Unit 1The Three-Part Test for Continuity

Concept 4.128Built for May 2027 AP Exam

The Three-Part Test for Continuity

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The three-part test is the version of continuity that gets written on the page. It asks for the value $f(a)$ , the two-sided limit $\lim_{x \to a} f(x)$ , and the comparison between them, in that order, in three separate statements. Each statement is checkable on its own, which is why the rubric reads them separately. On a piecewise definition the test becomes an equation: set the two one-sided limits equal and solve for whatever parameter is loose. The failure mode is not arithmetic, it is omission. A response that computes the limit and then announces continuity has skipped the sentence that carries the point.

On the AP Exam

This is the highest-frequency justification in Unit 1, because continuity has to be established before almost anything else can be claimed. Multiple choice compresses the test into one question: which condition fails, what parameter repairs the seam, which of four functions is continuous at a point. Free response expands it into a scoring line where the three statements are read separately, and partial credit is real: two of three statements can earn partial credit that a single sentence cannot.

What you’ll see: Piecewise definitions with a loose parameter, functions defined separately at a single point, and prompts that ask you to "justify your answer" without naming the test.
What you’ll do: Write the three statements in order, give each one a number, and close with the verdict. For a parameter, build one equation from the two one-sided limits.
Memory hook: Value, limit, compare — then say the word. The verdict comes last because it is a conclusion, not evidence.
Use it on the exam: For MCQs: If the question asks which condition fails, compute the limit before reading the choices. The value is given, the limit is the work, and the comparison is the answer. For FRQs: Number the three statements. Readers grade by line, and three labelled lines make the three points visible even when the arithmetic has a slip in it.

Concept Review

Three Statements, Three Purposes

The test is not a definition to recite. It is a sequence of three claims, each of which can fail on its own.

Statement	What it establishes	How it is written	Typical omission
One	The value exists and is a real number	" $f(2) = 3$ "	Assumed rather than stated
Two	The two-sided limit exists	" $\lim_{x \to 2^{-}} f(x) = \lim_{x \to 2^{+}} f(x) = 4$ , so $\lim_{x \to 2} f(x) = 4$ "	One side computed and doubled
Three	The two agree	"Since $4 \neq 3$ , the conditions for continuity fail"	The comparison, skipped after the limit is found

Only statement three is a comparison, and only statement three can be written in one short clause. Students spend their effort on statement two because it is the arithmetic, and then treat the conclusion as self-evident.

Running the Test on a Piecewise Definition

A piecewise definition is where the test earns its keep, because the value and the limit usually come from different lines of the definition.

Read the value from the branch that owns $a$ . The condition on that branch tells you whether the branch applies at $a$ itself.
Evaluate each one-sided limit on its own branch. Left branch for $x < a$ , right branch for $x > a$ . The value at $a$ is irrelevant to both.
Force the two limits together if a parameter is loose. Setting them equal is the whole equation. Solve it, then check that the resulting value also matches $f(a)$ .
Write the verdict with the numbers in it. "With $k = 1$ the limit is $5$ and $f(2) = 5$ , so $f$ is continuous at $2$ " is a finished answer.

Figure 1 — A seam with a gap the parameter has to close.

Caption. The left branch approaches the seam from below and stops at an open circle at height $4$ . The right branch starts at the filled dot at height $5$ . Continuity is the demand that these two heights coincide, so the test turns into an equation with the parameter in it.

Two branches meeting at x equals 2 with a vertical gap: the left branch ends at an open circle at height 4, the right branch begins at a filled dot at height 5

Three Corrections Worth Memorising

“Both branches are polynomials, so the function is continuous at the seam.”
- Correct: each branch is continuous on its own domain. The seam belongs to neither, and continuity there is exactly what the test is asked to decide.
“I proved the limit exists, so continuity follows.”
- Correct: a limit is statement two. Continuity is the equality in statement three, and it can fail while the limit is perfectly finite.
“A function is continuous where it is defined.”
- Correct: that sentence describes a function continuous on its domain, which is a weaker claim than continuity at a point. The exam asks about named points.

Topic-Specific Study Method

The Three-Statement Protocol

Borrowed idea: Continuous. That is the word the reader is looking for, and it is earned by three statements in a fixed order. First the value, then the two-sided limit, then the comparison between them, and the comparison is the sentence most responses omit because the conclusion already feels obvious. When a piecewise definition hands you a parameter, the test turns into an equation, and the equation is always built from the two one-sided limits evaluated on the branches that own them. Write the three lines even when the answer is plain. The rubric reads lines, not intentions.

Number the statements before writing them. Three bullets, numbered one to three, in the order the test specifies. Numbering forces each claim to stand on its own and makes a missing comparison visible to you before it is visible to the reader.
Put a number on every line. " $f(2) = 3$ ", " $\lim = 4$ ", " $4 \neq 3$ ". A line with no number in it is a claim, and claims are not scored.
For a parameter, write the equation first, then solve. "Continuity requires $4 + k = 5$ " comes before " $k = 1$ ". Solving first and asserting continuity later inverts the evidence.
Close with the verdict and its scope. Say continuous or not, and say at which point. A verdict about the wrong point is the cheapest possible error and it happens under time pressure.

What the item asks	Statements you write	Where the marks are
"Is $f$ continuous at $a$ ?"	Value, limit, comparison	Each statement is a separate point
"Find $k$ so that $f$ is continuous"	Two one-sided limits, then the equation	The equation earns the method point
"Justify your answer"	The same three, prose form	The comparison sentence is mandatory
"Where is $f$ discontinuous?"	The test at each candidate point	Naming the failing statement per point
"Is $f$ continuous on $[0,3]$ ?"	Pointwise test plus a statement about the interval	The interval claim needs every point, not a sample

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \begin{cases} x^{2} + 1, & x < 1 \cr a, & x = 1 \cr 2x, & x > 1 \end{cases}$ . Find $a$ so that $f$ is continuous at $x = 1$ .

Standard Execution:

Statement one. $f(1) = a$ , which exists for every $a$ .

Statement two. From the left, $1 + 1 = 2$ ; from the right, $2(1) = 2$ . The two sides agree, so the limit is $2$ .

Statement three. Continuity requires $a = 2$ . With $a = 2$ all three conditions hold.

Exam Variation (Context / Parameter Shift): Now let $f(x) = \begin{cases} x^{2} + 1, & x < 1 \cr a, & x = 1 \cr 3x - 1, & x > 1 \end{cases}$ . Find $a$ , or explain why no value works.

Solution: The left limit is still $2$ , but the right limit becomes $3(1) - 1 = 2$ as well. So the limit is $2$ again and $a = 2$ still works. Change the right branch to $3x$ and the right limit becomes $3$ ; then the one-sided limits disagree, the two-sided limit does not exist, and no value of $a$ can make the function continuous at $1$ .

Common Traps & Failure Signals:

The unnamed comparison (Signal: a response that computes a limit and then writes "therefore continuous"): The comparison is the sentence being scored. Write it with both numbers visible.
The single-sided limit (Signal: one branch evaluated and the result used for both sides): At a seam, both branches must be evaluated, even when the definition looks symmetric.
The wrong-branch value (Signal: $f(a)$ read from the branch with the strict inequality): The branch that decides $f(a)$ is the one whose condition includes equality.

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

Comparison Omission: You end at the limit and call it continuity.
Seam Assumption: Because each branch is a familiar continuous function, the seam gets treated as continuous by inheritance, which is the one place the test actually has work to do.
Template Drift: You know the three statements but write them out of order, so the comparison appears before the limit exists and the argument reads backwards to a reader.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three scored lines inside a multi-part item
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Value, limit, and equality, each written out
FRQ Years Appeared: 2019 (Q6d)

Practice Examples

MCQ 1

Which Justification Is Complete

A function satisfies $f(3) = 5$ and $\lim_{x \to 3} f(x) = 5$ . Which sentence is a complete justification that $f$ is continuous at $x = 3$ ?

MCQ 2

Solving the Seam

Let $f(x) = \begin{cases} x^{2} - 1, & x < 2 \cr 3x - k, & x \ge 2 \end{cases}$ For which value of $k$ is $f$ continuous at $x = 2$ ?

MCQ 3

Which Function Is Continuous There

Which function is continuous at $x = 0$ ?

FRQ 4

The Test Written Out

Let $f(x) = \begin{cases} 2x + 1, & x < 2 \cr 5, & x = 2 \cr x^{2} + b, & x > 2 \end{cases}$

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

(b) Determine the value of $b$ for which $f$ is continuous at $x = 2$ , and verify all three conditions for that value.

(c) Using the value of $b$ from part (b), is $f$ continuous at $x = 1$ ? Justify with the three statements.

(d) A student writes: "At $x = 2$ the left branch gives $5$ and $f(2) = 5$ , so the function is continuous there." Identify what the student failed to check.

Cheatsheet

The Test, Line by Line

	What you write	What it proves	If it fails
One	" $f(a) =$ (a number)"	The function is defined at the point	The function is not continuous, whatever the limit does
Two	Both one-sided limits, then the two-sided limit	The limit exists	The point is a jump or an infinite discontinuity
Three	"Since $\lim = f(a)$ , the function is continuous"	The equality that defines continuity	The point is a removable discontinuity with a wrong value
Parameter	"Continuity requires $L_{-} = L_{+}$ "	The equation that fixes the unknown	No value works if the two sides cannot be made equal

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Unit 1The Three-Part Test for Continuity

Concept 4.128Built for May 2027 AP Exam

The Three-Part Test for Continuity

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: Piecewise definitions with a loose parameter, functions defined separately at a single point, and prompts that ask you to "justify your answer" without naming the test.
What you’ll do: Write the three statements in order, give each one a number, and close with the verdict. For a parameter, build one equation from the two one-sided limits.
Memory hook: Value, limit, compare — then say the word. The verdict comes last because it is a conclusion, not evidence.
Use it on the exam: For MCQs: If the question asks which condition fails, compute the limit before reading the choices. The value is given, the limit is the work, and the comparison is the answer. For FRQs: Number the three statements. Readers grade by line, and three labelled lines make the three points visible even when the arithmetic has a slip in it.

Concept Review

Three Statements, Three Purposes

The test is not a definition to recite. It is a sequence of three claims, each of which can fail on its own.

Statement	What it establishes	How it is written	Typical omission
One	The value exists and is a real number	" $f(2) = 3$ "	Assumed rather than stated
Two	The two-sided limit exists	" $\lim_{x \to 2^{-}} f(x) = \lim_{x \to 2^{+}} f(x) = 4$ , so $\lim_{x \to 2} f(x) = 4$ "	One side computed and doubled
Three	The two agree	"Since $4 \neq 3$ , the conditions for continuity fail"	The comparison, skipped after the limit is found

Running the Test on a Piecewise Definition

A piecewise definition is where the test earns its keep, because the value and the limit usually come from different lines of the definition.

Read the value from the branch that owns $a$ . The condition on that branch tells you whether the branch applies at $a$ itself.
Evaluate each one-sided limit on its own branch. Left branch for $x < a$ , right branch for $x > a$ . The value at $a$ is irrelevant to both.
Force the two limits together if a parameter is loose. Setting them equal is the whole equation. Solve it, then check that the resulting value also matches $f(a)$ .
Write the verdict with the numbers in it. "With $k = 1$ the limit is $5$ and $f(2) = 5$ , so $f$ is continuous at $2$ " is a finished answer.

Figure 1 — A seam with a gap the parameter has to close.

Two branches meeting at x equals 2 with a vertical gap: the left branch ends at an open circle at height 4, the right branch begins at a filled dot at height 5

Three Corrections Worth Memorising

“Both branches are polynomials, so the function is continuous at the seam.”
- Correct: each branch is continuous on its own domain. The seam belongs to neither, and continuity there is exactly what the test is asked to decide.
“I proved the limit exists, so continuity follows.”
- Correct: a limit is statement two. Continuity is the equality in statement three, and it can fail while the limit is perfectly finite.
“A function is continuous where it is defined.”
- Correct: that sentence describes a function continuous on its domain, which is a weaker claim than continuity at a point. The exam asks about named points.

Topic-Specific Study Method

The Three-Statement Protocol

Number the statements before writing them. Three bullets, numbered one to three, in the order the test specifies. Numbering forces each claim to stand on its own and makes a missing comparison visible to you before it is visible to the reader.
Put a number on every line. " $f(2) = 3$ ", " $\lim = 4$ ", " $4 \neq 3$ ". A line with no number in it is a claim, and claims are not scored.
For a parameter, write the equation first, then solve. "Continuity requires $4 + k = 5$ " comes before " $k = 1$ ". Solving first and asserting continuity later inverts the evidence.
Close with the verdict and its scope. Say continuous or not, and say at which point. A verdict about the wrong point is the cheapest possible error and it happens under time pressure.

What the item asks	Statements you write	Where the marks are
"Is $f$ continuous at $a$ ?"	Value, limit, comparison	Each statement is a separate point
"Find $k$ so that $f$ is continuous"	Two one-sided limits, then the equation	The equation earns the method point
"Justify your answer"	The same three, prose form	The comparison sentence is mandatory
"Where is $f$ discontinuous?"	The test at each candidate point	Naming the failing statement per point
"Is $f$ continuous on $[0,3]$ ?"	Pointwise test plus a statement about the interval	The interval claim needs every point, not a sample

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \begin{cases} x^{2} + 1, & x < 1 \cr a, & x = 1 \cr 2x, & x > 1 \end{cases}$ . Find $a$ so that $f$ is continuous at $x = 1$ .

Standard Execution:

Statement one. $f(1) = a$ , which exists for every $a$ .

Statement two. From the left, $1 + 1 = 2$ ; from the right, $2(1) = 2$ . The two sides agree, so the limit is $2$ .

Statement three. Continuity requires $a = 2$ . With $a = 2$ all three conditions hold.

Exam Variation (Context / Parameter Shift): Now let $f(x) = \begin{cases} x^{2} + 1, & x < 1 \cr a, & x = 1 \cr 3x - 1, & x > 1 \end{cases}$ . Find $a$ , or explain why no value works.

Solution: The left limit is still $2$ , but the right limit becomes $3(1) - 1 = 2$ as well. So the limit is $2$ again and $a = 2$ still works. Change the right branch to $3x$ and the right limit becomes $3$ ; then the one-sided limits disagree, the two-sided limit does not exist, and no value of $a$ can make the function continuous at $1$ .

Common Traps & Failure Signals:

The unnamed comparison (Signal: a response that computes a limit and then writes "therefore continuous"): The comparison is the sentence being scored. Write it with both numbers visible.
The single-sided limit (Signal: one branch evaluated and the result used for both sides): At a seam, both branches must be evaluated, even when the definition looks symmetric.
The wrong-branch value (Signal: $f(a)$ read from the branch with the strict inequality): The branch that decides $f(a)$ is the one whose condition includes equality.

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

Comparison Omission: You end at the limit and call it continuity.
Seam Assumption: Because each branch is a familiar continuous function, the seam gets treated as continuous by inheritance, which is the one place the test actually has work to do.
Template Drift: You know the three statements but write them out of order, so the comparison appears before the limit exists and the argument reads backwards to a reader.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three scored lines inside a multi-part item
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Value, limit, and equality, each written out
FRQ Years Appeared: 2019 (Q6d)

Practice Examples

MCQ 1

Which Justification Is Complete

A function satisfies $f(3) = 5$ and $\lim_{x \to 3} f(x) = 5$ . Which sentence is a complete justification that $f$ is continuous at $x = 3$ ?

MCQ 2

Solving the Seam

Let $f(x) = \begin{cases} x^{2} - 1, & x < 2 \cr 3x - k, & x \ge 2 \end{cases}$ For which value of $k$ is $f$ continuous at $x = 2$ ?

MCQ 3

Which Function Is Continuous There

Which function is continuous at $x = 0$ ?

FRQ 4

The Test Written Out

Let $f(x) = \begin{cases} 2x + 1, & x < 2 \cr 5, & x = 2 \cr x^{2} + b, & x > 2 \end{cases}$

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

(b) Determine the value of $b$ for which $f$ is continuous at $x = 2$ , and verify all three conditions for that value.

(c) Using the value of $b$ from part (b), is $f$ continuous at $x = 1$ ? Justify with the three statements.

(d) A student writes: "At $x = 2$ the left branch gives $5$ and $f(2) = 5$ , so the function is continuous there." Identify what the student failed to check.

Cheatsheet

The Test, Line by Line

	What you write	What it proves	If it fails
One	" $f(a) =$ (a number)"	The function is defined at the point	The function is not continuous, whatever the limit does
Two	Both one-sided limits, then the two-sided limit	The limit exists	The point is a jump or an infinite discontinuity
Three	"Since $\lim = f(a)$ , the function is continuous"	The equality that defines continuity	The point is a removable discontinuity with a wrong value
Parameter	"Continuity requires $L_{-} = L_{+}$ "	The equation that fixes the unknown	No value works if the two sides cannot be made equal

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Start Topic Practice Next Topic: 1.16 Continuity on a Closed Interval

Unit 1The Three-Part Test for Continuity

Concept 4.128Built for May 2027 AP Exam

The Three-Part Test for Continuity

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: Piecewise definitions with a loose parameter, functions defined separately at a single point, and prompts that ask you to "justify your answer" without naming the test.
What you’ll do: Write the three statements in order, give each one a number, and close with the verdict. For a parameter, build one equation from the two one-sided limits.
Memory hook: Value, limit, compare — then say the word. The verdict comes last because it is a conclusion, not evidence.
Use it on the exam: For MCQs: If the question asks which condition fails, compute the limit before reading the choices. The value is given, the limit is the work, and the comparison is the answer. For FRQs: Number the three statements. Readers grade by line, and three labelled lines make the three points visible even when the arithmetic has a slip in it.

Concept Review

Three Statements, Three Purposes

The test is not a definition to recite. It is a sequence of three claims, each of which can fail on its own.

Statement	What it establishes	How it is written	Typical omission
One	The value exists and is a real number	" $f(2) = 3$ "	Assumed rather than stated
Two	The two-sided limit exists	" $\lim_{x \to 2^{-}} f(x) = \lim_{x \to 2^{+}} f(x) = 4$ , so $\lim_{x \to 2} f(x) = 4$ "	One side computed and doubled
Three	The two agree	"Since $4 \neq 3$ , the conditions for continuity fail"	The comparison, skipped after the limit is found

Running the Test on a Piecewise Definition

A piecewise definition is where the test earns its keep, because the value and the limit usually come from different lines of the definition.

Read the value from the branch that owns $a$ . The condition on that branch tells you whether the branch applies at $a$ itself.
Evaluate each one-sided limit on its own branch. Left branch for $x < a$ , right branch for $x > a$ . The value at $a$ is irrelevant to both.
Force the two limits together if a parameter is loose. Setting them equal is the whole equation. Solve it, then check that the resulting value also matches $f(a)$ .
Write the verdict with the numbers in it. "With $k = 1$ the limit is $5$ and $f(2) = 5$ , so $f$ is continuous at $2$ " is a finished answer.

Figure 1 — A seam with a gap the parameter has to close.

Two branches meeting at x equals 2 with a vertical gap: the left branch ends at an open circle at height 4, the right branch begins at a filled dot at height 5

Three Corrections Worth Memorising

“Both branches are polynomials, so the function is continuous at the seam.”
- Correct: each branch is continuous on its own domain. The seam belongs to neither, and continuity there is exactly what the test is asked to decide.
“I proved the limit exists, so continuity follows.”
- Correct: a limit is statement two. Continuity is the equality in statement three, and it can fail while the limit is perfectly finite.
“A function is continuous where it is defined.”
- Correct: that sentence describes a function continuous on its domain, which is a weaker claim than continuity at a point. The exam asks about named points.

Topic-Specific Study Method

The Three-Statement Protocol

Number the statements before writing them. Three bullets, numbered one to three, in the order the test specifies. Numbering forces each claim to stand on its own and makes a missing comparison visible to you before it is visible to the reader.
Put a number on every line. " $f(2) = 3$ ", " $\lim = 4$ ", " $4 \neq 3$ ". A line with no number in it is a claim, and claims are not scored.
For a parameter, write the equation first, then solve. "Continuity requires $4 + k = 5$ " comes before " $k = 1$ ". Solving first and asserting continuity later inverts the evidence.
Close with the verdict and its scope. Say continuous or not, and say at which point. A verdict about the wrong point is the cheapest possible error and it happens under time pressure.

What the item asks	Statements you write	Where the marks are
"Is $f$ continuous at $a$ ?"	Value, limit, comparison	Each statement is a separate point
"Find $k$ so that $f$ is continuous"	Two one-sided limits, then the equation	The equation earns the method point
"Justify your answer"	The same three, prose form	The comparison sentence is mandatory
"Where is $f$ discontinuous?"	The test at each candidate point	Naming the failing statement per point
"Is $f$ continuous on $[0,3]$ ?"	Pointwise test plus a statement about the interval	The interval claim needs every point, not a sample

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \begin{cases} x^{2} + 1, & x < 1 \cr a, & x = 1 \cr 2x, & x > 1 \end{cases}$ . Find $a$ so that $f$ is continuous at $x = 1$ .

Standard Execution:

Statement one. $f(1) = a$ , which exists for every $a$ .

Statement two. From the left, $1 + 1 = 2$ ; from the right, $2(1) = 2$ . The two sides agree, so the limit is $2$ .

Statement three. Continuity requires $a = 2$ . With $a = 2$ all three conditions hold.

Exam Variation (Context / Parameter Shift): Now let $f(x) = \begin{cases} x^{2} + 1, & x < 1 \cr a, & x = 1 \cr 3x - 1, & x > 1 \end{cases}$ . Find $a$ , or explain why no value works.

Solution: The left limit is still $2$ , but the right limit becomes $3(1) - 1 = 2$ as well. So the limit is $2$ again and $a = 2$ still works. Change the right branch to $3x$ and the right limit becomes $3$ ; then the one-sided limits disagree, the two-sided limit does not exist, and no value of $a$ can make the function continuous at $1$ .

Common Traps & Failure Signals:

The unnamed comparison (Signal: a response that computes a limit and then writes "therefore continuous"): The comparison is the sentence being scored. Write it with both numbers visible.
The single-sided limit (Signal: one branch evaluated and the result used for both sides): At a seam, both branches must be evaluated, even when the definition looks symmetric.
The wrong-branch value (Signal: $f(a)$ read from the branch with the strict inequality): The branch that decides $f(a)$ is the one whose condition includes equality.

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

Comparison Omission: You end at the limit and call it continuity.
Seam Assumption: Because each branch is a familiar continuous function, the seam gets treated as continuous by inheritance, which is the one place the test actually has work to do.
Template Drift: You know the three statements but write them out of order, so the comparison appears before the limit exists and the argument reads backwards to a reader.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three scored lines inside a multi-part item
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Value, limit, and equality, each written out
FRQ Years Appeared: 2019 (Q6d)

Practice Examples

MCQ 1

Which Justification Is Complete

A function satisfies $f(3) = 5$ and $\lim_{x \to 3} f(x) = 5$ . Which sentence is a complete justification that $f$ is continuous at $x = 3$ ?

MCQ 2

Solving the Seam

Let $f(x) = \begin{cases} x^{2} - 1, & x < 2 \cr 3x - k, & x \ge 2 \end{cases}$ For which value of $k$ is $f$ continuous at $x = 2$ ?

MCQ 3

Which Function Is Continuous There

Which function is continuous at $x = 0$ ?

FRQ 4

The Test Written Out

Let $f(x) = \begin{cases} 2x + 1, & x < 2 \cr 5, & x = 2 \cr x^{2} + b, & x > 2 \end{cases}$

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

(b) Determine the value of $b$ for which $f$ is continuous at $x = 2$ , and verify all three conditions for that value.

(c) Using the value of $b$ from part (b), is $f$ continuous at $x = 1$ ? Justify with the three statements.

(d) A student writes: "At $x = 2$ the left branch gives $5$ and $f(2) = 5$ , so the function is continuous there." Identify what the student failed to check.

Cheatsheet

The Test, Line by Line

	What you write	What it proves	If it fails
One	" $f(a) =$ (a number)"	The function is defined at the point	The function is not continuous, whatever the limit does
Two	Both one-sided limits, then the two-sided limit	The limit exists	The point is a jump or an infinite discontinuity
Three	"Since $\lim = f(a)$ , the function is continuous"	The equality that defines continuity	The point is a removable discontinuity with a wrong value
Parameter	"Continuity requires $L_{-} = L_{+}$ "	The equation that fixes the unknown	No value works if the two sides cannot be made equal

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