Unit 1Types of Discontinuities

Concept 4.106Built 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 a point where continuity fails, and the exam only recognises three kinds. Removable means the two-sided limit exists but the value does not match it, or does not exist at all: the hole can be plugged. Jump means both one-sided limits are finite and different: there is no single height to plug. Infinite means at least one side runs off to $\pm\infty$ : there is no height at all. The classification is decided by the one-sided limits, never by how the picture looks. A gap on a graph can be any of the three, and an undefined point is not automatically removable. Read the two limits, compare them, and the label follows.

On the AP Exam

Classification is a reading task with a scoring tail. Multiple choice gives you a graph or a formula and asks which label applies, or asks at which $x$ -values a given type occurs. Free response asks for the classification plus the limit evidence behind it, and the evidence is the point: "there is a hole" earns nothing, while " $\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) = 1$ but $f(1) = 2$ " earns the point.

What you’ll see: Rational functions with a cancelable factor, rational functions with a stubborn denominator factor, piecewise definitions whose branches disagree at the seam, and graphs with a visible break at a labelled $x$ -value.
What you’ll do: Compute or read both one-sided limits, compare them, and then check whether the value at the point matches. The first comparison names the type; the second decides whether the point is a hole or a misplaced dot.
Memory hook: Two limits, then the dot. Classify from the limits, then look at the value to see which flavour of removable you have.
Use it on the exam: For MCQs: Compute both one-sided limits before reading the choices. If they agree, the answer is removable and the value of $f(a)$ only tells you whether the dot is misplaced; if they disagree and both are finite, it is a jump; if either is unbounded, it is infinite. For FRQs: Quote the limits with their values in one sentence, then name the type. A justification that names the type without the two limits is a claim, not a justification, and the rubric scores the evidence.

Concept Review

The Three Labels and What Separates Them

Only the one-sided limits matter for the label. The value $f(a)$ matters for a second, narrower question: whether a removable discontinuity is a hole or a displaced point.

Type	Left and right limits	Two-sided limit	Is $f(a)$ involved in the label	Repair
Removable	Equal and finite	Exists	No. A mismatched or missing $f(a)$ is the usual cause.	Redefine $f(a)$ to equal the limit
Jump	Finite but unequal	Does not exist	No. $f(a)$ may be defined on either branch.	No repair exists
Infinite	At least one is $\pm\infty$	Does not exist	No. $f(a)$ is usually undefined, and irrelevant.	No repair exists

The column that decides everything is the first one. Students who classify from the picture rather than from the limits get the removable and infinite cases backwards roughly half the time, because both look like a break in the drawing.

Classifying in Three Moves

Confirm there is something to classify. Check the three continuity conditions at $a$ : $f(a)$ defined, the limit exists, and the two agree. If all three hold, stop; there is no discontinuity.
Compute both one-sided limits. Factor or conjugate if needed. If either side is unbounded, the label is infinite and you are done; the sign tells you which way each branch leaves.
Compare the two finite values. Equal means removable, unequal means jump. Then, and only then, look at $f(a)$ : if it exists and differs from the limit, the discontinuity is a displaced dot rather than a hole.

Figure 1 — Three breaks in one graph.

Caption. At $x = 1$ both sides approach height $1$ while the filled dot sits at height $2$ : removable, with a displaced value. At $x = 2$ the two sides land on heights $1$ and $-1$ : jump. At $x = 3.7$ the curve leaves the frame on both sides: infinite. Three different labels, one picture, and the picture is not what decides between them.

One graph showing three discontinuities: a hole at x equals 1 with a displaced filled dot at height 2, a jump at x equals 2 between heights 1 and minus 1, and a vertical asymptote at x equals 3.7

Three Corrections Worth Memorising

“The point is undefined, so it is a hole.”
- Correct: a hole requires equal finite one-sided limits. An undefined point with an unbounded side is an infinite discontinuity, and the exam builds distractors on exactly that confusion.
“The two branches do not meet, so it is a jump.”
- Correct: a jump requires both one-sided limits to be finite. If either runs to $\pm\infty$ , the label is infinite no matter how the drawing looks.
“ $f(a)$ is defined, so the function is continuous at $a$ .”
- Correct: defined is one of three conditions. A defined value that disagrees with the limit is still a discontinuity, and naming it removable is the correct call.

Topic-Specific Study Method

Two Limits, Then the Dot

Borrowed idea: Classify with the limits. Look at the dot last. Write both one-sided values on the page before you name anything. Equal finite values means removable. Unequal finite values means jump. Any unbounded side means infinite. Then check the dot. Do it in that order every time, and the picture stops being able to fool you.

Write the two one-sided limits as a pair. One line, both values, before any label. " $\lim_{x \to 1^{-}} f = 1$ , $\lim_{x \to 1^{+}} f = 1$ ." The pair is the evidence the rubric wants and the discipline that prevents picture-based guessing.
Apply the decision rule mechanically. Equal and finite, removable. Unequal and finite, jump. Any $\pm\infty$ , infinite. No exceptions, no case-by-case reasoning at the point.
Check $f(a)$ only after the label is fixed. If the label is removable and $f(a)$ disagrees with the limit, say so explicitly: the function has a removable discontinuity with a displaced value.
Name the repair. Removable discontinuities can be repaired by redefining one point; the other two cannot be repaired at all. Saying which is which is often the final part of an FRQ.

What the page shows	The two one-sided limits	Label	What to write
Rational with a cancelled factor	Equal finite	Removable	The pair, then "removable, since the two-sided limit exists"
Rational with a surviving $(x-a)$ downstairs	One or both unbounded	Infinite	The pair with signs, then "infinite, since the left side tends to $-\infty$ "
Piecewise with different formulas at the seam	Finite but unequal	Jump	Both values, then "jump, since the one-sided limits differ"
Absolute value quotient at its corner	Finite but unequal	Jump	Both values, then the label
Piecewise with matching formulas	Equal finite, matching value	Continuous	"No discontinuity; all three conditions hold"

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \dfrac{x^{2} - 1}{x - 1}$ for $x \neq 1$ and $f(1) = 3$ . Classify the discontinuity at $x = 1$ .

Standard Execution:

One-sided limits. For $x \neq 1$ , $f(x) = x + 1$ , so both one-sided limits equal $2$ .

Label. The two-sided limit exists and equals $2$ , so the discontinuity is removable.

Check the dot. $f(1) = 3 \neq 2$ , so the value is displaced. Redefining $f(1) = 2$ repairs it.

Exam Variation (Context / Parameter Shift): Let $g(x) = \begin{cases} x + 1, & x < 2 \cr 5 - x, & x \ge 2 \end{cases}$ . Classify the discontinuity at $x = 2$ , and state whether any single redefinition of $g(2)$ could remove it.

Solution: From the left, $\lim_{x \to 2^{-}} g(x) = 3$ . From the right, $\lim_{x \to 2^{+}} g(x) = 3$ . Both sides agree, so the discontinuity is removable despite the two branches looking different. Since $g(2) = 5 - 2 = 3$ already matches the limit, $g$ is in fact continuous at $2$ ; there is nothing to repair. A second version with $5 - 2x$ in place of $5 - x$ would give a right-hand limit of $1$ and a genuine jump.

Common Traps & Failure Signals:

The undefined-means-hole reflex (Signal: the item names a point where the function is undefined and one side is unbounded): Check the sides before labelling. Undefined plus unbounded is infinite, not removable.
The gap-means-jump reflex (Signal: a vertical asymptote drawn with a visible gap): A jump needs two finite heights. A gap into infinity is infinite.
The defined-means-continuous reflex (Signal: the item supplies $f(a)$ and asks about continuity): Defined is one condition out of three. Compare the limit to the value before you conclude.

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

Picture Classification: You name the type from the drawing.
Value Priority: You checked $f(a)$ first and let it decide the label.
Limit-Pair Omission: Because the label feels obvious from the graph, the two one-sided limits never get written down, and on a free response the classification arrives without the evidence the rubric requires, so the point for justification is lost even though the label itself is right.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No scored FRQ part in any released exam; breaks get named inside graph-reading items
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Removable, jump, and infinite causes named from a graph
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Displaced Dot

Let $f(x) = \dfrac{x^{2} - 1}{x - 1}$ for $x \neq 1$ , and let $f(1) = 3$ . Which statement describes the discontinuity of $f$ at $x = 1$ ?

MCQ 2

The Unbounded Side

Let $h(x) = \dfrac{3}{(x-2)^{2}}$ . Which statement describes the behaviour of $h$ at $x = 2$ ?

MCQ 3

Reading Two Branches

Let $k(x) = \begin{cases} 2x - 1, & x < 1 \cr x^{2} + 2, & x \ge 1 \end{cases}$ Which statement describes $k$ at $x = 1$ ?

FRQ 4

Three Points, Three Verdicts

Let $f(x) = \begin{cases} \dfrac{x^{2}-4}{x-2}, & x < 2 \cr 3, & x = 2 \cr \dfrac{1}{x-2}, & x > 2 \end{cases}$

(a) Evaluate $\lim_{x \to 2^{-}} f(x)$ and $\lim_{x \to 2^{+}} f(x)$ .

(b) Classify the behaviour of $f$ at $x = 2$ and justify the classification from your limits.

(c) The function $g(x) = \dfrac{x^{2}-9}{x-3}$ is undefined at $x = 3$ . Classify its discontinuity there and state the redefinition that removes it.

(d) A student writes: "Both functions have a gap at the point, so both discontinuities are removable." Identify the error.

Cheatsheet

Naming the Break

	Label	Two-sided limit	Repair by redefining one point
Equal and finite	Removable	Exists	Yes
Finite but unequal	Jump	Does not exist	No
At least one $\pm\infty$	Infinite	Does not exist	No
Equal and finite, matching $f(a)$	Continuous	Exists	Nothing to repair
One side finite, the other unbounded	Infinite	Does not exist	No

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Unit 1Types of Discontinuities

Concept 4.106Built 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

What you’ll see: Rational functions with a cancelable factor, rational functions with a stubborn denominator factor, piecewise definitions whose branches disagree at the seam, and graphs with a visible break at a labelled $x$ -value.
What you’ll do: Compute or read both one-sided limits, compare them, and then check whether the value at the point matches. The first comparison names the type; the second decides whether the point is a hole or a misplaced dot.
Memory hook: Two limits, then the dot. Classify from the limits, then look at the value to see which flavour of removable you have.
Use it on the exam: For MCQs: Compute both one-sided limits before reading the choices. If they agree, the answer is removable and the value of $f(a)$ only tells you whether the dot is misplaced; if they disagree and both are finite, it is a jump; if either is unbounded, it is infinite. For FRQs: Quote the limits with their values in one sentence, then name the type. A justification that names the type without the two limits is a claim, not a justification, and the rubric scores the evidence.

Concept Review

The Three Labels and What Separates Them

Only the one-sided limits matter for the label. The value $f(a)$ matters for a second, narrower question: whether a removable discontinuity is a hole or a displaced point.

Type	Left and right limits	Two-sided limit	Is $f(a)$ involved in the label	Repair
Removable	Equal and finite	Exists	No. A mismatched or missing $f(a)$ is the usual cause.	Redefine $f(a)$ to equal the limit
Jump	Finite but unequal	Does not exist	No. $f(a)$ may be defined on either branch.	No repair exists
Infinite	At least one is $\pm\infty$	Does not exist	No. $f(a)$ is usually undefined, and irrelevant.	No repair exists

Classifying in Three Moves

Confirm there is something to classify. Check the three continuity conditions at $a$ : $f(a)$ defined, the limit exists, and the two agree. If all three hold, stop; there is no discontinuity.
Compute both one-sided limits. Factor or conjugate if needed. If either side is unbounded, the label is infinite and you are done; the sign tells you which way each branch leaves.
Compare the two finite values. Equal means removable, unequal means jump. Then, and only then, look at $f(a)$ : if it exists and differs from the limit, the discontinuity is a displaced dot rather than a hole.

Figure 1 — Three breaks in one graph.

One graph showing three discontinuities: a hole at x equals 1 with a displaced filled dot at height 2, a jump at x equals 2 between heights 1 and minus 1, and a vertical asymptote at x equals 3.7

Three Corrections Worth Memorising

“The point is undefined, so it is a hole.”
- Correct: a hole requires equal finite one-sided limits. An undefined point with an unbounded side is an infinite discontinuity, and the exam builds distractors on exactly that confusion.
“The two branches do not meet, so it is a jump.”
- Correct: a jump requires both one-sided limits to be finite. If either runs to $\pm\infty$ , the label is infinite no matter how the drawing looks.
“ $f(a)$ is defined, so the function is continuous at $a$ .”
- Correct: defined is one of three conditions. A defined value that disagrees with the limit is still a discontinuity, and naming it removable is the correct call.

Topic-Specific Study Method

Two Limits, Then the Dot

Write the two one-sided limits as a pair. One line, both values, before any label. " $\lim_{x \to 1^{-}} f = 1$ , $\lim_{x \to 1^{+}} f = 1$ ." The pair is the evidence the rubric wants and the discipline that prevents picture-based guessing.
Apply the decision rule mechanically. Equal and finite, removable. Unequal and finite, jump. Any $\pm\infty$ , infinite. No exceptions, no case-by-case reasoning at the point.
Check $f(a)$ only after the label is fixed. If the label is removable and $f(a)$ disagrees with the limit, say so explicitly: the function has a removable discontinuity with a displaced value.
Name the repair. Removable discontinuities can be repaired by redefining one point; the other two cannot be repaired at all. Saying which is which is often the final part of an FRQ.

What the page shows	The two one-sided limits	Label	What to write
Rational with a cancelled factor	Equal finite	Removable	The pair, then "removable, since the two-sided limit exists"
Rational with a surviving $(x-a)$ downstairs	One or both unbounded	Infinite	The pair with signs, then "infinite, since the left side tends to $-\infty$ "
Piecewise with different formulas at the seam	Finite but unequal	Jump	Both values, then "jump, since the one-sided limits differ"
Absolute value quotient at its corner	Finite but unequal	Jump	Both values, then the label
Piecewise with matching formulas	Equal finite, matching value	Continuous	"No discontinuity; all three conditions hold"

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \dfrac{x^{2} - 1}{x - 1}$ for $x \neq 1$ and $f(1) = 3$ . Classify the discontinuity at $x = 1$ .

Standard Execution:

One-sided limits. For $x \neq 1$ , $f(x) = x + 1$ , so both one-sided limits equal $2$ .

Label. The two-sided limit exists and equals $2$ , so the discontinuity is removable.

Check the dot. $f(1) = 3 \neq 2$ , so the value is displaced. Redefining $f(1) = 2$ repairs it.

Solution: From the left, $\lim_{x \to 2^{-}} g(x) = 3$ . From the right, $\lim_{x \to 2^{+}} g(x) = 3$ . Both sides agree, so the discontinuity is removable despite the two branches looking different. Since $g(2) = 5 - 2 = 3$ already matches the limit, $g$ is in fact continuous at $2$ ; there is nothing to repair. A second version with $5 - 2x$ in place of $5 - x$ would give a right-hand limit of $1$ and a genuine jump.

Common Traps & Failure Signals:

The undefined-means-hole reflex (Signal: the item names a point where the function is undefined and one side is unbounded): Check the sides before labelling. Undefined plus unbounded is infinite, not removable.
The gap-means-jump reflex (Signal: a vertical asymptote drawn with a visible gap): A jump needs two finite heights. A gap into infinity is infinite.
The defined-means-continuous reflex (Signal: the item supplies $f(a)$ and asks about continuity): Defined is one condition out of three. Compare the limit to the value before you conclude.

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

Picture Classification: You name the type from the drawing.
Value Priority: You checked $f(a)$ first and let it decide the label.
Limit-Pair Omission: Because the label feels obvious from the graph, the two one-sided limits never get written down, and on a free response the classification arrives without the evidence the rubric requires, so the point for justification is lost even though the label itself is right.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No scored FRQ part in any released exam; breaks get named inside graph-reading items
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Removable, jump, and infinite causes named from a graph
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Displaced Dot

Let $f(x) = \dfrac{x^{2} - 1}{x - 1}$ for $x \neq 1$ , and let $f(1) = 3$ . Which statement describes the discontinuity of $f$ at $x = 1$ ?

MCQ 2

The Unbounded Side

Let $h(x) = \dfrac{3}{(x-2)^{2}}$ . Which statement describes the behaviour of $h$ at $x = 2$ ?

MCQ 3

Reading Two Branches

Let $k(x) = \begin{cases} 2x - 1, & x < 1 \cr x^{2} + 2, & x \ge 1 \end{cases}$ Which statement describes $k$ at $x = 1$ ?

FRQ 4

Three Points, Three Verdicts

Let $f(x) = \begin{cases} \dfrac{x^{2}-4}{x-2}, & x < 2 \cr 3, & x = 2 \cr \dfrac{1}{x-2}, & x > 2 \end{cases}$

(a) Evaluate $\lim_{x \to 2^{-}} f(x)$ and $\lim_{x \to 2^{+}} f(x)$ .

(b) Classify the behaviour of $f$ at $x = 2$ and justify the classification from your limits.

(c) The function $g(x) = \dfrac{x^{2}-9}{x-3}$ is undefined at $x = 3$ . Classify its discontinuity there and state the redefinition that removes it.

(d) A student writes: "Both functions have a gap at the point, so both discontinuities are removable." Identify the error.

Cheatsheet

Naming the Break

	Label	Two-sided limit	Repair by redefining one point
Equal and finite	Removable	Exists	Yes
Finite but unequal	Jump	Does not exist	No
At least one $\pm\infty$	Infinite	Does not exist	No
Equal and finite, matching $f(a)$	Continuous	Exists	Nothing to repair
One side finite, the other unbounded	Infinite	Does not exist	No

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Unit 1Types of Discontinuities

Concept 4.106Built 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

What you’ll see: Rational functions with a cancelable factor, rational functions with a stubborn denominator factor, piecewise definitions whose branches disagree at the seam, and graphs with a visible break at a labelled $x$ -value.
What you’ll do: Compute or read both one-sided limits, compare them, and then check whether the value at the point matches. The first comparison names the type; the second decides whether the point is a hole or a misplaced dot.
Memory hook: Two limits, then the dot. Classify from the limits, then look at the value to see which flavour of removable you have.
Use it on the exam: For MCQs: Compute both one-sided limits before reading the choices. If they agree, the answer is removable and the value of $f(a)$ only tells you whether the dot is misplaced; if they disagree and both are finite, it is a jump; if either is unbounded, it is infinite. For FRQs: Quote the limits with their values in one sentence, then name the type. A justification that names the type without the two limits is a claim, not a justification, and the rubric scores the evidence.

Concept Review

The Three Labels and What Separates Them

Only the one-sided limits matter for the label. The value $f(a)$ matters for a second, narrower question: whether a removable discontinuity is a hole or a displaced point.

Type	Left and right limits	Two-sided limit	Is $f(a)$ involved in the label	Repair
Removable	Equal and finite	Exists	No. A mismatched or missing $f(a)$ is the usual cause.	Redefine $f(a)$ to equal the limit
Jump	Finite but unequal	Does not exist	No. $f(a)$ may be defined on either branch.	No repair exists
Infinite	At least one is $\pm\infty$	Does not exist	No. $f(a)$ is usually undefined, and irrelevant.	No repair exists

Classifying in Three Moves

Confirm there is something to classify. Check the three continuity conditions at $a$ : $f(a)$ defined, the limit exists, and the two agree. If all three hold, stop; there is no discontinuity.
Compute both one-sided limits. Factor or conjugate if needed. If either side is unbounded, the label is infinite and you are done; the sign tells you which way each branch leaves.
Compare the two finite values. Equal means removable, unequal means jump. Then, and only then, look at $f(a)$ : if it exists and differs from the limit, the discontinuity is a displaced dot rather than a hole.

Figure 1 — Three breaks in one graph.

One graph showing three discontinuities: a hole at x equals 1 with a displaced filled dot at height 2, a jump at x equals 2 between heights 1 and minus 1, and a vertical asymptote at x equals 3.7

Three Corrections Worth Memorising

“The point is undefined, so it is a hole.”
- Correct: a hole requires equal finite one-sided limits. An undefined point with an unbounded side is an infinite discontinuity, and the exam builds distractors on exactly that confusion.
“The two branches do not meet, so it is a jump.”
- Correct: a jump requires both one-sided limits to be finite. If either runs to $\pm\infty$ , the label is infinite no matter how the drawing looks.
“ $f(a)$ is defined, so the function is continuous at $a$ .”
- Correct: defined is one of three conditions. A defined value that disagrees with the limit is still a discontinuity, and naming it removable is the correct call.

Topic-Specific Study Method

Two Limits, Then the Dot

Write the two one-sided limits as a pair. One line, both values, before any label. " $\lim_{x \to 1^{-}} f = 1$ , $\lim_{x \to 1^{+}} f = 1$ ." The pair is the evidence the rubric wants and the discipline that prevents picture-based guessing.
Apply the decision rule mechanically. Equal and finite, removable. Unequal and finite, jump. Any $\pm\infty$ , infinite. No exceptions, no case-by-case reasoning at the point.
Check $f(a)$ only after the label is fixed. If the label is removable and $f(a)$ disagrees with the limit, say so explicitly: the function has a removable discontinuity with a displaced value.
Name the repair. Removable discontinuities can be repaired by redefining one point; the other two cannot be repaired at all. Saying which is which is often the final part of an FRQ.

What the page shows	The two one-sided limits	Label	What to write
Rational with a cancelled factor	Equal finite	Removable	The pair, then "removable, since the two-sided limit exists"
Rational with a surviving $(x-a)$ downstairs	One or both unbounded	Infinite	The pair with signs, then "infinite, since the left side tends to $-\infty$ "
Piecewise with different formulas at the seam	Finite but unequal	Jump	Both values, then "jump, since the one-sided limits differ"
Absolute value quotient at its corner	Finite but unequal	Jump	Both values, then the label
Piecewise with matching formulas	Equal finite, matching value	Continuous	"No discontinuity; all three conditions hold"

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \dfrac{x^{2} - 1}{x - 1}$ for $x \neq 1$ and $f(1) = 3$ . Classify the discontinuity at $x = 1$ .

Standard Execution:

One-sided limits. For $x \neq 1$ , $f(x) = x + 1$ , so both one-sided limits equal $2$ .

Label. The two-sided limit exists and equals $2$ , so the discontinuity is removable.

Check the dot. $f(1) = 3 \neq 2$ , so the value is displaced. Redefining $f(1) = 2$ repairs it.

Solution: From the left, $\lim_{x \to 2^{-}} g(x) = 3$ . From the right, $\lim_{x \to 2^{+}} g(x) = 3$ . Both sides agree, so the discontinuity is removable despite the two branches looking different. Since $g(2) = 5 - 2 = 3$ already matches the limit, $g$ is in fact continuous at $2$ ; there is nothing to repair. A second version with $5 - 2x$ in place of $5 - x$ would give a right-hand limit of $1$ and a genuine jump.

Common Traps & Failure Signals:

The undefined-means-hole reflex (Signal: the item names a point where the function is undefined and one side is unbounded): Check the sides before labelling. Undefined plus unbounded is infinite, not removable.
The gap-means-jump reflex (Signal: a vertical asymptote drawn with a visible gap): A jump needs two finite heights. A gap into infinity is infinite.
The defined-means-continuous reflex (Signal: the item supplies $f(a)$ and asks about continuity): Defined is one condition out of three. Compare the limit to the value before you conclude.

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

Picture Classification: You name the type from the drawing.
Value Priority: You checked $f(a)$ first and let it decide the label.
Limit-Pair Omission: Because the label feels obvious from the graph, the two one-sided limits never get written down, and on a free response the classification arrives without the evidence the rubric requires, so the point for justification is lost even though the label itself is right.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No scored FRQ part in any released exam; breaks get named inside graph-reading items
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Removable, jump, and infinite causes named from a graph
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Displaced Dot

Let $f(x) = \dfrac{x^{2} - 1}{x - 1}$ for $x \neq 1$ , and let $f(1) = 3$ . Which statement describes the discontinuity of $f$ at $x = 1$ ?

MCQ 2

The Unbounded Side

Let $h(x) = \dfrac{3}{(x-2)^{2}}$ . Which statement describes the behaviour of $h$ at $x = 2$ ?

MCQ 3

Reading Two Branches

Let $k(x) = \begin{cases} 2x - 1, & x < 1 \cr x^{2} + 2, & x \ge 1 \end{cases}$ Which statement describes $k$ at $x = 1$ ?

FRQ 4

Three Points, Three Verdicts

Let $f(x) = \begin{cases} \dfrac{x^{2}-4}{x-2}, & x < 2 \cr 3, & x = 2 \cr \dfrac{1}{x-2}, & x > 2 \end{cases}$

(a) Evaluate $\lim_{x \to 2^{-}} f(x)$ and $\lim_{x \to 2^{+}} f(x)$ .

(b) Classify the behaviour of $f$ at $x = 2$ and justify the classification from your limits.

(c) The function $g(x) = \dfrac{x^{2}-9}{x-3}$ is undefined at $x = 3$ . Classify its discontinuity there and state the redefinition that removes it.

(d) A student writes: "Both functions have a gap at the point, so both discontinuities are removable." Identify the error.

Cheatsheet

Naming the Break

	Label	Two-sided limit	Repair by redefining one point
Equal and finite	Removable	Exists	Yes
Finite but unequal	Jump	Does not exist	No
At least one $\pm\infty$	Infinite	Does not exist	No
Equal and finite, matching $f(a)$	Continuous	Exists	Nothing to repair
One side finite, the other unbounded	Infinite	Does not exist	No

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