Unit 1Definition of Continuity

Concept 4.270Built for May 2027 AP Exam

Definition of Continuity

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The definition of continuity at a point is a three-clause sentence, and every point lost on this topic is traceable to one clause being skipped. The limit must exist. The function must be defined at that point. The two must be the same number. Written compactly, that is $\lim_{x \to a} f(x) = f(a)$ , but the compact form hides two presuppositions, and the exam rewards students who make them explicit. An equivalent formulation uses one-sided limits: $f$ is continuous at $a$ exactly when the left-hand limit, the right-hand limit, and the value all coincide. That version is often faster to apply, because it converts a definition into a comparison of three computable numbers.

On the AP Exam

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

What you’ll see: A statement to evaluate as true or false, a piecewise function with constants to determine, a request to justify continuity at a stated point, or a multiple-choice item asking which condition is equivalent to the definition.
What you’ll do: Write the three clauses, evaluate each, and compare. When constants are present, set the one-sided limits equal and solve the resulting equations.
Memory hook: Limit, value, equality — in that order, every time.
Use it on the exam: For MCQs: If the question asks which condition is equivalent, remember that only the full equality qualifies. Any single clause is necessary but not sufficient. For FRQs: The three-clause display is worth writing in full. It costs one line and it is frequently the difference between the justification point and no point. On released BC papers: Across the 2017–2026 BC papers, the definition is scored explicitly in exactly one place. On 2022 Q4 the stem supplies a rate function and the part requires the Intermediate Value Theorem; the released scoring notes state that a response must establish that $r'$ is continuous before invoking the theorem, and they award the point for saying so, on the grounds that $r'$ is differentiable and differentiability implies continuity. That is the definition doing scored work: not a graph to look at, but a hypothesis that must be produced. The other stems that mention continuity hand it to you as a given, which is why they are counted under the previous topic rather than here.

Concept Review

The Definition as a Sequence of Gates

Figure 1 — Three gates in order, with the failure each one produces.

Caption. The gates are not independent. Gate three can only be reached after gates one and two have been passed, which is why a response that asserts the equality without producing the two numbers has not actually tested anything.

$\boxed{\lim_{x \to a} f(x) = f(a)}$

Read literally, this single equality contains three assertions:

The limit exists and is a finite number $L$ .
The value exists, meaning $a$ belongs to the domain of $f$ .
The limit equals the value, meaning $L = f(a)$ .

None of the three implies another. A hole satisfies clause one and fails clause two. A displaced point satisfies clauses one and two and fails clause three. A jump fails clause one outright.

A three-gate decision chart: does the limit exist, is the value defined, do they agree, each gate with a failure label to its right and a pass bar below

The One-Sided Formulation

An equivalent and often faster statement:

$f \text{ is continuous at } a \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$

This version turns the definition into three numbers to compute. It also handles endpoints gracefully: at a domain endpoint, only the available side appears, which is exactly the one-sided definition used in the previous topic.

Continuity Preserved by Operations

The definition is applied point by point, but you almost never apply it by hand to a complicated expression, because of two closure theorems:

Arithmetic: if $f$ and $g$ are continuous at $a$ , then $f + g$ , $f - g$ , $f \cdot g$ , and $f/g$ (provided $g(a) \neq 0$ ) are continuous at $a$ .
Composition: if $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ , then $f \circ g$ is continuous at $a$ .

Together with the continuity of the elementary families, these two theorems let you certify continuity of a long expression by checking a handful of denominators and radicands.

What the Definition Rules Out

The three clauses exist to exclude three specific failures, and each exclusion is worth naming:

A function with a hole fails clause two: the limit exists but there is nothing to compare it to.
A function with a displaced point fails clause three: both numbers exist and they disagree.
A function with a jump or asymptote fails clause one: there is no two-sided limit to begin with.

Naming the excluded failure is what turns a verdict into a justification.

Common Misconception Check

Myth 1: "Continuity means the graph has no holes."
- Truth: A hole is only one of three failures. A displaced point has no hole in the naive sense yet still breaks continuity.
Myth 2: "If $f$ is defined at $a$ and the limit exists, continuity follows."
- Truth: Clause three is a separate requirement. The two numbers must be equal, not merely both present.
Myth 3: "The definition only applies at interior points."
- Truth: It applies wherever the function is defined, with one-sided versions at endpoints. Restricting it to interiors is what produces the wrong verdict at a closed interval's ends.

Topic-Specific Study Method

Write the Three Lines, Not the One

Borrowed idea: The definition is three clauses and nothing more. Write them in order, every time. Limit, value, equality. Most students can produce the equality and not the two facts it presupposes, which is why they declare continuity at a point where the function is undefined. Run the gates in sequence and the definition stops feeling like a sentence to memorise and starts behaving like a procedure that never fails you.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
"Is $f$ continuous at $a$ ?"	A formula with $a$ in or near the domain	Which clause fails, if any?	Three lines: limit, value, comparison
Two constants to find	A piecewise with the constants in the branches	What makes the pieces meet?	Left limit, right limit, value; solve the system
"Which statement is equivalent?"	Four candidate conditions	Is the condition sufficient or merely necessary?	Only the full equality is equivalent
Justify a substitution	A limit of a composite	Which continuity theorem applies?	Name arithmetic or composition, then the family
Endpoint of a closed interval	$\sqrt{x - c}$ or a closed bracket	Is only one side required?	Interior limit equals value

Executable Action Steps: ① Write clause one as an actual limit computation with a value. " $f(x) \to 4$ as $x \to 2$ " is a statement you can check; "the limit exists" is a claim you can only repeat. ② Write clause two as an evaluation. Substitute $a$ into the formula and record whether the result is a number. If it is not, stop: clause three cannot be reached. ③ Write clause three as a comparison of two visible numbers. Place the limit above the value and write $=$ or $\neq$ between them; the symbol is the verdict.

Training & Error Diagnosis

Core Archetype (Simplest Form): Determine whether $f(x) = \dfrac{x^2 - 5x + 6}{x - 2}$ is continuous at $x = 2$ .

Standard Execution:

The limit: $\frac{(x-2)(x-3)}{x-2} = x - 3 \to -1$ as $x \to 2$ . The limit exists and equals $-1$ .

The value: $f(2)$ is undefined, since the denominator vanishes there.

Clause two fails, so clause three cannot hold.

The function is not continuous at $2$ ; the break is removable, repaired by defining $f(2) = -1$ .

Exam Variation (Context / Parameter Shift): 1. Two Constants. Find $a$ and $b$ so that $f(x) = \begin{cases} ax + b & x < 1 \cr 4 & x = 1 \cr x^2 + a & x > 1 \end{cases}$ is continuous at $x = 1$ .

Solution: The left-hand limit is $a + b$ , the right-hand limit is $1 + a$ , and the value is $4$ . Continuity requires $a + b = 4$ and $1 + a = 4$ , so $a = 3$ and $b = 1$ .

2. Continuity at a Domain Endpoint. Is $f(x) = \sqrt{4 - x}$ continuous at $x = 4$ ?

Solution: Only the left-hand limit exists, and $\lim_{x \to 4^-}\sqrt{4 - x} = 0 = f(4)$ . The function is continuous from the left at $4$ , and therefore continuous at $4$ in the endpoint sense.

Common Traps & Failure Signals:

The one-line verdict (Signal: your response is "continuous" with no supporting numbers): The definition is a comparison, and a comparison needs both quantities displayed.
The uncomputed value (Signal: you never substituted $a$ into the formula): Clause two is exactly that substitution, and skipping it is how a hole gets declared continuous.
The endpoint overreach (Signal: you wrote DNE for a one-sided continuity question): At a domain endpoint, the unavailable side is not evidence against continuity.

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

Clause Skipped: You asserted the equality without producing either number.
- Presuppositions Hidden: Because the compact form of the definition conceals two prior requirements, you treated the equality as a single fact to recognise rather than as a comparison of two computed quantities, and the clause you never wrote is precisely the one your function violated.
- Order Ignored: You checked the value before the limit, so a nonexistent limit was never noticed and the verdict rested on two numbers that never should have been compared.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; One part where continuity must be produced as a sentence before the theorem can be applied
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; A single part, and the only one where the rubric scores the continuity sentence itself; the theorem inside it is counted on the statement page
FRQ Years Appeared: 2022 (Q4b)

Practice Examples

MCQ 1

Which Clause Fails?

Let $f(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1} & x \neq 1 \cr 3 & x = 1 \end{cases}$ . At $x = 1$ , which clause of the continuity definition fails?

MCQ 2

Two Constants

For what values of $a$ and $b$ is $f(x) = \begin{cases} ax + b & x < 1 \cr 4 & x = 1 \cr x^2 + a & x > 1 \end{cases}$ continuous at $x = 1$ ?

MCQ 3

Necessary or Equivalent?

Which statement is equivalent to " $f$ is continuous at $x = a$ "?

FRQ 4

Applying the Definition Twice

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

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

(b) Find the value of $k$ that makes $f$ continuous at $x = 2$ , and verify all three clauses explicitly.

(c) Suppose instead that $k = 5$ . Identify which clause fails and classify the resulting discontinuity.

Cheatsheet

Core Reference

	Formal Statement	Fails When	Typical Break
1. Limit exists	$\lim_{x \to a} f(x) = L$ , finite	One-sided limits differ, or a side is unbounded	Jump, infinite, oscillatory
2. Value exists	$f(a)$ is a number	$a$ outside the domain	Hole
3. Limit equals value	$L = f(a)$	The two numbers differ	Displaced point
One-sided form	$\lim_{x \to a^-} = \lim_{x \to a^+} = f(a)$	Any of the three differs	All of the above
At an endpoint	Only the interior side is required	The interior limit differs from the value	Endpoint break
Preserved by	Sums, products, quotients, compositions	A denominator vanishes	Inherited break

Sentences That Score

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Start Topic Practice Next Topic: 1.16 Continuity on Intervals

Unit 1Definition of Continuity

Concept 4.270Built for May 2027 AP Exam

Definition of Continuity

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 statement to evaluate as true or false, a piecewise function with constants to determine, a request to justify continuity at a stated point, or a multiple-choice item asking which condition is equivalent to the definition.
What you’ll do: Write the three clauses, evaluate each, and compare. When constants are present, set the one-sided limits equal and solve the resulting equations.
Memory hook: Limit, value, equality — in that order, every time.
Use it on the exam: For MCQs: If the question asks which condition is equivalent, remember that only the full equality qualifies. Any single clause is necessary but not sufficient. For FRQs: The three-clause display is worth writing in full. It costs one line and it is frequently the difference between the justification point and no point. On released BC papers: Across the 2017–2026 BC papers, the definition is scored explicitly in exactly one place. On 2022 Q4 the stem supplies a rate function and the part requires the Intermediate Value Theorem; the released scoring notes state that a response must establish that $r'$ is continuous before invoking the theorem, and they award the point for saying so, on the grounds that $r'$ is differentiable and differentiability implies continuity. That is the definition doing scored work: not a graph to look at, but a hypothesis that must be produced. The other stems that mention continuity hand it to you as a given, which is why they are counted under the previous topic rather than here.

Concept Review

The Definition as a Sequence of Gates

Figure 1 — Three gates in order, with the failure each one produces.

$\boxed{\lim_{x \to a} f(x) = f(a)}$

Read literally, this single equality contains three assertions:

The limit exists and is a finite number $L$ .
The value exists, meaning $a$ belongs to the domain of $f$ .
The limit equals the value, meaning $L = f(a)$ .

None of the three implies another. A hole satisfies clause one and fails clause two. A displaced point satisfies clauses one and two and fails clause three. A jump fails clause one outright.

A three-gate decision chart: does the limit exist, is the value defined, do they agree, each gate with a failure label to its right and a pass bar below

The One-Sided Formulation

An equivalent and often faster statement:

$f \text{ is continuous at } a \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$

Continuity Preserved by Operations

The definition is applied point by point, but you almost never apply it by hand to a complicated expression, because of two closure theorems:

Arithmetic: if $f$ and $g$ are continuous at $a$ , then $f + g$ , $f - g$ , $f \cdot g$ , and $f/g$ (provided $g(a) \neq 0$ ) are continuous at $a$ .
Composition: if $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ , then $f \circ g$ is continuous at $a$ .

Together with the continuity of the elementary families, these two theorems let you certify continuity of a long expression by checking a handful of denominators and radicands.

What the Definition Rules Out

The three clauses exist to exclude three specific failures, and each exclusion is worth naming:

A function with a hole fails clause two: the limit exists but there is nothing to compare it to.
A function with a displaced point fails clause three: both numbers exist and they disagree.
A function with a jump or asymptote fails clause one: there is no two-sided limit to begin with.

Naming the excluded failure is what turns a verdict into a justification.

Common Misconception Check

Myth 1: "Continuity means the graph has no holes."
- Truth: A hole is only one of three failures. A displaced point has no hole in the naive sense yet still breaks continuity.
Myth 2: "If $f$ is defined at $a$ and the limit exists, continuity follows."
- Truth: Clause three is a separate requirement. The two numbers must be equal, not merely both present.
Myth 3: "The definition only applies at interior points."
- Truth: It applies wherever the function is defined, with one-sided versions at endpoints. Restricting it to interiors is what produces the wrong verdict at a closed interval's ends.

Topic-Specific Study Method

Write the Three Lines, Not the One

Borrowed idea: The definition is three clauses and nothing more. Write them in order, every time. Limit, value, equality. Most students can produce the equality and not the two facts it presupposes, which is why they declare continuity at a point where the function is undefined. Run the gates in sequence and the definition stops feeling like a sentence to memorise and starts behaving like a procedure that never fails you.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
"Is $f$ continuous at $a$ ?"	A formula with $a$ in or near the domain	Which clause fails, if any?	Three lines: limit, value, comparison
Two constants to find	A piecewise with the constants in the branches	What makes the pieces meet?	Left limit, right limit, value; solve the system
"Which statement is equivalent?"	Four candidate conditions	Is the condition sufficient or merely necessary?	Only the full equality is equivalent
Justify a substitution	A limit of a composite	Which continuity theorem applies?	Name arithmetic or composition, then the family
Endpoint of a closed interval	$\sqrt{x - c}$ or a closed bracket	Is only one side required?	Interior limit equals value

Executable Action Steps: ① Write clause one as an actual limit computation with a value. " $f(x) \to 4$ as $x \to 2$ " is a statement you can check; "the limit exists" is a claim you can only repeat. ② Write clause two as an evaluation. Substitute $a$ into the formula and record whether the result is a number. If it is not, stop: clause three cannot be reached. ③ Write clause three as a comparison of two visible numbers. Place the limit above the value and write $=$ or $\neq$ between them; the symbol is the verdict.

Training & Error Diagnosis

Core Archetype (Simplest Form): Determine whether $f(x) = \dfrac{x^2 - 5x + 6}{x - 2}$ is continuous at $x = 2$ .

Standard Execution:

The limit: $\frac{(x-2)(x-3)}{x-2} = x - 3 \to -1$ as $x \to 2$ . The limit exists and equals $-1$ .

The value: $f(2)$ is undefined, since the denominator vanishes there.

Clause two fails, so clause three cannot hold.

The function is not continuous at $2$ ; the break is removable, repaired by defining $f(2) = -1$ .

Solution: The left-hand limit is $a + b$ , the right-hand limit is $1 + a$ , and the value is $4$ . Continuity requires $a + b = 4$ and $1 + a = 4$ , so $a = 3$ and $b = 1$ .

2. Continuity at a Domain Endpoint. Is $f(x) = \sqrt{4 - x}$ continuous at $x = 4$ ?

Solution: Only the left-hand limit exists, and $\lim_{x \to 4^-}\sqrt{4 - x} = 0 = f(4)$ . The function is continuous from the left at $4$ , and therefore continuous at $4$ in the endpoint sense.

Common Traps & Failure Signals:

The one-line verdict (Signal: your response is "continuous" with no supporting numbers): The definition is a comparison, and a comparison needs both quantities displayed.
The uncomputed value (Signal: you never substituted $a$ into the formula): Clause two is exactly that substitution, and skipping it is how a hole gets declared continuous.
The endpoint overreach (Signal: you wrote DNE for a one-sided continuity question): At a domain endpoint, the unavailable side is not evidence against continuity.

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

Clause Skipped: You asserted the equality without producing either number.
- Presuppositions Hidden: Because the compact form of the definition conceals two prior requirements, you treated the equality as a single fact to recognise rather than as a comparison of two computed quantities, and the clause you never wrote is precisely the one your function violated.
- Order Ignored: You checked the value before the limit, so a nonexistent limit was never noticed and the verdict rested on two numbers that never should have been compared.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; One part where continuity must be produced as a sentence before the theorem can be applied
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; A single part, and the only one where the rubric scores the continuity sentence itself; the theorem inside it is counted on the statement page
FRQ Years Appeared: 2022 (Q4b)

Practice Examples

MCQ 1

Which Clause Fails?

Let $f(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1} & x \neq 1 \cr 3 & x = 1 \end{cases}$ . At $x = 1$ , which clause of the continuity definition fails?

MCQ 2

Two Constants

For what values of $a$ and $b$ is $f(x) = \begin{cases} ax + b & x < 1 \cr 4 & x = 1 \cr x^2 + a & x > 1 \end{cases}$ continuous at $x = 1$ ?

MCQ 3

Necessary or Equivalent?

Which statement is equivalent to " $f$ is continuous at $x = a$ "?

FRQ 4

Applying the Definition Twice

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

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

(b) Find the value of $k$ that makes $f$ continuous at $x = 2$ , and verify all three clauses explicitly.

(c) Suppose instead that $k = 5$ . Identify which clause fails and classify the resulting discontinuity.

Cheatsheet

Core Reference

	Formal Statement	Fails When	Typical Break
1. Limit exists	$\lim_{x \to a} f(x) = L$ , finite	One-sided limits differ, or a side is unbounded	Jump, infinite, oscillatory
2. Value exists	$f(a)$ is a number	$a$ outside the domain	Hole
3. Limit equals value	$L = f(a)$	The two numbers differ	Displaced point
One-sided form	$\lim_{x \to a^-} = \lim_{x \to a^+} = f(a)$	Any of the three differs	All of the above
At an endpoint	Only the interior side is required	The interior limit differs from the value	Endpoint break
Preserved by	Sums, products, quotients, compositions	A denominator vanishes	Inherited break

Sentences That Score

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Start Topic Practice Next Topic: 1.16 Continuity on Intervals

Unit 1Definition of Continuity

Concept 4.270Built for May 2027 AP Exam

Definition of Continuity

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 statement to evaluate as true or false, a piecewise function with constants to determine, a request to justify continuity at a stated point, or a multiple-choice item asking which condition is equivalent to the definition.
What you’ll do: Write the three clauses, evaluate each, and compare. When constants are present, set the one-sided limits equal and solve the resulting equations.
Memory hook: Limit, value, equality — in that order, every time.
Use it on the exam: For MCQs: If the question asks which condition is equivalent, remember that only the full equality qualifies. Any single clause is necessary but not sufficient. For FRQs: The three-clause display is worth writing in full. It costs one line and it is frequently the difference between the justification point and no point. On released BC papers: Across the 2017–2026 BC papers, the definition is scored explicitly in exactly one place. On 2022 Q4 the stem supplies a rate function and the part requires the Intermediate Value Theorem; the released scoring notes state that a response must establish that $r'$ is continuous before invoking the theorem, and they award the point for saying so, on the grounds that $r'$ is differentiable and differentiability implies continuity. That is the definition doing scored work: not a graph to look at, but a hypothesis that must be produced. The other stems that mention continuity hand it to you as a given, which is why they are counted under the previous topic rather than here.

Concept Review

The Definition as a Sequence of Gates

Figure 1 — Three gates in order, with the failure each one produces.

$\boxed{\lim_{x \to a} f(x) = f(a)}$

Read literally, this single equality contains three assertions:

The limit exists and is a finite number $L$ .
The value exists, meaning $a$ belongs to the domain of $f$ .
The limit equals the value, meaning $L = f(a)$ .

None of the three implies another. A hole satisfies clause one and fails clause two. A displaced point satisfies clauses one and two and fails clause three. A jump fails clause one outright.

A three-gate decision chart: does the limit exist, is the value defined, do they agree, each gate with a failure label to its right and a pass bar below

The One-Sided Formulation

An equivalent and often faster statement:

$f \text{ is continuous at } a \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$

Continuity Preserved by Operations

The definition is applied point by point, but you almost never apply it by hand to a complicated expression, because of two closure theorems:

Arithmetic: if $f$ and $g$ are continuous at $a$ , then $f + g$ , $f - g$ , $f \cdot g$ , and $f/g$ (provided $g(a) \neq 0$ ) are continuous at $a$ .
Composition: if $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ , then $f \circ g$ is continuous at $a$ .

Together with the continuity of the elementary families, these two theorems let you certify continuity of a long expression by checking a handful of denominators and radicands.

What the Definition Rules Out

The three clauses exist to exclude three specific failures, and each exclusion is worth naming:

A function with a hole fails clause two: the limit exists but there is nothing to compare it to.
A function with a displaced point fails clause three: both numbers exist and they disagree.
A function with a jump or asymptote fails clause one: there is no two-sided limit to begin with.

Naming the excluded failure is what turns a verdict into a justification.

Common Misconception Check

Myth 1: "Continuity means the graph has no holes."
- Truth: A hole is only one of three failures. A displaced point has no hole in the naive sense yet still breaks continuity.
Myth 2: "If $f$ is defined at $a$ and the limit exists, continuity follows."
- Truth: Clause three is a separate requirement. The two numbers must be equal, not merely both present.
Myth 3: "The definition only applies at interior points."
- Truth: It applies wherever the function is defined, with one-sided versions at endpoints. Restricting it to interiors is what produces the wrong verdict at a closed interval's ends.

Topic-Specific Study Method

Write the Three Lines, Not the One

Borrowed idea: The definition is three clauses and nothing more. Write them in order, every time. Limit, value, equality. Most students can produce the equality and not the two facts it presupposes, which is why they declare continuity at a point where the function is undefined. Run the gates in sequence and the definition stops feeling like a sentence to memorise and starts behaving like a procedure that never fails you.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
"Is $f$ continuous at $a$ ?"	A formula with $a$ in or near the domain	Which clause fails, if any?	Three lines: limit, value, comparison
Two constants to find	A piecewise with the constants in the branches	What makes the pieces meet?	Left limit, right limit, value; solve the system
"Which statement is equivalent?"	Four candidate conditions	Is the condition sufficient or merely necessary?	Only the full equality is equivalent
Justify a substitution	A limit of a composite	Which continuity theorem applies?	Name arithmetic or composition, then the family
Endpoint of a closed interval	$\sqrt{x - c}$ or a closed bracket	Is only one side required?	Interior limit equals value

Executable Action Steps: ① Write clause one as an actual limit computation with a value. " $f(x) \to 4$ as $x \to 2$ " is a statement you can check; "the limit exists" is a claim you can only repeat. ② Write clause two as an evaluation. Substitute $a$ into the formula and record whether the result is a number. If it is not, stop: clause three cannot be reached. ③ Write clause three as a comparison of two visible numbers. Place the limit above the value and write $=$ or $\neq$ between them; the symbol is the verdict.

Training & Error Diagnosis

Core Archetype (Simplest Form): Determine whether $f(x) = \dfrac{x^2 - 5x + 6}{x - 2}$ is continuous at $x = 2$ .

Standard Execution:

The limit: $\frac{(x-2)(x-3)}{x-2} = x - 3 \to -1$ as $x \to 2$ . The limit exists and equals $-1$ .

The value: $f(2)$ is undefined, since the denominator vanishes there.

Clause two fails, so clause three cannot hold.

The function is not continuous at $2$ ; the break is removable, repaired by defining $f(2) = -1$ .

Solution: The left-hand limit is $a + b$ , the right-hand limit is $1 + a$ , and the value is $4$ . Continuity requires $a + b = 4$ and $1 + a = 4$ , so $a = 3$ and $b = 1$ .

2. Continuity at a Domain Endpoint. Is $f(x) = \sqrt{4 - x}$ continuous at $x = 4$ ?

Solution: Only the left-hand limit exists, and $\lim_{x \to 4^-}\sqrt{4 - x} = 0 = f(4)$ . The function is continuous from the left at $4$ , and therefore continuous at $4$ in the endpoint sense.

Common Traps & Failure Signals:

The one-line verdict (Signal: your response is "continuous" with no supporting numbers): The definition is a comparison, and a comparison needs both quantities displayed.
The uncomputed value (Signal: you never substituted $a$ into the formula): Clause two is exactly that substitution, and skipping it is how a hole gets declared continuous.
The endpoint overreach (Signal: you wrote DNE for a one-sided continuity question): At a domain endpoint, the unavailable side is not evidence against continuity.

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

Clause Skipped: You asserted the equality without producing either number.
- Presuppositions Hidden: Because the compact form of the definition conceals two prior requirements, you treated the equality as a single fact to recognise rather than as a comparison of two computed quantities, and the clause you never wrote is precisely the one your function violated.
- Order Ignored: You checked the value before the limit, so a nonexistent limit was never noticed and the verdict rested on two numbers that never should have been compared.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; One part where continuity must be produced as a sentence before the theorem can be applied
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; A single part, and the only one where the rubric scores the continuity sentence itself; the theorem inside it is counted on the statement page
FRQ Years Appeared: 2022 (Q4b)

Practice Examples

MCQ 1

Which Clause Fails?

Let $f(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1} & x \neq 1 \cr 3 & x = 1 \end{cases}$ . At $x = 1$ , which clause of the continuity definition fails?

MCQ 2

Two Constants

For what values of $a$ and $b$ is $f(x) = \begin{cases} ax + b & x < 1 \cr 4 & x = 1 \cr x^2 + a & x > 1 \end{cases}$ continuous at $x = 1$ ?

MCQ 3

Necessary or Equivalent?

Which statement is equivalent to " $f$ is continuous at $x = a$ "?

FRQ 4

Applying the Definition Twice

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

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

(b) Find the value of $k$ that makes $f$ continuous at $x = 2$ , and verify all three clauses explicitly.

(c) Suppose instead that $k = 5$ . Identify which clause fails and classify the resulting discontinuity.

Cheatsheet

Core Reference

	Formal Statement	Fails When	Typical Break
1. Limit exists	$\lim_{x \to a} f(x) = L$ , finite	One-sided limits differ, or a side is unbounded	Jump, infinite, oscillatory
2. Value exists	$f(a)$ is a number	$a$ outside the domain	Hole
3. Limit equals value	$L = f(a)$	The two numbers differ	Displaced point
One-sided form	$\lim_{x \to a^-} = \lim_{x \to a^+} = f(a)$	Any of the three differs	All of the above
At an endpoint	Only the interior side is required	The interior limit differs from the value	Endpoint break
Preserved by	Sums, products, quotients, compositions	A denominator vanishes	Inherited break

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