparent asymptote into a removable hole, and that single check changes the entire answer. ③ Say which clause of the definition fails, not just the type. "Removable because the limit exists and the value does not" is a complete diagnosis; "removable" alone is a label.

Training & Error Diagnosis

Core Archetype (Simplest Form): Classify the discontinuity of $f(x) = \dfrac{x - 1}{x^2 - 1}$ at $x = 1$ .

Standard Execution:

The denominator vanishes at $x = 1$ and so does the numerator, so the type is not yet determined.

Factor: $f(x) = \frac{x - 1}{(x-1)(x+1)} = \frac{1}{x+1}$ for $x \neq 1$ .

Both one-sided limits equal $\frac{1}{2}$ .

The limit exists and the value does not, so the discontinuity is removable, repaired by defining $f(1) = \frac{1}{2}$ .

Exam Variation (Context / Parameter Shift): 1. A Jump. Classify the discontinuity of $g(x) = \dfrac{\lvert x - 2 \rvert}{x - 2}$ at $x = 2$ .

Solution: For $x < 2$ the quotient is $-1$ ; for $x > 2$ it is $1$ . Both one-sided limits exist and differ, so the break is a jump of height $2$ , and the two-sided limit does not exist.

2. An Infinite Break With Matching Signs. Classify the discontinuity of $h(x) = \dfrac{1}{(x-3)^2}$ at $x = 3$ .

Solution: The denominator is positive on both sides and approaches $0$ , so both one-sided limits are $+\infty$ . The break is infinite, and writing $+\infty$ for both sides is more informative than writing DNE alone.

Common Traps & Failure Signals:

The automatic asymptote (Signal: you named the type from the denominator alone): A vanishing denominator with a vanishing numerator is a candidate for a hole, not an asymptote.
The mislabeled jump (Signal: you wrote DNE without giving either one-sided value): A jump is identified by its two levels, and reporting both is the point of the classification.
The oscillation called infinite (Signal: you used $\infty$ for a bounded function that failed to settle): Unbounded and unsettled are different failures, and the notation differs.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Repair Attempted: You redefined a function at a jump.
- Misclassified Break: Because you read the denominator's zero as proof of an asymptote rather than checking whether the numerator vanished too, you reached for a repair that cannot work and then reported a limit that does not exist.
- One Side Read: You examined the left-hand behavior and generalized to both sides without checking whether they agreed.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No part asks you to name a discontinuity; the classification does diagnostic work instead
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Zero parts, since the label is a means to a repair rather than an answer
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Removable or Infinite?

The function $f(x) = \dfrac{x^2 - 1}{x - 1}$ has a discontinuity at $x = 1$ . Which type is it?

MCQ 2

A Jump

The function $g(x) = \dfrac{\lvert x \rvert}{x}$ for $x \neq 0$ has which type of discontinuity at $x = 0$ ?

MCQ 3

Same Side Both Ways

The function $h(x) = \dfrac{3}{(x - 2)^2}$ has which type of discontinuity at $x = 2$ ?

FRQ 4

Classifying Three Breaks

Consider the following functions at the indicated points.

$f(x) = \frac{x^2 - 4}{x - 2} \text{ at } x = 2, \qquad g(x) = \frac{x + 3}{x - 1} \text{ at } x = 1, \qquad h(x) = \begin{cases} x + 1 & x < 0 \cr x^2 + 2 & x \ge 0 \end{cases} \text{ at } x = 0$

(a) Classify the discontinuity of $f$ at $x = 2$ , and state whether it is removable.

(b) Classify the discontinuity of $g$ at $x = 1$ , giving both one-sided limits.

(c) Classify the discontinuity of $h$ at $x = 0$ , giving both one-sided limits and the function value.

Cheatsheet

Core Reference

	Removable	Jump	Infinite	Oscillatory
Left-hand limit	$L$	$L_1$	$\pm\infty$	Does not exist
Right-hand limit	$L$	$L_2 \neq L_1$	$\pm\infty$	Does not exist
$f(a)$	Missing or $\neq L$	Any value	Not finite	Any value
Two-sided limit	Equals $L$	Does not exist	Does not exist	Does not exist
Repairable	Yes, set $f(a) = L$	No	No	No
Typical source	A cancelled factor	A piecewise join	A zero denominator without cancellation	$\sin(1/x)$ near $0$

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Start Topic Practice Next Topic: 1.15 Definition of Continuity

Unit 1Types of Discontinuities

Concept 4.275Built for May 2027 AP Exam

Types of Discontinuities

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus BC tests this skill through the representations and reasoning described below.

What you’ll see: A graph with a visible break to classify, a rational or piecewise formula whose behavior at a point must be diagnosed, or a table of values that shows one column heading in two directions.
What you’ll do: Compute the two one-sided limits, compare them, and let the comparison name the break. Only then decide whether the discontinuity is removable.
Memory hook: Two sides, one verdict: same, different, or infinite.
Use it on the exam: For MCQs: Classify from the graph in seconds. Agreeing sides with a hole is removable; disagreeing sides is a jump; an arrow along an asymptote is infinite. For FRQs: State the one-sided limits explicitly before naming the type. A classification without the supporting limits is an assertion, and it is regularly denied credit. On released BC papers: Classifying a discontinuity has not appeared as a scored free-response part across the 2017–2026 BC papers. The topic lives in the multiple-choice sections, where a graph or a formula is presented and four labels are offered, and it also does the diagnostic work inside free-response parts that ask about a vertical asymptote without naming the classification. The count is zero, and the payoff is indirect: knowing that a jump cannot be repaired prevents the wrong repair in every part that touches a piecewise function.

Concept Review

Four Signatures

Figure 1 — The three classic breaks, each with its diagnostic signature.

Type	Left-Hand Limit	Right-Hand Limit	Value at $a$	Repairable?
Removable	$L$	$L$	Missing, or $\neq L$	Yes, by defining $f(a) = L$
Jump	$L_1$	$L_2 \neq L_1$	Any value or none	No
Infinite	$\pm\infty$ or DNE	$\pm\infty$ or DNE	Not finite	No
Oscillatory	Does not exist	Does not exist	Any value or none	No

Three small panels showing a removable hole, a jump between two one-sided levels, and a vertical asymptote, each labelled with its diagnostic signature

Reading the Type Off a Formula

For a rational expression, the first diagnostic is whether the numerator also vanishes at the point:

Denominator zero, numerator nonzero gives an infinite discontinuity, with the sign determined by the sign of each side.
Both zero does not yet decide anything. Factor or rationalize; if the obstruction cancels, the break is removable; if it does not, the break is infinite.
Neither zero gives continuity at that point, not a discontinuity.

For a piecewise definition, the diagnostic is the comparison of the two branch limits at the join. Equal limits with a mismatched value is removable; unequal limits is a jump.

Oscillation, the Fourth Type

Why the Classification Matters

The name of the break determines the correct response:

Removable: report the limit, and if asked, note that redefining the function at $a$ produces a continuous extension.
Jump: report that the two-sided limit does not exist, and give both one-sided values. There is no repair.
Infinite: report that the limit does not exist because the function is unbounded, and state the behavior on each side using $\infty$ or $-\infty$ notation.
Oscillatory: report that the limit does not exist because the function fails to settle, and say so in words rather than with an infinity symbol.

Common Misconception Check

Myth 1: "Every discontinuity can be removed by defining the missing value."
- Truth: Only a removable one. A jump has two competing limit values, and no single definition satisfies both.
Myth 2: "A zero in the denominator always means a vertical asymptote."
- Truth: It means the function is undefined there. Whether the graph runs to infinity depends on the numerator, and a common factor produces a hole instead.
Myth 3: "If the limit does not exist, the discontinuity is infinite."
- Truth: Oscillation also produces a failed limit with no unbounded behavior. The distinction is whether either side runs away or merely refuses to settle.

Topic-Specific Study Method

Classify Before You Compute Anything Else

Borrowed idea: Name the break and the remedy follows. A hole wants a definition. A jump wants honesty: the limit genuinely does not exist, and no redefinition repairs it. An asymptote wants a different question, because the function is not small there at all. Classify first. Every repair you will attempt depends on the classification being right, and the classification is a ten-second reading of two one-sided limits, so the order of operations matters more than the algebra that follows.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
Graph with a hole	Two branches meeting an open circle	Do the sides agree, and is the value missing?	Both one-sided limits, then compare to the plotted point
Graph with a step	Two levels joined by a vertical gap	Do the sides disagree, and by how much?	Report $L_1$ and $L_2$ separately, then DNE
Graph with an asymptote	A branch leaving the frame	Which side diverges, and with what sign?	Describe each side with $\pm\infty$
Rational formula	Denominator zero at $a$	Does the numerator vanish too?	Check the numerator before naming the type
Piecewise formula	Two expressions joined at $c$	Are the branch limits equal?	Compare the two one-sided limits at $c$

Executable Action Steps: ① Write both one-sided limits before naming anything. The type is a function of those two numbers, and naming first invites the wrong category. ② Check the numerator whenever the denominator vanishes. A shared factor converts an apparent asymptote into a removable hole, and that single check changes the entire answer. ③ Say which clause of the definition fails, not just the type. "Removable because the limit exists and the value does not" is a complete diagnosis; "removable" alone is a label.

Training & Error Diagnosis

Core Archetype (Simplest Form): Classify the discontinuity of $f(x) = \dfrac{x - 1}{x^2 - 1}$ at $x = 1$ .

Standard Execution:

The denominator vanishes at $x = 1$ and so does the numerator, so the type is not yet determined.

Factor: $f(x) = \frac{x - 1}{(x-1)(x+1)} = \frac{1}{x+1}$ for $x \neq 1$ .

Both one-sided limits equal $\frac{1}{2}$ .

The limit exists and the value does not, so the discontinuity is removable, repaired by defining $f(1) = \frac{1}{2}$ .

Exam Variation (Context / Parameter Shift): 1. A Jump. Classify the discontinuity of $g(x) = \dfrac{\lvert x - 2 \rvert}{x - 2}$ at $x = 2$ .

Solution: For $x < 2$ the quotient is $-1$ ; for $x > 2$ it is $1$ . Both one-sided limits exist and differ, so the break is a jump of height $2$ , and the two-sided limit does not exist.

2. An Infinite Break With Matching Signs. Classify the discontinuity of $h(x) = \dfrac{1}{(x-3)^2}$ at $x = 3$ .

Solution: The denominator is positive on both sides and approaches $0$ , so both one-sided limits are $+\infty$ . The break is infinite, and writing $+\infty$ for both sides is more informative than writing DNE alone.

Common Traps & Failure Signals:

The automatic asymptote (Signal: you named the type from the denominator alone): A vanishing denominator with a vanishing numerator is a candidate for a hole, not an asymptote.
The mislabeled jump (Signal: you wrote DNE without giving either one-sided value): A jump is identified by its two levels, and reporting both is the point of the classification.
The oscillation called infinite (Signal: you used $\infty$ for a bounded function that failed to settle): Unbounded and unsettled are different failures, and the notation differs.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Repair Attempted: You redefined a function at a jump.
- Misclassified Break: Because you read the denominator's zero as proof of an asymptote rather than checking whether the numerator vanished too, you reached for a repair that cannot work and then reported a limit that does not exist.
- One Side Read: You examined the left-hand behavior and generalized to both sides without checking whether they agreed.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No part asks you to name a discontinuity; the classification does diagnostic work instead
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Zero parts, since the label is a means to a repair rather than an answer
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Removable or Infinite?

The function $f(x) = \dfrac{x^2 - 1}{x - 1}$ has a discontinuity at $x = 1$ . Which type is it?

MCQ 2

A Jump

The function $g(x) = \dfrac{\lvert x \rvert}{x}$ for $x \neq 0$ has which type of discontinuity at $x = 0$ ?

MCQ 3

Same Side Both Ways

The function $h(x) = \dfrac{3}{(x - 2)^2}$ has which type of discontinuity at $x = 2$ ?

FRQ 4

Classifying Three Breaks

Consider the following functions at the indicated points.

$f(x) = \frac{x^2 - 4}{x - 2} \text{ at } x = 2, \qquad g(x) = \frac{x + 3}{x - 1} \text{ at } x = 1, \qquad h(x) = \begin{cases} x + 1 & x < 0 \cr x^2 + 2 & x \ge 0 \end{cases} \text{ at } x = 0$

(a) Classify the discontinuity of $f$ at $x = 2$ , and state whether it is removable.

(b) Classify the discontinuity of $g$ at $x = 1$ , giving both one-sided limits.

(c) Classify the discontinuity of $h$ at $x = 0$ , giving both one-sided limits and the function value.

Cheatsheet

Core Reference

	Removable	Jump	Infinite	Oscillatory
Left-hand limit	$L$	$L_1$	$\pm\infty$	Does not exist
Right-hand limit	$L$	$L_2 \neq L_1$	$\pm\infty$	Does not exist
$f(a)$	Missing or $\neq L$	Any value	Not finite	Any value
Two-sided limit	Equals $L$	Does not exist	Does not exist	Does not exist
Repairable	Yes, set $f(a) = L$	No	No	No
Typical source	A cancelled factor	A piecewise join	A zero denominator without cancellation	$\sin(1/x)$ near $0$

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Start Topic Practice Next Topic: 1.15 Definition of Continuity

Unit 1Types of Discontinuities

Concept 4.275Built for May 2027 AP Exam

Types of Discontinuities

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus BC tests this skill through the representations and reasoning described below.

What you’ll see: A graph with a visible break to classify, a rational or piecewise formula whose behavior at a point must be diagnosed, or a table of values that shows one column heading in two directions.
What you’ll do: Compute the two one-sided limits, compare them, and let the comparison name the break. Only then decide whether the discontinuity is removable.
Memory hook: Two sides, one verdict: same, different, or infinite.
Use it on the exam: For MCQs: Classify from the graph in seconds. Agreeing sides with a hole is removable; disagreeing sides is a jump; an arrow along an asymptote is infinite. For FRQs: State the one-sided limits explicitly before naming the type. A classification without the supporting limits is an assertion, and it is regularly denied credit. On released BC papers: Classifying a discontinuity has not appeared as a scored free-response part across the 2017–2026 BC papers. The topic lives in the multiple-choice sections, where a graph or a formula is presented and four labels are offered, and it also does the diagnostic work inside free-response parts that ask about a vertical asymptote without naming the classification. The count is zero, and the payoff is indirect: knowing that a jump cannot be repaired prevents the wrong repair in every part that touches a piecewise function.

Concept Review

Four Signatures

Figure 1 — The three classic breaks, each with its diagnostic signature.

Type	Left-Hand Limit	Right-Hand Limit	Value at $a$	Repairable?
Removable	$L$	$L$	Missing, or $\neq L$	Yes, by defining $f(a) = L$
Jump	$L_1$	$L_2 \neq L_1$	Any value or none	No
Infinite	$\pm\infty$ or DNE	$\pm\infty$ or DNE	Not finite	No
Oscillatory	Does not exist	Does not exist	Any value or none	No

Three small panels showing a removable hole, a jump between two one-sided levels, and a vertical asymptote, each labelled with its diagnostic signature

Reading the Type Off a Formula

For a rational expression, the first diagnostic is whether the numerator also vanishes at the point:

Denominator zero, numerator nonzero gives an infinite discontinuity, with the sign determined by the sign of each side.
Both zero does not yet decide anything. Factor or rationalize; if the obstruction cancels, the break is removable; if it does not, the break is infinite.
Neither zero gives continuity at that point, not a discontinuity.

For a piecewise definition, the diagnostic is the comparison of the two branch limits at the join. Equal limits with a mismatched value is removable; unequal limits is a jump.

Oscillation, the Fourth Type

Why the Classification Matters

The name of the break determines the correct response:

Removable: report the limit, and if asked, note that redefining the function at $a$ produces a continuous extension.
Jump: report that the two-sided limit does not exist, and give both one-sided values. There is no repair.
Infinite: report that the limit does not exist because the function is unbounded, and state the behavior on each side using $\infty$ or $-\infty$ notation.
Oscillatory: report that the limit does not exist because the function fails to settle, and say so in words rather than with an infinity symbol.

Common Misconception Check

Myth 1: "Every discontinuity can be removed by defining the missing value."
- Truth: Only a removable one. A jump has two competing limit values, and no single definition satisfies both.
Myth 2: "A zero in the denominator always means a vertical asymptote."
- Truth: It means the function is undefined there. Whether the graph runs to infinity depends on the numerator, and a common factor produces a hole instead.
Myth 3: "If the limit does not exist, the discontinuity is infinite."
- Truth: Oscillation also produces a failed limit with no unbounded behavior. The distinction is whether either side runs away or merely refuses to settle.

Topic-Specific Study Method

Classify Before You Compute Anything Else

Borrowed idea: Name the break and the remedy follows. A hole wants a definition. A jump wants honesty: the limit genuinely does not exist, and no redefinition repairs it. An asymptote wants a different question, because the function is not small there at all. Classify first. Every repair you will attempt depends on the classification being right, and the classification is a ten-second reading of two one-sided limits, so the order of operations matters more than the algebra that follows.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
Graph with a hole	Two branches meeting an open circle	Do the sides agree, and is the value missing?	Both one-sided limits, then compare to the plotted point
Graph with a step	Two levels joined by a vertical gap	Do the sides disagree, and by how much?	Report $L_1$ and $L_2$ separately, then DNE
Graph with an asymptote	A branch leaving the frame	Which side diverges, and with what sign?	Describe each side with $\pm\infty$
Rational formula	Denominator zero at $a$	Does the numerator vanish too?	Check the numerator before naming the type
Piecewise formula	Two expressions joined at $c$	Are the branch limits equal?	Compare the two one-sided limits at $c$

Executable Action Steps: ① Write both one-sided limits before naming anything. The type is a function of those two numbers, and naming first invites the wrong category. ② Check the numerator whenever the denominator vanishes. A shared factor converts an apparent asymptote into a removable hole, and that single check changes the entire answer. ③ Say which clause of the definition fails, not just the type. "Removable because the limit exists and the value does not" is a complete diagnosis; "removable" alone is a label.

Training & Error Diagnosis

Core Archetype (Simplest Form): Classify the discontinuity of $f(x) = \dfrac{x - 1}{x^2 - 1}$ at $x = 1$ .

Standard Execution:

The denominator vanishes at $x = 1$ and so does the numerator, so the type is not yet determined.

Factor: $f(x) = \frac{x - 1}{(x-1)(x+1)} = \frac{1}{x+1}$ for $x \neq 1$ .

Both one-sided limits equal $\frac{1}{2}$ .

The limit exists and the value does not, so the discontinuity is removable, repaired by defining $f(1) = \frac{1}{2}$ .

Exam Variation (Context / Parameter Shift): 1. A Jump. Classify the discontinuity of $g(x) = \dfrac{\lvert x - 2 \rvert}{x - 2}$ at $x = 2$ .

Solution: For $x < 2$ the quotient is $-1$ ; for $x > 2$ it is $1$ . Both one-sided limits exist and differ, so the break is a jump of height $2$ , and the two-sided limit does not exist.

2. An Infinite Break With Matching Signs. Classify the discontinuity of $h(x) = \dfrac{1}{(x-3)^2}$ at $x = 3$ .

Solution: The denominator is positive on both sides and approaches $0$ , so both one-sided limits are $+\infty$ . The break is infinite, and writing $+\infty$ for both sides is more informative than writing DNE alone.

Common Traps & Failure Signals:

The automatic asymptote (Signal: you named the type from the denominator alone): A vanishing denominator with a vanishing numerator is a candidate for a hole, not an asymptote.
The mislabeled jump (Signal: you wrote DNE without giving either one-sided value): A jump is identified by its two levels, and reporting both is the point of the classification.
The oscillation called infinite (Signal: you used $\infty$ for a bounded function that failed to settle): Unbounded and unsettled are different failures, and the notation differs.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Repair Attempted: You redefined a function at a jump.
- Misclassified Break: Because you read the denominator's zero as proof of an asymptote rather than checking whether the numerator vanished too, you reached for a repair that cannot work and then reported a limit that does not exist.
- One Side Read: You examined the left-hand behavior and generalized to both sides without checking whether they agreed.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No part asks you to name a discontinuity; the classification does diagnostic work instead
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Zero parts, since the label is a means to a repair rather than an answer
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Removable or Infinite?

The function $f(x) = \dfrac{x^2 - 1}{x - 1}$ has a discontinuity at $x = 1$ . Which type is it?

MCQ 2

A Jump

The function $g(x) = \dfrac{\lvert x \rvert}{x}$ for $x \neq 0$ has which type of discontinuity at $x = 0$ ?

MCQ 3

Same Side Both Ways

The function $h(x) = \dfrac{3}{(x - 2)^2}$ has which type of discontinuity at $x = 2$ ?

FRQ 4

Classifying Three Breaks

Consider the following functions at the indicated points.

$f(x) = \frac{x^2 - 4}{x - 2} \text{ at } x = 2, \qquad g(x) = \frac{x + 3}{x - 1} \text{ at } x = 1, \qquad h(x) = \begin{cases} x + 1 & x < 0 \cr x^2 + 2 & x \ge 0 \end{cases} \text{ at } x = 0$

(a) Classify the discontinuity of $f$ at $x = 2$ , and state whether it is removable.

(b) Classify the discontinuity of $g$ at $x = 1$ , giving both one-sided limits.

(c) Classify the discontinuity of $h$ at $x = 0$ , giving both one-sided limits and the function value.

Cheatsheet

Core Reference

	Removable	Jump	Infinite	Oscillatory
Left-hand limit	$L$	$L_1$	$\pm\infty$	Does not exist
Right-hand limit	$L$	$L_2 \neq L_1$	$\pm\infty$	Does not exist
$f(a)$	Missing or $\neq L$	Any value	Not finite	Any value
Two-sided limit	Equals $L$	Does not exist	Does not exist	Does not exist
Repairable	Yes, set $f(a) = L$	No	No	No
Typical source	A cancelled factor	A piecewise join	A zero denominator without cancellation	$\sin(1/x)$ near $0$

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