Unit 1Definition of Continuity at a Point

Concept 4.65Built for May 2027 AP Exam

Definition of Continuity at a Point

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Continuity at a point is a three-part claim, not a one-part one. The function has to be defined there, the two-sided limit has to exist there, and the two have to be the same number. All three, or the function is not continuous. That is the whole definition, and almost every lost point on this topic comes from checking only the first part: a defined value is treated as a licence to declare continuity, when it is only the first of three gates. On the exam the definition arrives as a justification, not as a definition question, which means the three conditions have to be stated and, more importantly, verified with the actual limit.

On the AP Exam

Free response uses this definition as a scored template. A continuity justification is worth its point only when all three conditions appear and the limit is computed rather than asserted. Multiple choice tests the definition backwards: it hands you a function with one condition broken and asks which one, or supplies a parameter and asks what value repairs the third condition. Both formats reward the same habit, which is checking in order and refusing to skip the middle gate.

What you’ll see: Piecewise definitions at their seam, rational functions with a cancelled factor, functions with a stated value at a single point, and parameters to be solved for.
What you’ll do: Evaluate the three conditions in order, name the one that fails, and either classify the failure or find the parameter that repairs it.
Memory hook: Value, limit, match. Three words, and the third is the one that gets skipped.
Use it on the exam: For MCQs: Find the limit first, then compare it to the stated value. If the item asks which condition fails, the answer is usually the third, because the first two are hard to break without also breaking it visibly. For FRQs: Write the three conditions as three numbered statements with the numbers substituted in. "The limit is $4$ and $f(2) = 3$ , so the third condition fails" is worth a point; "the graph has a hole" is worth none.

Concept Review

The Definition, Stated Precisely

For a function $f$ and a point $a$ in an open interval around which $f$ is defined, $f$ is continuous at $a$ exactly when all three hold.

Condition	Statement	What breaking it looks like
One	$f(a)$ exists and is a real number.	A hole with no point drawn, or a value excluded from the domain.
Two	$\lim_{x \to a} f(x)$ exists.	Two different finite side values, or an unbounded side.
Three	$\lim_{x \to a} f(x) = f(a)$ .	A filled dot sitting at the wrong height while the curve approaches another.

The definition is an equation, $\lim_{x \to a} f(x) = f(a)$ , decorated with two existence conditions. When students say "the function is defined there, so it is continuous", they are quoting half of condition two and none of condition three.

The Three Gates

Work the gates in order. The order is not decorative: a failure at gate one makes gate two meaningless, and a failure at gate two makes gate three unanswerable.

Gate one: is there a value? Look for $f(a)$ in the definition. A piecewise function may assign it, a rational function may lose it to a cancelled factor.
Gate two: does the limit exist? Compute both one-sided limits. They must agree and both must be finite.
Gate three: do the numbers match? Compare the limit to $f(a)$ . This is the only gate that needs both previous answers, and it is the one the exam builds distractors around.

Figure 1 — Three gates, in series.

Caption. The third gate is shaded because it is where most responses stop being justifications. Gates one and two can both pass while the function is still discontinuous, and that combination is exactly what a misplaced dot produces.

Three numbered gates in series: a value exists at the point, the two-sided limit exists, and the two are equal

Three Corrections Worth Memorising

“ $f(a)$ is defined, so $f$ is continuous at $a$ .”
- Correct: defined is gate one. A value that disagrees with the limit fails gate three, and the function is discontinuous.
“The limit exists, so $f$ is continuous at $a$ .”
- Correct: the limit existing is gate two. Without a matching value, there is still a discontinuity, and the only discontinuity that a limit can repair is a removable one.
“Continuity is a property of the whole graph.”
- Correct: the definition above is local. It concerns one point and the behaviour immediately around it, and the exam asks about named points far more often than about intervals.

Topic-Specific Study Method

The Three-Gate Check

Borrowed idea: Check three things, in order. First, is there a value at the point? Second, do the two sides agree? Third, do those two answers match? Most students stop at the first one. Write the three answers on the page. Do it before you judge. The check takes ten seconds. Guessing takes longer.

Write the value first. One line: " $f(2) = 3$ " or " $f(2)$ undefined". If it is undefined, gates two and three cannot save the function, though the limit may still be worth computing for a later part.
Write the two one-sided limits as a pair. Then combine them into the two-sided limit. If they disagree, stop: the function is discontinuous and the classification work begins.
Compare the two numbers explicitly. Put the limit and the value side by side on the page. The comparison is the justification, and writing it as an inequality or a " $\neq$ " is faster than describing it in words.
State the verdict in the rubric's language. "Since $\lim_{x \to 2} f(x) = 4 \neq 3 = f(2)$ , $f$ is not continuous at $x = 2$ ." One sentence, three conditions accounted for.

What the item gives you	Gate that fails	What you must write
A piecewise rule with a stated value at the seam	Three, usually	The limit, the value, and the mismatch
A rational function with a cancelled factor and no point value	One	The limit, then "undefined at $a$ , so not continuous"
A rational function with a surviving factor downstairs	Two	The one-sided limits with signs
A parameter $k$ to be found	Three	The limit, then solve limit $= k$
A function continuous everywhere	None	All three conditions with the numbers filled in

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \dfrac{x^{2}-4}{x-2}$ for $x \neq 2$ and $f(2) = 3$ . Is $f$ continuous at $x = 2$ ?

Standard Execution:

Gate one. $f(2) = 3$ , so the value exists.

Gate two. For $x \neq 2$ , $f(x) = x + 2$ , so both one-sided limits equal $4$ and the two-sided limit is $4$ .

Gate three. $4 \neq 3$ , so the third condition fails. The function is not continuous at $2$ ; the discontinuity is removable, and redefining $f(2) = 4$ repairs it.

Exam Variation (Context / Parameter Shift): Let $g(x) = \begin{cases} x^{2} + k, & x < 2 \cr 3x - 1, & x \ge 2 \end{cases}$ . Find the value of $k$ that makes $g$ continuous at $x = 2$ .

Solution: Gate one holds for every $k$ , since $g(2) = 5$ . Gate two requires the two sides to agree: from the left, $\lim_{x \to 2^{-}} g(x) = 4 + k$ ; from the right, $\lim_{x \to 2^{+}} g(x) = 5$ . Setting $4 + k = 5$ gives $k = 1$ , and then the limit equals $g(2) = 5$ , so gate three holds as well.

Common Traps & Failure Signals:

Value-only verdict (Signal: the word "continuous" appears before any limit is computed): Defined is one gate. Write the limit before the verdict.
Limit-only verdict (Signal: a limit computed with no reference to $f(a)$ ): A limit that exists proves nothing about continuity on its own; the comparison is the point.
Parameter solved against the wrong side (Signal: a piecewise $k$ set equal to the branch containing $k$ ): Choose the branch that does not contain the unknown, then solve.

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

Gate Skipping: You check the value and stop.
Definition Truncation: You remembered "defined and the limit exists" and dropped the equality, which is the only clause that connects the two halves of the definition and the only clause the rubric asks you to verify explicitly.
Parameter Drift: You solve for $k$ using the wrong branch at the seam.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; As a justification line, rarely as a definition question
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Three conditions checked in order at a named point
FRQ Years Appeared: 2019 (Q6d)

Practice Examples

MCQ 1

Which Gate Fails

Let $f(x) = \dfrac{x^{2}-4}{x-2}$ for $x \neq 2$ and $f(2) = 3$ . Which condition for continuity at $x = 2$ fails?

MCQ 2

The Parameter That Repairs the Seam

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

MCQ 3

Reading the Definition Backwards

A function $f$ satisfies $\lim_{x \to 4} f(x) = 7$ and $f(4) = 7$ . Which statement must be true?

FRQ 4

Continuity, Proved and Repaired

Let $f(x) = \begin{cases} \dfrac{x^{2} - 9}{x - 3}, & x < 3 \cr k, & x = 3 \cr 2x - 1, & x > 3 \end{cases}$

(a) Evaluate $\lim_{x \to 3^{-}} f(x)$ and $\lim_{x \to 3^{+}} f(x)$ .

(b) Determine whether $\lim_{x \to 3} f(x)$ exists, and justify your answer.

(c) Determine whether any value of $k$ makes $f$ continuous at $x = 3$ , and justify your conclusion.

(d) A student writes: "Since $k$ can be chosen to make the limit equal the value, $f$ is continuous at $x = 3$ for every $k$ ." Identify the error.

Cheatsheet

Checking Continuity at a Point

	Gate one	Gate two	Gate three	Verdict
Polynomial at any real $a$	Passes	Passes	Passes	Continuous
Rational with cancelled factor, no value assigned	Fails	Passes	Cannot be checked	Not continuous; removable
Rational with cancelled factor, value displaced	Passes	Passes	Fails	Not continuous; removable
Piecewise with unequal finite sides	Passes	Fails	Cannot be checked	Not continuous; jump
Piecewise with equal sides and matching value	Passes	Passes	Passes	Continuous

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Start Topic Practice Next Topic: 1.15 The Three-Part Test for Continuity

Unit 1Definition of Continuity at a Point

Concept 4.65Built for May 2027 AP Exam

Definition of Continuity at a Point

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: Piecewise definitions at their seam, rational functions with a cancelled factor, functions with a stated value at a single point, and parameters to be solved for.
What you’ll do: Evaluate the three conditions in order, name the one that fails, and either classify the failure or find the parameter that repairs it.
Memory hook: Value, limit, match. Three words, and the third is the one that gets skipped.
Use it on the exam: For MCQs: Find the limit first, then compare it to the stated value. If the item asks which condition fails, the answer is usually the third, because the first two are hard to break without also breaking it visibly. For FRQs: Write the three conditions as three numbered statements with the numbers substituted in. "The limit is $4$ and $f(2) = 3$ , so the third condition fails" is worth a point; "the graph has a hole" is worth none.

Concept Review

The Definition, Stated Precisely

For a function $f$ and a point $a$ in an open interval around which $f$ is defined, $f$ is continuous at $a$ exactly when all three hold.

Condition	Statement	What breaking it looks like
One	$f(a)$ exists and is a real number.	A hole with no point drawn, or a value excluded from the domain.
Two	$\lim_{x \to a} f(x)$ exists.	Two different finite side values, or an unbounded side.
Three	$\lim_{x \to a} f(x) = f(a)$ .	A filled dot sitting at the wrong height while the curve approaches another.

The Three Gates

Work the gates in order. The order is not decorative: a failure at gate one makes gate two meaningless, and a failure at gate two makes gate three unanswerable.

Gate one: is there a value? Look for $f(a)$ in the definition. A piecewise function may assign it, a rational function may lose it to a cancelled factor.
Gate two: does the limit exist? Compute both one-sided limits. They must agree and both must be finite.
Gate three: do the numbers match? Compare the limit to $f(a)$ . This is the only gate that needs both previous answers, and it is the one the exam builds distractors around.

Figure 1 — Three gates, in series.

Three numbered gates in series: a value exists at the point, the two-sided limit exists, and the two are equal

Three Corrections Worth Memorising

“ $f(a)$ is defined, so $f$ is continuous at $a$ .”
- Correct: defined is gate one. A value that disagrees with the limit fails gate three, and the function is discontinuous.
“The limit exists, so $f$ is continuous at $a$ .”
- Correct: the limit existing is gate two. Without a matching value, there is still a discontinuity, and the only discontinuity that a limit can repair is a removable one.
“Continuity is a property of the whole graph.”
- Correct: the definition above is local. It concerns one point and the behaviour immediately around it, and the exam asks about named points far more often than about intervals.

Topic-Specific Study Method

The Three-Gate Check

Write the value first. One line: " $f(2) = 3$ " or " $f(2)$ undefined". If it is undefined, gates two and three cannot save the function, though the limit may still be worth computing for a later part.
Write the two one-sided limits as a pair. Then combine them into the two-sided limit. If they disagree, stop: the function is discontinuous and the classification work begins.
Compare the two numbers explicitly. Put the limit and the value side by side on the page. The comparison is the justification, and writing it as an inequality or a " $\neq$ " is faster than describing it in words.
State the verdict in the rubric's language. "Since $\lim_{x \to 2} f(x) = 4 \neq 3 = f(2)$ , $f$ is not continuous at $x = 2$ ." One sentence, three conditions accounted for.

What the item gives you	Gate that fails	What you must write
A piecewise rule with a stated value at the seam	Three, usually	The limit, the value, and the mismatch
A rational function with a cancelled factor and no point value	One	The limit, then "undefined at $a$ , so not continuous"
A rational function with a surviving factor downstairs	Two	The one-sided limits with signs
A parameter $k$ to be found	Three	The limit, then solve limit $= k$
A function continuous everywhere	None	All three conditions with the numbers filled in

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \dfrac{x^{2}-4}{x-2}$ for $x \neq 2$ and $f(2) = 3$ . Is $f$ continuous at $x = 2$ ?

Standard Execution:

Gate one. $f(2) = 3$ , so the value exists.

Gate two. For $x \neq 2$ , $f(x) = x + 2$ , so both one-sided limits equal $4$ and the two-sided limit is $4$ .

Gate three. $4 \neq 3$ , so the third condition fails. The function is not continuous at $2$ ; the discontinuity is removable, and redefining $f(2) = 4$ repairs it.

Exam Variation (Context / Parameter Shift): Let $g(x) = \begin{cases} x^{2} + k, & x < 2 \cr 3x - 1, & x \ge 2 \end{cases}$ . Find the value of $k$ that makes $g$ continuous at $x = 2$ .

Solution: Gate one holds for every $k$ , since $g(2) = 5$ . Gate two requires the two sides to agree: from the left, $\lim_{x \to 2^{-}} g(x) = 4 + k$ ; from the right, $\lim_{x \to 2^{+}} g(x) = 5$ . Setting $4 + k = 5$ gives $k = 1$ , and then the limit equals $g(2) = 5$ , so gate three holds as well.

Common Traps & Failure Signals:

Value-only verdict (Signal: the word "continuous" appears before any limit is computed): Defined is one gate. Write the limit before the verdict.
Limit-only verdict (Signal: a limit computed with no reference to $f(a)$ ): A limit that exists proves nothing about continuity on its own; the comparison is the point.
Parameter solved against the wrong side (Signal: a piecewise $k$ set equal to the branch containing $k$ ): Choose the branch that does not contain the unknown, then solve.

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

Gate Skipping: You check the value and stop.
Definition Truncation: You remembered "defined and the limit exists" and dropped the equality, which is the only clause that connects the two halves of the definition and the only clause the rubric asks you to verify explicitly.
Parameter Drift: You solve for $k$ using the wrong branch at the seam.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; As a justification line, rarely as a definition question
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Three conditions checked in order at a named point
FRQ Years Appeared: 2019 (Q6d)

Practice Examples

MCQ 1

Which Gate Fails

Let $f(x) = \dfrac{x^{2}-4}{x-2}$ for $x \neq 2$ and $f(2) = 3$ . Which condition for continuity at $x = 2$ fails?

MCQ 2

The Parameter That Repairs the Seam

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

MCQ 3

Reading the Definition Backwards

A function $f$ satisfies $\lim_{x \to 4} f(x) = 7$ and $f(4) = 7$ . Which statement must be true?

FRQ 4

Continuity, Proved and Repaired

Let $f(x) = \begin{cases} \dfrac{x^{2} - 9}{x - 3}, & x < 3 \cr k, & x = 3 \cr 2x - 1, & x > 3 \end{cases}$

(a) Evaluate $\lim_{x \to 3^{-}} f(x)$ and $\lim_{x \to 3^{+}} f(x)$ .

(b) Determine whether $\lim_{x \to 3} f(x)$ exists, and justify your answer.

(c) Determine whether any value of $k$ makes $f$ continuous at $x = 3$ , and justify your conclusion.

(d) A student writes: "Since $k$ can be chosen to make the limit equal the value, $f$ is continuous at $x = 3$ for every $k$ ." Identify the error.

Cheatsheet

Checking Continuity at a Point

	Gate one	Gate two	Gate three	Verdict
Polynomial at any real $a$	Passes	Passes	Passes	Continuous
Rational with cancelled factor, no value assigned	Fails	Passes	Cannot be checked	Not continuous; removable
Rational with cancelled factor, value displaced	Passes	Passes	Fails	Not continuous; removable
Piecewise with unequal finite sides	Passes	Fails	Cannot be checked	Not continuous; jump
Piecewise with equal sides and matching value	Passes	Passes	Passes	Continuous

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Unit 1Definition of Continuity at a Point

Concept 4.65Built for May 2027 AP Exam

Definition of Continuity at a Point

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: Piecewise definitions at their seam, rational functions with a cancelled factor, functions with a stated value at a single point, and parameters to be solved for.
What you’ll do: Evaluate the three conditions in order, name the one that fails, and either classify the failure or find the parameter that repairs it.
Memory hook: Value, limit, match. Three words, and the third is the one that gets skipped.
Use it on the exam: For MCQs: Find the limit first, then compare it to the stated value. If the item asks which condition fails, the answer is usually the third, because the first two are hard to break without also breaking it visibly. For FRQs: Write the three conditions as three numbered statements with the numbers substituted in. "The limit is $4$ and $f(2) = 3$ , so the third condition fails" is worth a point; "the graph has a hole" is worth none.

Concept Review

The Definition, Stated Precisely

For a function $f$ and a point $a$ in an open interval around which $f$ is defined, $f$ is continuous at $a$ exactly when all three hold.

Condition	Statement	What breaking it looks like
One	$f(a)$ exists and is a real number.	A hole with no point drawn, or a value excluded from the domain.
Two	$\lim_{x \to a} f(x)$ exists.	Two different finite side values, or an unbounded side.
Three	$\lim_{x \to a} f(x) = f(a)$ .	A filled dot sitting at the wrong height while the curve approaches another.

The Three Gates

Work the gates in order. The order is not decorative: a failure at gate one makes gate two meaningless, and a failure at gate two makes gate three unanswerable.

Gate one: is there a value? Look for $f(a)$ in the definition. A piecewise function may assign it, a rational function may lose it to a cancelled factor.
Gate two: does the limit exist? Compute both one-sided limits. They must agree and both must be finite.
Gate three: do the numbers match? Compare the limit to $f(a)$ . This is the only gate that needs both previous answers, and it is the one the exam builds distractors around.

Figure 1 — Three gates, in series.

Three numbered gates in series: a value exists at the point, the two-sided limit exists, and the two are equal

Three Corrections Worth Memorising

“ $f(a)$ is defined, so $f$ is continuous at $a$ .”
- Correct: defined is gate one. A value that disagrees with the limit fails gate three, and the function is discontinuous.
“The limit exists, so $f$ is continuous at $a$ .”
- Correct: the limit existing is gate two. Without a matching value, there is still a discontinuity, and the only discontinuity that a limit can repair is a removable one.
“Continuity is a property of the whole graph.”
- Correct: the definition above is local. It concerns one point and the behaviour immediately around it, and the exam asks about named points far more often than about intervals.

Topic-Specific Study Method

The Three-Gate Check

Write the value first. One line: " $f(2) = 3$ " or " $f(2)$ undefined". If it is undefined, gates two and three cannot save the function, though the limit may still be worth computing for a later part.
Write the two one-sided limits as a pair. Then combine them into the two-sided limit. If they disagree, stop: the function is discontinuous and the classification work begins.
Compare the two numbers explicitly. Put the limit and the value side by side on the page. The comparison is the justification, and writing it as an inequality or a " $\neq$ " is faster than describing it in words.
State the verdict in the rubric's language. "Since $\lim_{x \to 2} f(x) = 4 \neq 3 = f(2)$ , $f$ is not continuous at $x = 2$ ." One sentence, three conditions accounted for.

What the item gives you	Gate that fails	What you must write
A piecewise rule with a stated value at the seam	Three, usually	The limit, the value, and the mismatch
A rational function with a cancelled factor and no point value	One	The limit, then "undefined at $a$ , so not continuous"
A rational function with a surviving factor downstairs	Two	The one-sided limits with signs
A parameter $k$ to be found	Three	The limit, then solve limit $= k$
A function continuous everywhere	None	All three conditions with the numbers filled in

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = \dfrac{x^{2}-4}{x-2}$ for $x \neq 2$ and $f(2) = 3$ . Is $f$ continuous at $x = 2$ ?

Standard Execution:

Gate one. $f(2) = 3$ , so the value exists.

Gate two. For $x \neq 2$ , $f(x) = x + 2$ , so both one-sided limits equal $4$ and the two-sided limit is $4$ .

Gate three. $4 \neq 3$ , so the third condition fails. The function is not continuous at $2$ ; the discontinuity is removable, and redefining $f(2) = 4$ repairs it.

Exam Variation (Context / Parameter Shift): Let $g(x) = \begin{cases} x^{2} + k, & x < 2 \cr 3x - 1, & x \ge 2 \end{cases}$ . Find the value of $k$ that makes $g$ continuous at $x = 2$ .

Solution: Gate one holds for every $k$ , since $g(2) = 5$ . Gate two requires the two sides to agree: from the left, $\lim_{x \to 2^{-}} g(x) = 4 + k$ ; from the right, $\lim_{x \to 2^{+}} g(x) = 5$ . Setting $4 + k = 5$ gives $k = 1$ , and then the limit equals $g(2) = 5$ , so gate three holds as well.

Common Traps & Failure Signals:

Value-only verdict (Signal: the word "continuous" appears before any limit is computed): Defined is one gate. Write the limit before the verdict.
Limit-only verdict (Signal: a limit computed with no reference to $f(a)$ ): A limit that exists proves nothing about continuity on its own; the comparison is the point.
Parameter solved against the wrong side (Signal: a piecewise $k$ set equal to the branch containing $k$ ): Choose the branch that does not contain the unknown, then solve.

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

Gate Skipping: You check the value and stop.
Definition Truncation: You remembered "defined and the limit exists" and dropped the equality, which is the only clause that connects the two halves of the definition and the only clause the rubric asks you to verify explicitly.
Parameter Drift: You solve for $k$ using the wrong branch at the seam.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; As a justification line, rarely as a definition question
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Three conditions checked in order at a named point
FRQ Years Appeared: 2019 (Q6d)

Practice Examples

MCQ 1

Which Gate Fails

Let $f(x) = \dfrac{x^{2}-4}{x-2}$ for $x \neq 2$ and $f(2) = 3$ . Which condition for continuity at $x = 2$ fails?

MCQ 2

The Parameter That Repairs the Seam

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

MCQ 3

Reading the Definition Backwards

A function $f$ satisfies $\lim_{x \to 4} f(x) = 7$ and $f(4) = 7$ . Which statement must be true?

FRQ 4

Continuity, Proved and Repaired

Let $f(x) = \begin{cases} \dfrac{x^{2} - 9}{x - 3}, & x < 3 \cr k, & x = 3 \cr 2x - 1, & x > 3 \end{cases}$

(a) Evaluate $\lim_{x \to 3^{-}} f(x)$ and $\lim_{x \to 3^{+}} f(x)$ .

(b) Determine whether $\lim_{x \to 3} f(x)$ exists, and justify your answer.

(c) Determine whether any value of $k$ makes $f$ continuous at $x = 3$ , and justify your conclusion.

(d) A student writes: "Since $k$ can be chosen to make the limit equal the value, $f$ is continuous at $x = 3$ for every $k$ ." Identify the error.

Cheatsheet

Checking Continuity at a Point

	Gate one	Gate two	Gate three	Verdict
Polynomial at any real $a$	Passes	Passes	Passes	Continuous
Rational with cancelled factor, no value assigned	Fails	Passes	Cannot be checked	Not continuous; removable
Rational with cancelled factor, value displaced	Passes	Passes	Fails	Not continuous; removable
Piecewise with unequal finite sides	Passes	Fails	Cannot be checked	Not continuous; jump
Piecewise with equal sides and matching value	Passes	Passes	Passes	Continuous

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