Unit 1Continuity

Concept 4.131Built for May 2027 AP Exam

Continuity

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Continuity at a point is the statement that three separate facts hold at once: the two-sided limit exists, the function is defined there, and the two agree. Nothing more, nothing less. That definition is worth memorizing in its exact three-clause form, because every discontinuity you will ever diagnose is a violation of one specific clause, and the clause tells you which repair is available. Continuity is also the reason direct substitution works, which makes this topic the quiet foundation under most of Unit 1. A function assembled from polynomials, rationals, radicals, trigonometric, exponential, and logarithmic pieces is continuous everywhere it is defined, and every exception traces back to a division by zero or a point outside a domain.

On the AP Exam

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

What you’ll see: A piecewise function with an unknown constant, a graph with a visible break, a question asking whether a given function is continuous at a stated point, or a stem that requires you to justify why substitution is legal.
What you’ll do: Test the three clauses in order, name the clause that fails when it fails, and when a parameter is present, set the one-sided limits equal and solve.
Memory hook: Limit exists, value exists, and the two are the same number.
Use it on the exam: For MCQs: Test in order and stop at the first failure. If the limit does not exist, the other two clauses are irrelevant and checking them wastes time. For FRQs: Write the three conditions as three statements. A response that asserts continuity without naming the conditions typically loses the justification point even when the conclusion is right. On released BC papers: Continuity almost never gets a part to itself; it arrives as the hypothesis the rest of the question consumes. The released stems hand it to you as an adjective attached to a function before any question is asked: "the continuous function $f$ is defined on the closed interval" on 2017 Q1 and 2021 Q4, "the graph of the continuous function $g$ , the derivative of the function $f$ " on 2018 Q3, and "the continuous function $f$ is not explicitly given" on 2018 Q2, where the word is doing real work because no formula follows it. That assumption appears on 2017 Q1, 2018 Q2, 2018 Q3, 2019 Q3, 2021 Q4, and 2025 Q4, and in each case the continuity is not the question but the licence that makes the question answerable. The one exam where continuity had to be produced rather than assumed is counted under the definition topic, because there the rubric scored the sentence itself.

Concept Review

Three Clauses, One Idea

Figure 1 — The curve reaches the point and passes through it.

Caption. Continuity is not a property of the drawing style. It is the coincidence of three quantities, and the picture is merely what that coincidence looks like when you plot it.

The formal definition, in the order you should test it:

$\lim_{x \to a} f(x) \text{ exists}, \qquad f(a) \text{ exists}, \qquad \lim_{x \to a} f(x) = f(a)$

The third clause presupposes the first two, which is why the compact form $\lim_{x \to a} f(x) = f(a)$ is often written as the whole definition. On an exam, however, writing it out in three parts is what demonstrates that you know why the equality matters: each part can fail on its own.

A smooth curve passing through a filled point above a with dashed guides to both axes, beside three numbered condition cards for continuity

Continuity of the Elementary Functions

Every function you meet in this course is continuous on its domain, and knowing the domain is therefore most of the work:

Family	Continuous Where	The Only Obstruction
Polynomial	Everywhere	None
Rational	Denominator nonzero	Zeros of the denominator
Radical, even index	Radicand positive, one-sided at zero	Radicand negative
Radical, odd index	Everywhere	None
Trigonometric	Everywhere they are defined	Zeros of $\cos$ for $\tan$ and $\sec$
Exponential	Everywhere	None
Logarithmic	Argument positive	Argument zero or negative

Continuity also survives the usual operations. Sums, differences, products, quotients (where the denominator is nonzero), and compositions of continuous functions are continuous. That closure is what lets you assert continuity of a complicated expression after checking a handful of points.

Continuity From One Side

At the endpoint of a closed interval, a two-sided limit is unavailable, so the definition adapts:

Continuous from the right at $a$ : $\lim_{x \to a^+} f(x) = f(a)$ .
Continuous from the left at $a$ : $\lim_{x \to a^-} f(x) = f(a)$ .

A function is continuous at an interior point exactly when it is continuous from both sides there. At an endpoint, only the available side is required, which is why $\sqrt{x}$ is continuous at $0$ without having a two-sided limit there.

What Continuity Buys

Continuity is not decoration. Three consequences recur:

Substitution becomes legal. If $f$ is continuous at $a$ , then $\lim_{x \to a} f(x) = f(a)$ , which is the theorem behind every direct substitution you perform.
Limits pass through compositions. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ , then $\lim_{x \to a} f(g(x)) = f(g(a))$ .
Existence theorems apply. The Intermediate Value Theorem requires continuity on a closed interval, and so does the Extreme Value Theorem. Without continuity, neither has anything to say.

Common Misconception Check

Myth 1: "A function is continuous if you can draw it without lifting your pencil."
- Truth: That is a description, not a test. A function with a hole at a single point looks drawable at a glance and fails clause two.
Myth 2: "If the limit exists, the function is continuous there."
- Truth: The limit may exist while the value is missing or displaced. A removable discontinuity satisfies clause one and fails clause three.
Myth 3: "Continuity is a property of the whole function."
- Truth: Continuity is asserted point by point. A function can be continuous everywhere except one input, and that single exception is usually where the exam question lives.

Topic-Specific Study Method

Run the Checklist, In Order

Borrowed idea: Picture first. A function is continuous at a point when the curve you would draw there has no interruption, and every formal condition simply names one way an interruption could occur. The formal test is short: the limit must exist, the value must exist, and they must agree. Those three clauses are the three ways a graph can break. Learn them as a checklist rather than as a definition, and continuity stops being a vocabulary word and becomes a procedure you can run in fifteen seconds without hesitating.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
Piecewise with an unknown constant	Two formulas joined at $x = c$	Which value of the constant makes the pieces meet?	Compute both one-sided limits, set them equal, solve
Graph with a visible break	A jump or a hole in the sketch	Which clause fails?	Check the limit first; if it exists, check the value
"Is $f$ continuous at $x = 2$ ?"	A rational or radical expression	Does the domain include $2$ and do the pieces agree?	Evaluate the denominator, then the limit, then compare
Justify a substitution	A limit of an elementary function	Why is substitution legal?	Name the family and the continuity theorem
Endpoint of an interval	A radical or a closed interval	Is only one side required?	Use the one-sided definition and say so

Executable Action Steps: ① Test clause one before clause two. If the two-sided limit does not exist, continuity fails immediately and no further checking is needed or credited. ② Evaluate the function value by substitution, then compare it to the limit. Writing both numbers on adjacent lines makes the comparison impossible to fumble. ③ When a parameter appears, solve the equality of the two one-sided limits rather than guessing. The condition is $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$ , and that equation determines the constant.

Training & Error Diagnosis

Core Archetype (Simplest Form): Determine whether $f(x) = \dfrac{x^2 - 4}{x - 2}$ is continuous at $x = 2$ .

Standard Execution:

$f(2)$ is undefined, because the denominator vanishes there. Clause two fails.

The limit still exists: $\frac{(x-2)(x+2)}{x-2} \to 4$ as $x \to 2$ .

Since the value does not exist, the equality in clause three cannot be stated.

The function is not continuous at $2$ , and the failure is a removable discontinuity rather than a divergence.

Exam Variation (Context / Parameter Shift): 1. Solving for a Parameter. Find $k$ so that $f(x) = \begin{cases} kx + 1 & x \le 2 \cr x^2 - 1 & x > 2 \end{cases}$ is continuous at $x = 2$ .

Solution: The left-hand limit is $2k + 1$ and the right-hand limit is $3$ . Continuity requires $2k + 1 = 3$ , so $k = 1$ . Both limits also equal $f(2) = 3$ , which closes clause three.

2. A Radical at the Edge of Its Domain. Is $g(x) = \sqrt{x - 5}$ continuous at $x = 5$ ?

Solution: Only the right-hand limit exists, and $\lim_{x \to 5^+}\sqrt{x - 5} = 0 = g(5)$ . The function is continuous from the right at $5$ , which is the correct statement at a domain endpoint.

Common Traps & Failure Signals:

The value never checked (Signal: your work contains a limit and no $f(a)$ ): Two of the three clauses are about the value, and skipping them is the most common way to declare continuity falsely.
The limit never checked (Signal: you compared $f(a)$ to a number you assumed): Without the limit there is nothing to compare, and clause three is empty.
The two-sided test at an endpoint (Signal: you wrote DNE for a square root at its domain endpoint): At an endpoint, only the interior side is required.

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

Value Assumed: You computed the limit and never checked that f(a) exists.
- Point Ignored: Because you treated continuity as a property of the whole graph rather than of a single point, you asserted it at a location where the denominator vanished, and the rest of your work inherited a claim that was never true there.
- Checklist Compressed: You checked two of the three conditions and reported a verdict on the third.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Continuity arrives as the hypothesis other parts consume, not as the thing being asked for
10-Year Total Count: 6 of 9 Exams · 6 FRQ Questions; Six stems that hand continuity to you as an adjective on the function rather than as the thing to prove
FRQ Years Appeared: 2017 (Q1), 2018 (Q2), 2018 (Q3), 2019 (Q3), 2021 (Q4), 2025 (Q4)

Practice Examples

MCQ 1

A Removable Break

The function $f$ is defined by $f(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \neq 3$ . Which statement about $f$ at $x = 3$ is correct?

MCQ 2

Solving for a Constant

For what value of $k$ is $f(x) = \begin{cases} kx + 2 & x \le 1 \cr 3x^2 & x > 1 \end{cases}$ continuous at $x = 1$ ?

MCQ 3

Continuity of a Composite

Which function is continuous at $x = 0$ ?

FRQ 4

Testing and Repairing Continuity

Let $f(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1} & x < 1 \cr a & x = 1 \cr 2x + b & x > 1 \end{cases}$

(a) Determine whether $\lim_{x \to 1^-} f(x)$ exists, and find its value.

(b) Find the value of $b$ that makes $\lim_{x \to 1^+} f(x)$ equal to the left-hand limit from part (a).

(c) Using your value of $b$ , state whether $f$ is continuous at $x = 1$ when $a = 0$ , and identify which clause of the definition fails.

Cheatsheet

Core Reference

	Statement	How It Fails
1. Limit exists	$\lim_{x \to a} f(x)$ is a finite number	Jump, asymptote, or oscillation
2. Value exists	$f(a)$ is defined	$a$ outside the domain, or a hole
3. They agree	$\lim_{x \to a} f(x) = f(a)$	Value displaced from the limit
From the right	$\lim_{x \to a^+} f(x) = f(a)$	Needed at a left endpoint
From the left	$\lim_{x \to a^-} f(x) = f(a)$	Needed at a right endpoint
Elementary functions	Continuous on their domains	Only at excluded points

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Unit 1Continuity

Concept 4.131Built for May 2027 AP Exam

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 piecewise function with an unknown constant, a graph with a visible break, a question asking whether a given function is continuous at a stated point, or a stem that requires you to justify why substitution is legal.
What you’ll do: Test the three clauses in order, name the clause that fails when it fails, and when a parameter is present, set the one-sided limits equal and solve.
Memory hook: Limit exists, value exists, and the two are the same number.
Use it on the exam: For MCQs: Test in order and stop at the first failure. If the limit does not exist, the other two clauses are irrelevant and checking them wastes time. For FRQs: Write the three conditions as three statements. A response that asserts continuity without naming the conditions typically loses the justification point even when the conclusion is right. On released BC papers: Continuity almost never gets a part to itself; it arrives as the hypothesis the rest of the question consumes. The released stems hand it to you as an adjective attached to a function before any question is asked: "the continuous function $f$ is defined on the closed interval" on 2017 Q1 and 2021 Q4, "the graph of the continuous function $g$ , the derivative of the function $f$ " on 2018 Q3, and "the continuous function $f$ is not explicitly given" on 2018 Q2, where the word is doing real work because no formula follows it. That assumption appears on 2017 Q1, 2018 Q2, 2018 Q3, 2019 Q3, 2021 Q4, and 2025 Q4, and in each case the continuity is not the question but the licence that makes the question answerable. The one exam where continuity had to be produced rather than assumed is counted under the definition topic, because there the rubric scored the sentence itself.

Concept Review

Three Clauses, One Idea

Figure 1 — The curve reaches the point and passes through it.

Caption. Continuity is not a property of the drawing style. It is the coincidence of three quantities, and the picture is merely what that coincidence looks like when you plot it.

The formal definition, in the order you should test it:

$\lim_{x \to a} f(x) \text{ exists}, \qquad f(a) \text{ exists}, \qquad \lim_{x \to a} f(x) = f(a)$

A smooth curve passing through a filled point above a with dashed guides to both axes, beside three numbered condition cards for continuity

Continuity of the Elementary Functions

Every function you meet in this course is continuous on its domain, and knowing the domain is therefore most of the work:

Family	Continuous Where	The Only Obstruction
Polynomial	Everywhere	None
Rational	Denominator nonzero	Zeros of the denominator
Radical, even index	Radicand positive, one-sided at zero	Radicand negative
Radical, odd index	Everywhere	None
Trigonometric	Everywhere they are defined	Zeros of $\cos$ for $\tan$ and $\sec$
Exponential	Everywhere	None
Logarithmic	Argument positive	Argument zero or negative

Continuity From One Side

At the endpoint of a closed interval, a two-sided limit is unavailable, so the definition adapts:

Continuous from the right at $a$ : $\lim_{x \to a^+} f(x) = f(a)$ .
Continuous from the left at $a$ : $\lim_{x \to a^-} f(x) = f(a)$ .

What Continuity Buys

Continuity is not decoration. Three consequences recur:

Substitution becomes legal. If $f$ is continuous at $a$ , then $\lim_{x \to a} f(x) = f(a)$ , which is the theorem behind every direct substitution you perform.
Limits pass through compositions. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ , then $\lim_{x \to a} f(g(x)) = f(g(a))$ .
Existence theorems apply. The Intermediate Value Theorem requires continuity on a closed interval, and so does the Extreme Value Theorem. Without continuity, neither has anything to say.

Common Misconception Check

Myth 1: "A function is continuous if you can draw it without lifting your pencil."
- Truth: That is a description, not a test. A function with a hole at a single point looks drawable at a glance and fails clause two.
Myth 2: "If the limit exists, the function is continuous there."
- Truth: The limit may exist while the value is missing or displaced. A removable discontinuity satisfies clause one and fails clause three.
Myth 3: "Continuity is a property of the whole function."
- Truth: Continuity is asserted point by point. A function can be continuous everywhere except one input, and that single exception is usually where the exam question lives.

Topic-Specific Study Method

Run the Checklist, In Order

Borrowed idea: Picture first. A function is continuous at a point when the curve you would draw there has no interruption, and every formal condition simply names one way an interruption could occur. The formal test is short: the limit must exist, the value must exist, and they must agree. Those three clauses are the three ways a graph can break. Learn them as a checklist rather than as a definition, and continuity stops being a vocabulary word and becomes a procedure you can run in fifteen seconds without hesitating.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
Piecewise with an unknown constant	Two formulas joined at $x = c$	Which value of the constant makes the pieces meet?	Compute both one-sided limits, set them equal, solve
Graph with a visible break	A jump or a hole in the sketch	Which clause fails?	Check the limit first; if it exists, check the value
"Is $f$ continuous at $x = 2$ ?"	A rational or radical expression	Does the domain include $2$ and do the pieces agree?	Evaluate the denominator, then the limit, then compare
Justify a substitution	A limit of an elementary function	Why is substitution legal?	Name the family and the continuity theorem
Endpoint of an interval	A radical or a closed interval	Is only one side required?	Use the one-sided definition and say so

Executable Action Steps: ① Test clause one before clause two. If the two-sided limit does not exist, continuity fails immediately and no further checking is needed or credited. ② Evaluate the function value by substitution, then compare it to the limit. Writing both numbers on adjacent lines makes the comparison impossible to fumble. ③ When a parameter appears, solve the equality of the two one-sided limits rather than guessing. The condition is $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$ , and that equation determines the constant.

Training & Error Diagnosis

Core Archetype (Simplest Form): Determine whether $f(x) = \dfrac{x^2 - 4}{x - 2}$ is continuous at $x = 2$ .

Standard Execution:

$f(2)$ is undefined, because the denominator vanishes there. Clause two fails.

The limit still exists: $\frac{(x-2)(x+2)}{x-2} \to 4$ as $x \to 2$ .

Since the value does not exist, the equality in clause three cannot be stated.

The function is not continuous at $2$ , and the failure is a removable discontinuity rather than a divergence.

Exam Variation (Context / Parameter Shift): 1. Solving for a Parameter. Find $k$ so that $f(x) = \begin{cases} kx + 1 & x \le 2 \cr x^2 - 1 & x > 2 \end{cases}$ is continuous at $x = 2$ .

Solution: The left-hand limit is $2k + 1$ and the right-hand limit is $3$ . Continuity requires $2k + 1 = 3$ , so $k = 1$ . Both limits also equal $f(2) = 3$ , which closes clause three.

2. A Radical at the Edge of Its Domain. Is $g(x) = \sqrt{x - 5}$ continuous at $x = 5$ ?

Solution: Only the right-hand limit exists, and $\lim_{x \to 5^+}\sqrt{x - 5} = 0 = g(5)$ . The function is continuous from the right at $5$ , which is the correct statement at a domain endpoint.

Common Traps & Failure Signals:

The value never checked (Signal: your work contains a limit and no $f(a)$ ): Two of the three clauses are about the value, and skipping them is the most common way to declare continuity falsely.
The limit never checked (Signal: you compared $f(a)$ to a number you assumed): Without the limit there is nothing to compare, and clause three is empty.
The two-sided test at an endpoint (Signal: you wrote DNE for a square root at its domain endpoint): At an endpoint, only the interior side is required.

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

Value Assumed: You computed the limit and never checked that f(a) exists.
- Point Ignored: Because you treated continuity as a property of the whole graph rather than of a single point, you asserted it at a location where the denominator vanished, and the rest of your work inherited a claim that was never true there.
- Checklist Compressed: You checked two of the three conditions and reported a verdict on the third.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Continuity arrives as the hypothesis other parts consume, not as the thing being asked for
10-Year Total Count: 6 of 9 Exams · 6 FRQ Questions; Six stems that hand continuity to you as an adjective on the function rather than as the thing to prove
FRQ Years Appeared: 2017 (Q1), 2018 (Q2), 2018 (Q3), 2019 (Q3), 2021 (Q4), 2025 (Q4)

Practice Examples

MCQ 1

A Removable Break

The function $f$ is defined by $f(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \neq 3$ . Which statement about $f$ at $x = 3$ is correct?

MCQ 2

Solving for a Constant

For what value of $k$ is $f(x) = \begin{cases} kx + 2 & x \le 1 \cr 3x^2 & x > 1 \end{cases}$ continuous at $x = 1$ ?

MCQ 3

Continuity of a Composite

Which function is continuous at $x = 0$ ?

FRQ 4

Testing and Repairing Continuity

Let $f(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1} & x < 1 \cr a & x = 1 \cr 2x + b & x > 1 \end{cases}$

(a) Determine whether $\lim_{x \to 1^-} f(x)$ exists, and find its value.

(b) Find the value of $b$ that makes $\lim_{x \to 1^+} f(x)$ equal to the left-hand limit from part (a).

(c) Using your value of $b$ , state whether $f$ is continuous at $x = 1$ when $a = 0$ , and identify which clause of the definition fails.

Cheatsheet

Core Reference

	Statement	How It Fails
1. Limit exists	$\lim_{x \to a} f(x)$ is a finite number	Jump, asymptote, or oscillation
2. Value exists	$f(a)$ is defined	$a$ outside the domain, or a hole
3. They agree	$\lim_{x \to a} f(x) = f(a)$	Value displaced from the limit
From the right	$\lim_{x \to a^+} f(x) = f(a)$	Needed at a left endpoint
From the left	$\lim_{x \to a^-} f(x) = f(a)$	Needed at a right endpoint
Elementary functions	Continuous on their domains	Only at excluded points

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 1.14 Types of Discontinuities

Unit 1Continuity

Concept 4.131Built for May 2027 AP Exam

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 piecewise function with an unknown constant, a graph with a visible break, a question asking whether a given function is continuous at a stated point, or a stem that requires you to justify why substitution is legal.
What you’ll do: Test the three clauses in order, name the clause that fails when it fails, and when a parameter is present, set the one-sided limits equal and solve.
Memory hook: Limit exists, value exists, and the two are the same number.
Use it on the exam: For MCQs: Test in order and stop at the first failure. If the limit does not exist, the other two clauses are irrelevant and checking them wastes time. For FRQs: Write the three conditions as three statements. A response that asserts continuity without naming the conditions typically loses the justification point even when the conclusion is right. On released BC papers: Continuity almost never gets a part to itself; it arrives as the hypothesis the rest of the question consumes. The released stems hand it to you as an adjective attached to a function before any question is asked: "the continuous function $f$ is defined on the closed interval" on 2017 Q1 and 2021 Q4, "the graph of the continuous function $g$ , the derivative of the function $f$ " on 2018 Q3, and "the continuous function $f$ is not explicitly given" on 2018 Q2, where the word is doing real work because no formula follows it. That assumption appears on 2017 Q1, 2018 Q2, 2018 Q3, 2019 Q3, 2021 Q4, and 2025 Q4, and in each case the continuity is not the question but the licence that makes the question answerable. The one exam where continuity had to be produced rather than assumed is counted under the definition topic, because there the rubric scored the sentence itself.

Concept Review

Three Clauses, One Idea

Figure 1 — The curve reaches the point and passes through it.

Caption. Continuity is not a property of the drawing style. It is the coincidence of three quantities, and the picture is merely what that coincidence looks like when you plot it.

The formal definition, in the order you should test it:

$\lim_{x \to a} f(x) \text{ exists}, \qquad f(a) \text{ exists}, \qquad \lim_{x \to a} f(x) = f(a)$

A smooth curve passing through a filled point above a with dashed guides to both axes, beside three numbered condition cards for continuity

Continuity of the Elementary Functions

Every function you meet in this course is continuous on its domain, and knowing the domain is therefore most of the work:

Family	Continuous Where	The Only Obstruction
Polynomial	Everywhere	None
Rational	Denominator nonzero	Zeros of the denominator
Radical, even index	Radicand positive, one-sided at zero	Radicand negative
Radical, odd index	Everywhere	None
Trigonometric	Everywhere they are defined	Zeros of $\cos$ for $\tan$ and $\sec$
Exponential	Everywhere	None
Logarithmic	Argument positive	Argument zero or negative

Continuity From One Side

At the endpoint of a closed interval, a two-sided limit is unavailable, so the definition adapts:

Continuous from the right at $a$ : $\lim_{x \to a^+} f(x) = f(a)$ .
Continuous from the left at $a$ : $\lim_{x \to a^-} f(x) = f(a)$ .

What Continuity Buys

Continuity is not decoration. Three consequences recur:

Substitution becomes legal. If $f$ is continuous at $a$ , then $\lim_{x \to a} f(x) = f(a)$ , which is the theorem behind every direct substitution you perform.
Limits pass through compositions. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ , then $\lim_{x \to a} f(g(x)) = f(g(a))$ .
Existence theorems apply. The Intermediate Value Theorem requires continuity on a closed interval, and so does the Extreme Value Theorem. Without continuity, neither has anything to say.

Common Misconception Check

Myth 1: "A function is continuous if you can draw it without lifting your pencil."
- Truth: That is a description, not a test. A function with a hole at a single point looks drawable at a glance and fails clause two.
Myth 2: "If the limit exists, the function is continuous there."
- Truth: The limit may exist while the value is missing or displaced. A removable discontinuity satisfies clause one and fails clause three.
Myth 3: "Continuity is a property of the whole function."
- Truth: Continuity is asserted point by point. A function can be continuous everywhere except one input, and that single exception is usually where the exam question lives.

Topic-Specific Study Method

Run the Checklist, In Order

Borrowed idea: Picture first. A function is continuous at a point when the curve you would draw there has no interruption, and every formal condition simply names one way an interruption could occur. The formal test is short: the limit must exist, the value must exist, and they must agree. Those three clauses are the three ways a graph can break. Learn them as a checklist rather than as a definition, and continuity stops being a vocabulary word and becomes a procedure you can run in fifteen seconds without hesitating.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
Piecewise with an unknown constant	Two formulas joined at $x = c$	Which value of the constant makes the pieces meet?	Compute both one-sided limits, set them equal, solve
Graph with a visible break	A jump or a hole in the sketch	Which clause fails?	Check the limit first; if it exists, check the value
"Is $f$ continuous at $x = 2$ ?"	A rational or radical expression	Does the domain include $2$ and do the pieces agree?	Evaluate the denominator, then the limit, then compare
Justify a substitution	A limit of an elementary function	Why is substitution legal?	Name the family and the continuity theorem
Endpoint of an interval	A radical or a closed interval	Is only one side required?	Use the one-sided definition and say so

Executable Action Steps: ① Test clause one before clause two. If the two-sided limit does not exist, continuity fails immediately and no further checking is needed or credited. ② Evaluate the function value by substitution, then compare it to the limit. Writing both numbers on adjacent lines makes the comparison impossible to fumble. ③ When a parameter appears, solve the equality of the two one-sided limits rather than guessing. The condition is $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$ , and that equation determines the constant.

Training & Error Diagnosis

Core Archetype (Simplest Form): Determine whether $f(x) = \dfrac{x^2 - 4}{x - 2}$ is continuous at $x = 2$ .

Standard Execution:

$f(2)$ is undefined, because the denominator vanishes there. Clause two fails.

The limit still exists: $\frac{(x-2)(x+2)}{x-2} \to 4$ as $x \to 2$ .

Since the value does not exist, the equality in clause three cannot be stated.

The function is not continuous at $2$ , and the failure is a removable discontinuity rather than a divergence.

Exam Variation (Context / Parameter Shift): 1. Solving for a Parameter. Find $k$ so that $f(x) = \begin{cases} kx + 1 & x \le 2 \cr x^2 - 1 & x > 2 \end{cases}$ is continuous at $x = 2$ .

Solution: The left-hand limit is $2k + 1$ and the right-hand limit is $3$ . Continuity requires $2k + 1 = 3$ , so $k = 1$ . Both limits also equal $f(2) = 3$ , which closes clause three.

2. A Radical at the Edge of Its Domain. Is $g(x) = \sqrt{x - 5}$ continuous at $x = 5$ ?

Solution: Only the right-hand limit exists, and $\lim_{x \to 5^+}\sqrt{x - 5} = 0 = g(5)$ . The function is continuous from the right at $5$ , which is the correct statement at a domain endpoint.

Common Traps & Failure Signals:

The value never checked (Signal: your work contains a limit and no $f(a)$ ): Two of the three clauses are about the value, and skipping them is the most common way to declare continuity falsely.
The limit never checked (Signal: you compared $f(a)$ to a number you assumed): Without the limit there is nothing to compare, and clause three is empty.
The two-sided test at an endpoint (Signal: you wrote DNE for a square root at its domain endpoint): At an endpoint, only the interior side is required.

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

Value Assumed: You computed the limit and never checked that f(a) exists.
- Point Ignored: Because you treated continuity as a property of the whole graph rather than of a single point, you asserted it at a location where the denominator vanished, and the rest of your work inherited a claim that was never true there.
- Checklist Compressed: You checked two of the three conditions and reported a verdict on the third.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Continuity arrives as the hypothesis other parts consume, not as the thing being asked for
10-Year Total Count: 6 of 9 Exams · 6 FRQ Questions; Six stems that hand continuity to you as an adjective on the function rather than as the thing to prove
FRQ Years Appeared: 2017 (Q1), 2018 (Q2), 2018 (Q3), 2019 (Q3), 2021 (Q4), 2025 (Q4)

Practice Examples

MCQ 1

A Removable Break

The function $f$ is defined by $f(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \neq 3$ . Which statement about $f$ at $x = 3$ is correct?

MCQ 2

Solving for a Constant

For what value of $k$ is $f(x) = \begin{cases} kx + 2 & x \le 1 \cr 3x^2 & x > 1 \end{cases}$ continuous at $x = 1$ ?

MCQ 3

Continuity of a Composite

Which function is continuous at $x = 0$ ?

FRQ 4

Testing and Repairing Continuity

Let $f(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1} & x < 1 \cr a & x = 1 \cr 2x + b & x > 1 \end{cases}$

(a) Determine whether $\lim_{x \to 1^-} f(x)$ exists, and find its value.

(b) Find the value of $b$ that makes $\lim_{x \to 1^+} f(x)$ equal to the left-hand limit from part (a).

(c) Using your value of $b$ , state whether $f$ is continuous at $x = 1$ when $a = 0$ , and identify which clause of the definition fails.

Cheatsheet

Core Reference

	Statement	How It Fails
1. Limit exists	$\lim_{x \to a} f(x)$ is a finite number	Jump, asymptote, or oscillation
2. Value exists	$f(a)$ is defined	$a$ outside the domain, or a hole
3. They agree	$\lim_{x \to a} f(x) = f(a)$	Value displaced from the limit
From the right	$\lim_{x \to a^+} f(x) = f(a)$	Needed at a left endpoint
From the left	$\lim_{x \to a^-} f(x) = f(a)$	Needed at a right endpoint
Elementary functions	Continuous on their domains	Only at excluded points

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