Unit 1Redefining Functions

Concept 4.137Built for May 2027 AP Exam

Redefining Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

A removable discontinuity has a standard repair: compute the limit and define the function to take that value. The repair is not a guess and not a fudge, because the limit is the only number that can make all three clauses of the continuity definition true at once. What makes the procedure legitimate is that it changes the function at exactly one input and nowhere else, so every other limit in the problem is untouched. The same procedure fails completely at a jump, at an asymptote, and at an oscillation, because in those cases there is no single candidate value for the limit to supply.

On the AP Exam

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

What you’ll see: "Find the value of $k$ that makes $f$ continuous at $x = 2$ ", a function defined with a hole and a question about the continuous extension, or a two-part item asking first for a limit and then for the definition that removes the break.
What you’ll do: Compute the limit at the point, assign it as the value, and then verify continuity by checking that the new value matches both one-sided limits.
Memory hook: The limit is the only value that can close the gap.
Use it on the exam: For MCQs: The answer is always the limit. If a choice is anything else, it came from evaluating the original formula at the point, which is exactly what the hole prevents. For FRQs: Write the redefinition as its own line, " $f(2) = 4$ ", and then state that the function is now continuous there because the limit and the value agree. On released BC papers: Redefining a function to remove a discontinuity has not appeared as a scored free-response part in the 2017–2026 BC sets. The idea is nonetheless exam-relevant in a different guise: parts that introduce a function by an integral or by a graph implicitly rely on the continuous extension when they evaluate the function at an endpoint, and stems that ask for a limit of a difference quotient assume the value at the point has already been assigned consistently. So the count is zero for the mechanic, and the mechanic is what makes several other parts legible.

Concept Review

One Point, One Change

Figure 1 — The repair moves the point, not the curve.

Caption. The two panels differ in a single mark. The limit was $2$ before the repair and it is $2$ after, which is why the repair is invisible everywhere except at the point that was rewritten.

The procedure, in three steps:

Compute $\lim_{x \to a} f(x)$ , using whatever algebra the expression requires.
Define $f(a)$ to equal that limit.
Verify that the redefined function satisfies all three clauses at $a$ , and note that no other input was affected.

Two panels labelled before and after: the first shows a curve with an open circle at the point a, the second the same curve with the circle filled, joined by an arrow labelled define f of a equals L

Why the Repair Is Legal

Redefining a single value changes the function only at that input. Limits at every other point are computed from values nearby, so they are unchanged. Even the limit at $a$ itself is unchanged, because a limit at $a$ never reads $f(a)$ .

That is the entire justification, and it is worth stating explicitly on a free response: the repair works because the limit was already in place and only the value was missing or displaced.

When No Repair Exists

The procedure requires a limit to assign. Without one there is no candidate value, and the four cases sort themselves out cleanly:

Break	Does a limit exist to assign?	Repair available?
Removable	Yes, a finite $L$	Yes: define $f(a) = L$
Jump	No, the sides differ	No: no single value satisfies both
Infinite	No, a side is unbounded	No: no finite value to assign
Oscillatory	No, neither side settles	No: no candidate exists

A useful test for a multiple-choice item: if the two one-sided limits disagree, no redefinition can help, and every answer choice offering a value is wrong.

Continuous Extensions

When a function is defined on a punctured interval and its limit exists at the missing point, the redefined function is called the continuous extension of the original. This language appears in stems that ask you to "extend $f$ continuously" or to "remove the discontinuity", and the operation is the same: assign the limit.

An extension is only possible where the limit exists. A function with a jump on an interval cannot be extended continuously across that point, no matter how the value is chosen.

Common Misconception Check

Myth 1: "Any convenient value will repair a removable break."
- Truth: Only the limit works. Any other value leaves clause three failing, which is simply a different displaced point.
Myth 2: "Redefining the value also changes the limit."
- Truth: The limit at $a$ ignores $f(a)$ by definition. The repair changes nothing about the approach.
Myth 3: "Every discontinuity can be removed with the right definition."
- Truth: Only those where a finite two-sided limit exists. Jumps, asymptotes, and oscillations have no candidate value.

Topic-Specific Study Method

Compute, Assign, Verify

Borrowed idea: A repair is a definition, not a discovery. You are not finding the missing value; you are choosing it. That choice is legal only when the limit exists, because the limit is the only candidate that preserves continuity. So the procedure is always the same. Compute the limit. Assign it, then check that the jump has disappeared, and you will find that the new function agrees with the old one everywhere except at the single point you rewrote.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
"Find $k$ so that $f$ is continuous"	A piecewise with a constant at the point	Which value closes clause three?	Compute the limit, set $k$ equal to it
Hole in a rational function	A cancelled common factor	What is the continuous extension?	Limit by cancellation, then assign
Jump in a piecewise	Two branches meeting at different heights	Can any value repair it?	No: the one-sided limits differ
Displaced point	A branch that disagrees with the curve	Which number moves?	The value moves to the limit
"Extend $f$ continuously"	A function on a punctured domain	Where is the extension possible?	Only where the limit exists

Executable Action Steps: ① Compute the limit before touching the definition. The limit is the answer to every version of this question, so produce it first and let it drive the rest. ② Write the new definition as a separate line of the piecewise. " $f(2) = 4$ " belongs in the definition, not in a sentence of commentary, because that is what the grader is looking for. ③ Verify by comparing three numbers. The left limit, the right limit, and the new value should all be the same number; if they are, the repair is complete.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \dfrac{x^2 - 4}{x - 2}$ for $x \neq 2$ . Find the value of $f(2)$ that makes $f$ continuous at $x = 2$ .

Standard Execution:

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

Since the limit exists and equals $4$ , the only candidate value is $4$ .

Define $f(2) = 4$ . Then the limit, the value, and the equality all hold.

The function is continuous at $2$ , and no other input was changed.

Exam Variation (Context / Parameter Shift): 1. A Jump That Cannot Be Repaired. Let $g(x) = \frac{\lvert x - 1 \rvert}{x - 1}$ for $x \neq 1$ . Can $g(1)$ be defined to make $g$ continuous at $1$ ?

Solution: No. The left-hand limit is $-1$ and the right-hand limit is $1$ , so no single value can equal both. Every choice of $g(1)$ leaves the two-sided limit failing.

2. An Extension at an Endpoint. Let $h(x) = \frac{\sin x}{x}$ for $x \neq 0$ . What value at $0$ extends $h$ continuously?

Solution: The limit is $1$ by the squeeze theorem applied to the geometric inequality, so $h(0) = 1$ is the only continuous extension.

Common Traps & Failure Signals:

The substituted value (Signal: your answer equals the original formula evaluated at $a$ ): The formula is undefined or wrong at $a$ ; that is why the question exists.
The repair at a jump (Signal: you assigned one of the two one-sided limits): Both sides must agree before a repair is possible.
The claim that the limit changed (Signal: your justification says the new value makes the limit exist): The limit existed before the repair; only the value was missing.

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

Redefinition Refused: You declared a removable break permanent.
- Limit Unused: Because you invented a value without consulting the limit, the repaired function disagreed with the curve at the very point you were trying to fix, and the new break was identical in kind to the old one.
- Both Sides Assumed: You defined a value from one side only, so the repair succeeded on the half of the graph you examined and failed on the other.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No part asks for the redefinition; continuous extension is assumed rather than demanded
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Zero parts, though the mechanic is what makes several other parts legible
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Finding the Value

Let $f(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \neq 3$ . What value of $f(3)$ makes $f$ continuous at $x = 3$ ?

MCQ 2

When No Repair Is Possible

Let $g(x) = \dfrac{x + 2}{x - 1}$ for $x \neq 1$ . Which statement is correct?

MCQ 3

After the Repair

Let $f(x) = \dfrac{x^2 - 1}{x - 1}$ for $x \neq 1$ and $f(1) = 2$ . Which statement is true?

FRQ 4

Repairing and Refusing to Repair

Let $f(x) = \dfrac{x^2 + x - 6}{x - 2}$ for $x \neq 2$ , and let $g(x) = \dfrac{\lvert x - 2 \rvert}{x - 2}$ for $x \neq 2$ .

(a) Find $\lim_{x \to 2} f(x)$ and state the value of $f(2)$ that makes $f$ continuous at $x = 2$ .

(b) Determine whether any value of $g(2)$ makes $g$ continuous at $x = 2$ , and justify your answer with one-sided limits.

(c) Explain why the same procedure succeeds for $f$ and fails for $g$ , referring to the definition of continuity.

Cheatsheet

Core Reference

	The Limit	The Repair	Result
Hole in a rational	Finite, found by cancellation	Set $f(a) = L$	Continuous
Displaced point	Finite, reaches $L$	Move the value to $L$	Continuous
Jump	Does not exist	None possible	Stays discontinuous
Infinite break	Not finite	None possible	Stays discontinuous
Oscillation	Does not exist	None possible	Stays discontinuous
Missing endpoint of a domain	One-sided limit exists	Set the value to that limit	Continuous from that side

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Unit 1Redefining Functions

Concept 4.137Built for May 2027 AP Exam

Redefining Functions

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: "Find the value of $k$ that makes $f$ continuous at $x = 2$ ", a function defined with a hole and a question about the continuous extension, or a two-part item asking first for a limit and then for the definition that removes the break.
What you’ll do: Compute the limit at the point, assign it as the value, and then verify continuity by checking that the new value matches both one-sided limits.
Memory hook: The limit is the only value that can close the gap.
Use it on the exam: For MCQs: The answer is always the limit. If a choice is anything else, it came from evaluating the original formula at the point, which is exactly what the hole prevents. For FRQs: Write the redefinition as its own line, " $f(2) = 4$ ", and then state that the function is now continuous there because the limit and the value agree. On released BC papers: Redefining a function to remove a discontinuity has not appeared as a scored free-response part in the 2017–2026 BC sets. The idea is nonetheless exam-relevant in a different guise: parts that introduce a function by an integral or by a graph implicitly rely on the continuous extension when they evaluate the function at an endpoint, and stems that ask for a limit of a difference quotient assume the value at the point has already been assigned consistently. So the count is zero for the mechanic, and the mechanic is what makes several other parts legible.

Concept Review

One Point, One Change

Figure 1 — The repair moves the point, not the curve.

Caption. The two panels differ in a single mark. The limit was $2$ before the repair and it is $2$ after, which is why the repair is invisible everywhere except at the point that was rewritten.

The procedure, in three steps:

Compute $\lim_{x \to a} f(x)$ , using whatever algebra the expression requires.
Define $f(a)$ to equal that limit.
Verify that the redefined function satisfies all three clauses at $a$ , and note that no other input was affected.

Two panels labelled before and after: the first shows a curve with an open circle at the point a, the second the same curve with the circle filled, joined by an arrow labelled define f of a equals L

Why the Repair Is Legal

That is the entire justification, and it is worth stating explicitly on a free response: the repair works because the limit was already in place and only the value was missing or displaced.

When No Repair Exists

The procedure requires a limit to assign. Without one there is no candidate value, and the four cases sort themselves out cleanly:

Break	Does a limit exist to assign?	Repair available?
Removable	Yes, a finite $L$	Yes: define $f(a) = L$
Jump	No, the sides differ	No: no single value satisfies both
Infinite	No, a side is unbounded	No: no finite value to assign
Oscillatory	No, neither side settles	No: no candidate exists

A useful test for a multiple-choice item: if the two one-sided limits disagree, no redefinition can help, and every answer choice offering a value is wrong.

Continuous Extensions

An extension is only possible where the limit exists. A function with a jump on an interval cannot be extended continuously across that point, no matter how the value is chosen.

Common Misconception Check

Myth 1: "Any convenient value will repair a removable break."
- Truth: Only the limit works. Any other value leaves clause three failing, which is simply a different displaced point.
Myth 2: "Redefining the value also changes the limit."
- Truth: The limit at $a$ ignores $f(a)$ by definition. The repair changes nothing about the approach.
Myth 3: "Every discontinuity can be removed with the right definition."
- Truth: Only those where a finite two-sided limit exists. Jumps, asymptotes, and oscillations have no candidate value.

Topic-Specific Study Method

Compute, Assign, Verify

Borrowed idea: A repair is a definition, not a discovery. You are not finding the missing value; you are choosing it. That choice is legal only when the limit exists, because the limit is the only candidate that preserves continuity. So the procedure is always the same. Compute the limit. Assign it, then check that the jump has disappeared, and you will find that the new function agrees with the old one everywhere except at the single point you rewrote.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
"Find $k$ so that $f$ is continuous"	A piecewise with a constant at the point	Which value closes clause three?	Compute the limit, set $k$ equal to it
Hole in a rational function	A cancelled common factor	What is the continuous extension?	Limit by cancellation, then assign
Jump in a piecewise	Two branches meeting at different heights	Can any value repair it?	No: the one-sided limits differ
Displaced point	A branch that disagrees with the curve	Which number moves?	The value moves to the limit
"Extend $f$ continuously"	A function on a punctured domain	Where is the extension possible?	Only where the limit exists

Executable Action Steps: ① Compute the limit before touching the definition. The limit is the answer to every version of this question, so produce it first and let it drive the rest. ② Write the new definition as a separate line of the piecewise. " $f(2) = 4$ " belongs in the definition, not in a sentence of commentary, because that is what the grader is looking for. ③ Verify by comparing three numbers. The left limit, the right limit, and the new value should all be the same number; if they are, the repair is complete.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \dfrac{x^2 - 4}{x - 2}$ for $x \neq 2$ . Find the value of $f(2)$ that makes $f$ continuous at $x = 2$ .

Standard Execution:

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

Since the limit exists and equals $4$ , the only candidate value is $4$ .

Define $f(2) = 4$ . Then the limit, the value, and the equality all hold.

The function is continuous at $2$ , and no other input was changed.

Solution: No. The left-hand limit is $-1$ and the right-hand limit is $1$ , so no single value can equal both. Every choice of $g(1)$ leaves the two-sided limit failing.

2. An Extension at an Endpoint. Let $h(x) = \frac{\sin x}{x}$ for $x \neq 0$ . What value at $0$ extends $h$ continuously?

Solution: The limit is $1$ by the squeeze theorem applied to the geometric inequality, so $h(0) = 1$ is the only continuous extension.

Common Traps & Failure Signals:

The substituted value (Signal: your answer equals the original formula evaluated at $a$ ): The formula is undefined or wrong at $a$ ; that is why the question exists.
The repair at a jump (Signal: you assigned one of the two one-sided limits): Both sides must agree before a repair is possible.
The claim that the limit changed (Signal: your justification says the new value makes the limit exist): The limit existed before the repair; only the value was missing.

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

Redefinition Refused: You declared a removable break permanent.
- Limit Unused: Because you invented a value without consulting the limit, the repaired function disagreed with the curve at the very point you were trying to fix, and the new break was identical in kind to the old one.
- Both Sides Assumed: You defined a value from one side only, so the repair succeeded on the half of the graph you examined and failed on the other.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No part asks for the redefinition; continuous extension is assumed rather than demanded
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Zero parts, though the mechanic is what makes several other parts legible
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Finding the Value

Let $f(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \neq 3$ . What value of $f(3)$ makes $f$ continuous at $x = 3$ ?

MCQ 2

When No Repair Is Possible

Let $g(x) = \dfrac{x + 2}{x - 1}$ for $x \neq 1$ . Which statement is correct?

MCQ 3

After the Repair

Let $f(x) = \dfrac{x^2 - 1}{x - 1}$ for $x \neq 1$ and $f(1) = 2$ . Which statement is true?

FRQ 4

Repairing and Refusing to Repair

Let $f(x) = \dfrac{x^2 + x - 6}{x - 2}$ for $x \neq 2$ , and let $g(x) = \dfrac{\lvert x - 2 \rvert}{x - 2}$ for $x \neq 2$ .

(a) Find $\lim_{x \to 2} f(x)$ and state the value of $f(2)$ that makes $f$ continuous at $x = 2$ .

(b) Determine whether any value of $g(2)$ makes $g$ continuous at $x = 2$ , and justify your answer with one-sided limits.

(c) Explain why the same procedure succeeds for $f$ and fails for $g$ , referring to the definition of continuity.

Cheatsheet

Core Reference

	The Limit	The Repair	Result
Hole in a rational	Finite, found by cancellation	Set $f(a) = L$	Continuous
Displaced point	Finite, reaches $L$	Move the value to $L$	Continuous
Jump	Does not exist	None possible	Stays discontinuous
Infinite break	Not finite	None possible	Stays discontinuous
Oscillation	Does not exist	None possible	Stays discontinuous
Missing endpoint of a domain	One-sided limit exists	Set the value to that limit	Continuous from that side

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Unit 1Redefining Functions

Concept 4.137Built for May 2027 AP Exam

Redefining Functions

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: "Find the value of $k$ that makes $f$ continuous at $x = 2$ ", a function defined with a hole and a question about the continuous extension, or a two-part item asking first for a limit and then for the definition that removes the break.
What you’ll do: Compute the limit at the point, assign it as the value, and then verify continuity by checking that the new value matches both one-sided limits.
Memory hook: The limit is the only value that can close the gap.
Use it on the exam: For MCQs: The answer is always the limit. If a choice is anything else, it came from evaluating the original formula at the point, which is exactly what the hole prevents. For FRQs: Write the redefinition as its own line, " $f(2) = 4$ ", and then state that the function is now continuous there because the limit and the value agree. On released BC papers: Redefining a function to remove a discontinuity has not appeared as a scored free-response part in the 2017–2026 BC sets. The idea is nonetheless exam-relevant in a different guise: parts that introduce a function by an integral or by a graph implicitly rely on the continuous extension when they evaluate the function at an endpoint, and stems that ask for a limit of a difference quotient assume the value at the point has already been assigned consistently. So the count is zero for the mechanic, and the mechanic is what makes several other parts legible.

Concept Review

One Point, One Change

Figure 1 — The repair moves the point, not the curve.

Caption. The two panels differ in a single mark. The limit was $2$ before the repair and it is $2$ after, which is why the repair is invisible everywhere except at the point that was rewritten.

The procedure, in three steps:

Compute $\lim_{x \to a} f(x)$ , using whatever algebra the expression requires.
Define $f(a)$ to equal that limit.
Verify that the redefined function satisfies all three clauses at $a$ , and note that no other input was affected.

Two panels labelled before and after: the first shows a curve with an open circle at the point a, the second the same curve with the circle filled, joined by an arrow labelled define f of a equals L

Why the Repair Is Legal

That is the entire justification, and it is worth stating explicitly on a free response: the repair works because the limit was already in place and only the value was missing or displaced.

When No Repair Exists

The procedure requires a limit to assign. Without one there is no candidate value, and the four cases sort themselves out cleanly:

Break	Does a limit exist to assign?	Repair available?
Removable	Yes, a finite $L$	Yes: define $f(a) = L$
Jump	No, the sides differ	No: no single value satisfies both
Infinite	No, a side is unbounded	No: no finite value to assign
Oscillatory	No, neither side settles	No: no candidate exists

A useful test for a multiple-choice item: if the two one-sided limits disagree, no redefinition can help, and every answer choice offering a value is wrong.

Continuous Extensions

An extension is only possible where the limit exists. A function with a jump on an interval cannot be extended continuously across that point, no matter how the value is chosen.

Common Misconception Check

Myth 1: "Any convenient value will repair a removable break."
- Truth: Only the limit works. Any other value leaves clause three failing, which is simply a different displaced point.
Myth 2: "Redefining the value also changes the limit."
- Truth: The limit at $a$ ignores $f(a)$ by definition. The repair changes nothing about the approach.
Myth 3: "Every discontinuity can be removed with the right definition."
- Truth: Only those where a finite two-sided limit exists. Jumps, asymptotes, and oscillations have no candidate value.

Topic-Specific Study Method

Compute, Assign, Verify

Borrowed idea: A repair is a definition, not a discovery. You are not finding the missing value; you are choosing it. That choice is legal only when the limit exists, because the limit is the only candidate that preserves continuity. So the procedure is always the same. Compute the limit. Assign it, then check that the jump has disappeared, and you will find that the new function agrees with the old one everywhere except at the single point you rewrote.

Exam Scenario	What the Page Shows	What the Prompt Is Really Asking	Required Execution
"Find $k$ so that $f$ is continuous"	A piecewise with a constant at the point	Which value closes clause three?	Compute the limit, set $k$ equal to it
Hole in a rational function	A cancelled common factor	What is the continuous extension?	Limit by cancellation, then assign
Jump in a piecewise	Two branches meeting at different heights	Can any value repair it?	No: the one-sided limits differ
Displaced point	A branch that disagrees with the curve	Which number moves?	The value moves to the limit
"Extend $f$ continuously"	A function on a punctured domain	Where is the extension possible?	Only where the limit exists

Executable Action Steps: ① Compute the limit before touching the definition. The limit is the answer to every version of this question, so produce it first and let it drive the rest. ② Write the new definition as a separate line of the piecewise. " $f(2) = 4$ " belongs in the definition, not in a sentence of commentary, because that is what the grader is looking for. ③ Verify by comparing three numbers. The left limit, the right limit, and the new value should all be the same number; if they are, the repair is complete.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \dfrac{x^2 - 4}{x - 2}$ for $x \neq 2$ . Find the value of $f(2)$ that makes $f$ continuous at $x = 2$ .

Standard Execution:

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

Since the limit exists and equals $4$ , the only candidate value is $4$ .

Define $f(2) = 4$ . Then the limit, the value, and the equality all hold.

The function is continuous at $2$ , and no other input was changed.

Solution: No. The left-hand limit is $-1$ and the right-hand limit is $1$ , so no single value can equal both. Every choice of $g(1)$ leaves the two-sided limit failing.

2. An Extension at an Endpoint. Let $h(x) = \frac{\sin x}{x}$ for $x \neq 0$ . What value at $0$ extends $h$ continuously?

Solution: The limit is $1$ by the squeeze theorem applied to the geometric inequality, so $h(0) = 1$ is the only continuous extension.

Common Traps & Failure Signals:

The substituted value (Signal: your answer equals the original formula evaluated at $a$ ): The formula is undefined or wrong at $a$ ; that is why the question exists.
The repair at a jump (Signal: you assigned one of the two one-sided limits): Both sides must agree before a repair is possible.
The claim that the limit changed (Signal: your justification says the new value makes the limit exist): The limit existed before the repair; only the value was missing.

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

Redefinition Refused: You declared a removable break permanent.
- Limit Unused: Because you invented a value without consulting the limit, the repaired function disagreed with the curve at the very point you were trying to fix, and the new break was identical in kind to the old one.
- Both Sides Assumed: You defined a value from one side only, so the repair succeeded on the half of the graph you examined and failed on the other.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No part asks for the redefinition; continuous extension is assumed rather than demanded
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Zero parts, though the mechanic is what makes several other parts legible
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Finding the Value

Let $f(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \neq 3$ . What value of $f(3)$ makes $f$ continuous at $x = 3$ ?

MCQ 2

When No Repair Is Possible

Let $g(x) = \dfrac{x + 2}{x - 1}$ for $x \neq 1$ . Which statement is correct?

MCQ 3

After the Repair

Let $f(x) = \dfrac{x^2 - 1}{x - 1}$ for $x \neq 1$ and $f(1) = 2$ . Which statement is true?

FRQ 4

Repairing and Refusing to Repair

Let $f(x) = \dfrac{x^2 + x - 6}{x - 2}$ for $x \neq 2$ , and let $g(x) = \dfrac{\lvert x - 2 \rvert}{x - 2}$ for $x \neq 2$ .

(a) Find $\lim_{x \to 2} f(x)$ and state the value of $f(2)$ that makes $f$ continuous at $x = 2$ .

(b) Determine whether any value of $g(2)$ makes $g$ continuous at $x = 2$ , and justify your answer with one-sided limits.

(c) Explain why the same procedure succeeds for $f$ and fails for $g$ , referring to the definition of continuity.

Cheatsheet

Core Reference

	The Limit	The Repair	Result
Hole in a rational	Finite, found by cancellation	Set $f(a) = L$	Continuous
Displaced point	Finite, reaches $L$	Move the value to $L$	Continuous
Jump	Does not exist	None possible	Stays discontinuous
Infinite break	Not finite	None possible	Stays discontinuous
Oscillation	Does not exist	None possible	Stays discontinuous
Missing endpoint of a domain	One-sided limit exists	Set the value to that limit	Continuous from that side

Sentences That Score

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