Unit 1Direct Substitution

Concept 4.90Built for May 2027 AP Exam

Direct Substitution

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Direct substitution is the first move on almost every limit question, and it takes one line. Put the target value $x = a$ into the expression and compute $f(a)$ . If the result is a finite number, that number is the limit, and the justification is a single clause: the function is continuous there. If the result is not finite, substitution has failed as a method, not as an answer. Zero over zero means the expression can be rewritten. A nonzero number over zero means the two sides need separate work. The exam tests both halves of that habit, knowing when one line is enough and knowing when one line is a signal to stop and rewrite.

On the AP Exam

Substitution is not a topic the exam asks about. It is the move that opens nearly every limit item, so the points ride on how fast you recognise which of its outcomes you are looking at. Multiple choice plants the failure modes as distractors: an undefined value offered as an answer, a limit marked "does not exist" because the plug failed, a piecewise expression evaluated on the wrong branch. Free response rarely names the method, but the first line of a model solution usually is the substitution and its verdict.

What you’ll see: Polynomial, rational, radical, trigonometric, exponential, and piecewise expressions at a named input, often with one value chosen so the plug does not go through cleanly.
What you’ll do: Substitute, read the result, and either stop with a justification clause or switch methods on the strength of that result.
Memory hook: Plug, read, then decide. Three words, in that order. The plug is not the answer; it is the reading that tells you which method the item is actually about.
Use it on the exam: For MCQs: Substitute before you look at the choices. If the plug returns a number, the answer is on the page and no algebra is needed. If it returns zero over zero, you now know the item is a factoring or rationalising item, and you have saved the thirty seconds you would have spent guessing. For FRQs: Write the substitution and its verdict as a sentence, not as a bare number. "Since $f$ is a polynomial, $f$ is continuous at $x = 2$ , so $\lim_{x \to 2} f(x) = f(2) = 14$ " earns the justification. A bare $14$ does not.

Concept Review

What the Method Actually Claims

Direct substitution is a one-line application of a continuity theorem. It is not a computational trick that happens to work.

Term	What it means	Where it breaks
Direct substitution	Replacing the approaching value with the value itself and evaluating the function there.	It has no meaning unless $f(a)$ exists.
Continuity eligibility	For $f$ continuous at $a$ , $\lim_{x \to a} f(x) = f(a)$ by definition of continuity.	Discontinuous points, holes, jumps, and asymptotes all void it.
Domain eligibility	$a$ must be a real input of $f$ : no zero denominator, no negative under an even root, no logarithm of a non-positive number.	A value outside the domain has no $f(a)$ to report.

The two eligibilities are separate, and students collapse them into one. A function can be defined at $a$ and still be discontinuous there, which is exactly the case a piecewise expression constructs on purpose.

The Three-Question Gate

Ask these in order, and stop at the first no.

Is the expression built from continuous pieces? Polynomials, sine, cosine, exponentials, and logarithms on their domains are continuous everywhere they are defined. Sums, products, quotients, and composites of continuous functions stay continuous wherever they are defined.
Is the target value in the domain? Check the denominator, the radicand, and the logarithm argument before you write anything. This is a reading step, not an arithmetic step.
Does the plug return a finite number? If it does, that number is the limit and the work is over. If it does not, the form of the failure selects the next method.

Those three answers cover every limit item on the exam. The fourth case, where the function is continuous but the plug is still unhelpful, does not exist; that combination is what the zero-over-zero case actually is.

Figure 1 — Four things a single substitution can return.

Caption. Only the left column ends the question. The right column means the expression, not the limit, has to change before any number can be reported.

Four boxes: a nonzero numerator over a nonzero denominator gives the limit directly, zero over a nonzero denominator gives zero, a nonzero numerator over zero gives no finite value, and zero over zero is undetermined

Three Corrections Worth Memorising

“Substitution failed, so the limit does not exist.”
- Correct: substitution failed as a method. The limit may well exist, and for a rational function at a hole it usually does.
- Common error: reading a broken method as a broken limit.
“The function is undefined at $a$ , so the function is not continuous there.”
- Correct: undefined at $a$ does mean not continuous at $a$ , but the limit at $a$ does not care about $f(a)$ .
- Common error: importing the continuity verdict into the limit question.
“Zero over zero means the limit is zero.”
- Correct: zero over zero carries no information at all. It is the form that says the numerator and denominator both vanish, which is a fact about the expression, not about the limit.

Topic-Specific Study Method

The Plug-First Gate

Borrowed idea: When a limit looks like work, plug the number in first. Do it on paper, not in your head. Write the value under the expression. Most of the time the answer appears, and the page stays clean. The habit that earns points is the pause after the plug. Read what came back. A number ends the question. A broken result starts a different one, and the shape of the break tells you which method to pick up next.

Write the substitution as a line of work, not as a thought. Put $x = a$ under the expression and record the arithmetic. A plug you did mentally cannot be reviewed, and a wrong mental plug sends you into the wrong method.
Classify the result into one of four forms. Finite, zero, nonzero over zero, or zero over zero. Say the form out loud in your own words, because the classification is what selects the next move.
Switch methods only on evidence. Zero over zero sends you to factoring, rationalising, or a common denominator. A nonzero number over zero sends you to one-sided analysis. Neither sends you to "does not exist" as a first answer.
Close the item with a clause, not a number. One sentence naming continuity or naming the rewrite. That clause is the difference between a scored answer and a checked answer.

What the page shows	What your brain must notice	What you write	Where the marks go
A polynomial at an integer	Nothing blocks the plug	$f(a)$ and "continuous, so limit equals $f(a)$ "	Answer point only; the clause protects the justification point
A rational function that returns zero over zero	Numerator and denominator share the factor $(x-a)$	The rewrite, then the second plug	The rewrite line earns the method point
A radical that returns zero over zero	The conjugate is the only route out	Multiply by the conjugate over itself	The conjugate line must be written, not implied
A piecewise expression at its break	The side decides which branch applies	Two one-sided plugs, then a verdict	Each side is a separate scored step
A nonzero number over zero	The sign of each side, not the size	One-sided limits with signs, then "no two-sided limit"	The verdict without signs earns nothing

Training & Error Diagnosis

Core Archetype (Simplest Form): Evaluate $\lim_{x \to 2}\left(3x^{2} - x + 4\right)$ .

Standard Execution:

Check continuity. The expression is a polynomial, so it is continuous at every real input.

Substitute. $3(2)^{2} - 2 + 4 = 12 - 2 + 4$ .

Report. The value is $14$ , and the limit equals $14$ by continuity.

Exam Variation (Context / Parameter Shift): Let $f(x) = \begin{cases} 2x + 1, & x < 2 \cr x^{2} + 2, & x > 2 \end{cases}$ and let $k = f(2)$ be chosen later. Find $\lim_{x \to 2^{-}} f(x)$ and $\lim_{x \to 2^{+}} f(x)$ , then decide whether a value of $k$ can make $f$ continuous at $2$ .

Solution: The left branch applies to inputs below $2$ , so $\lim_{x \to 2^{-}} f(x) = 2(2) + 1 = 5$ . The right branch applies above $2$ , so $\lim_{x \to 2^{+}} f(x) = 2^{2} + 2 = 6$ . The two one-sided limits disagree, so no value of $k$ can repair continuity at $2$ ; a jump is not a hole.

Common Traps & Failure Signals:

The undefined answer (Signal: the plug produces a denominator of zero and a choice offers "undefined"): An undefined output is not a limit. Rewrite the expression and evaluate again.
The premature verdict (Signal: you are about to write "does not exist" on the first line): Zero over zero never licenses that verdict. The form has to be resolved first.
The wrong branch (Signal: a piecewise expression with a side condition in the limit): Substitute into the branch that owns the side, not the branch that looks easiest.

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

Value Confusion: You plug in and stop checking.
Method Lock: Because years of polynomial drills made substitution feel like a complete method, the moment a rational expression returns zero over zero the page keeps going in the same handwriting instead of switching to factoring, and the item that would have earned the answer point ends with an undefined value the rubric cannot credit.
Branch Blindness: You substitute into whichever piece looks familiar, and the side condition in the limit notation never gets read.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; The opening move of almost every limit item
10-Year Total Count: 2 of 9 Exams · 2 FRQ Parts; Loading a value, then handing off to algebra or one-sided analysis
FRQ Years Appeared: 2019 (Q3d), 2026 (Q6a)

Practice Examples

MCQ 1

The Plug That Returns Zero Over Zero

Evaluate $\lim_{x \to -1}\frac{x^{2} + 3x + 2}{x + 1}$

MCQ 2

Which Limit Is Legal as Written

For which of the following limits may direct substitution be applied immediately, with no rewriting?

MCQ 3

The Piecewise One-Sided Plug

Let $f(x) = \begin{cases} x^{2} - 4, & x < 3 \cr x + 1, & x \ge 3 \end{cases}$ Evaluate $\lim_{x \to 3^{-}} f(x)$ .

FRQ 4

One Expression, Four Verdicts

Let $g(x) = \dfrac{x^{2} - 9}{x - 3}$ for $x \neq 3$ , and let $h(x) = \dfrac{4}{(x-2)^{2}}$ .

(a) Evaluate $\lim_{x \to 3} g(x)$ and state why direct substitution cannot be used as written.

(b) The function $g$ has a removable discontinuity at $x = 3$ . Find the value $k$ that makes the extended function continuous at $3$ .

(c) Evaluate $\lim_{x \to 2} h(x)$ , justifying the sign of your answer.

(d) A student writes: " $\lim_{x \to 3} g(x)$ does not exist because $g(3)$ is undefined." Identify the error and correct the sentence.

Cheatsheet

Reading the Plug

	Verdict	Next action	Sentence to write
A finite number	The limit, settled	None	"Since $f$ is continuous at $a$ , the limit equals $f(a)$ ."
Zero over a nonzero number	The limit is zero	None	"The numerator tends to zero while the denominator does not, so the quotient tends to zero."
A nonzero number over zero	No finite limit	One-sided analysis with signs	"The denominator approaches zero through positive values, so the quotient grows without bound."
Zero over zero	Undetermined	Factor, rationalise, or combine	"The form is indeterminate, so the expression is rewritten before evaluation."
A negative value under an even root	Not a real limit	Check the domain near $a$	"For $x$ near $a$ the radicand is negative, so the function is not defined there."

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Unit 1Direct Substitution

Concept 4.90Built for May 2027 AP Exam

Direct Substitution

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: Polynomial, rational, radical, trigonometric, exponential, and piecewise expressions at a named input, often with one value chosen so the plug does not go through cleanly.
What you’ll do: Substitute, read the result, and either stop with a justification clause or switch methods on the strength of that result.
Memory hook: Plug, read, then decide. Three words, in that order. The plug is not the answer; it is the reading that tells you which method the item is actually about.
Use it on the exam: For MCQs: Substitute before you look at the choices. If the plug returns a number, the answer is on the page and no algebra is needed. If it returns zero over zero, you now know the item is a factoring or rationalising item, and you have saved the thirty seconds you would have spent guessing. For FRQs: Write the substitution and its verdict as a sentence, not as a bare number. "Since $f$ is a polynomial, $f$ is continuous at $x = 2$ , so $\lim_{x \to 2} f(x) = f(2) = 14$ " earns the justification. A bare $14$ does not.

Concept Review

What the Method Actually Claims

Direct substitution is a one-line application of a continuity theorem. It is not a computational trick that happens to work.

Term	What it means	Where it breaks
Direct substitution	Replacing the approaching value with the value itself and evaluating the function there.	It has no meaning unless $f(a)$ exists.
Continuity eligibility	For $f$ continuous at $a$ , $\lim_{x \to a} f(x) = f(a)$ by definition of continuity.	Discontinuous points, holes, jumps, and asymptotes all void it.
Domain eligibility	$a$ must be a real input of $f$ : no zero denominator, no negative under an even root, no logarithm of a non-positive number.	A value outside the domain has no $f(a)$ to report.

The Three-Question Gate

Ask these in order, and stop at the first no.

Is the expression built from continuous pieces? Polynomials, sine, cosine, exponentials, and logarithms on their domains are continuous everywhere they are defined. Sums, products, quotients, and composites of continuous functions stay continuous wherever they are defined.
Is the target value in the domain? Check the denominator, the radicand, and the logarithm argument before you write anything. This is a reading step, not an arithmetic step.
Does the plug return a finite number? If it does, that number is the limit and the work is over. If it does not, the form of the failure selects the next method.

Figure 1 — Four things a single substitution can return.

Caption. Only the left column ends the question. The right column means the expression, not the limit, has to change before any number can be reported.

Three Corrections Worth Memorising

“Substitution failed, so the limit does not exist.”
- Correct: substitution failed as a method. The limit may well exist, and for a rational function at a hole it usually does.
- Common error: reading a broken method as a broken limit.
“The function is undefined at $a$ , so the function is not continuous there.”
- Correct: undefined at $a$ does mean not continuous at $a$ , but the limit at $a$ does not care about $f(a)$ .
- Common error: importing the continuity verdict into the limit question.
“Zero over zero means the limit is zero.”
- Correct: zero over zero carries no information at all. It is the form that says the numerator and denominator both vanish, which is a fact about the expression, not about the limit.

Topic-Specific Study Method

The Plug-First Gate

Write the substitution as a line of work, not as a thought. Put $x = a$ under the expression and record the arithmetic. A plug you did mentally cannot be reviewed, and a wrong mental plug sends you into the wrong method.
Classify the result into one of four forms. Finite, zero, nonzero over zero, or zero over zero. Say the form out loud in your own words, because the classification is what selects the next move.
Switch methods only on evidence. Zero over zero sends you to factoring, rationalising, or a common denominator. A nonzero number over zero sends you to one-sided analysis. Neither sends you to "does not exist" as a first answer.
Close the item with a clause, not a number. One sentence naming continuity or naming the rewrite. That clause is the difference between a scored answer and a checked answer.

What the page shows	What your brain must notice	What you write	Where the marks go
A polynomial at an integer	Nothing blocks the plug	$f(a)$ and "continuous, so limit equals $f(a)$ "	Answer point only; the clause protects the justification point
A rational function that returns zero over zero	Numerator and denominator share the factor $(x-a)$	The rewrite, then the second plug	The rewrite line earns the method point
A radical that returns zero over zero	The conjugate is the only route out	Multiply by the conjugate over itself	The conjugate line must be written, not implied
A piecewise expression at its break	The side decides which branch applies	Two one-sided plugs, then a verdict	Each side is a separate scored step
A nonzero number over zero	The sign of each side, not the size	One-sided limits with signs, then "no two-sided limit"	The verdict without signs earns nothing

Training & Error Diagnosis

Core Archetype (Simplest Form): Evaluate $\lim_{x \to 2}\left(3x^{2} - x + 4\right)$ .

Standard Execution:

Check continuity. The expression is a polynomial, so it is continuous at every real input.

Substitute. $3(2)^{2} - 2 + 4 = 12 - 2 + 4$ .

Report. The value is $14$ , and the limit equals $14$ by continuity.

Solution: The left branch applies to inputs below $2$ , so $\lim_{x \to 2^{-}} f(x) = 2(2) + 1 = 5$ . The right branch applies above $2$ , so $\lim_{x \to 2^{+}} f(x) = 2^{2} + 2 = 6$ . The two one-sided limits disagree, so no value of $k$ can repair continuity at $2$ ; a jump is not a hole.

Common Traps & Failure Signals:

The undefined answer (Signal: the plug produces a denominator of zero and a choice offers "undefined"): An undefined output is not a limit. Rewrite the expression and evaluate again.
The premature verdict (Signal: you are about to write "does not exist" on the first line): Zero over zero never licenses that verdict. The form has to be resolved first.
The wrong branch (Signal: a piecewise expression with a side condition in the limit): Substitute into the branch that owns the side, not the branch that looks easiest.

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

Value Confusion: You plug in and stop checking.
Method Lock: Because years of polynomial drills made substitution feel like a complete method, the moment a rational expression returns zero over zero the page keeps going in the same handwriting instead of switching to factoring, and the item that would have earned the answer point ends with an undefined value the rubric cannot credit.
Branch Blindness: You substitute into whichever piece looks familiar, and the side condition in the limit notation never gets read.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; The opening move of almost every limit item
10-Year Total Count: 2 of 9 Exams · 2 FRQ Parts; Loading a value, then handing off to algebra or one-sided analysis
FRQ Years Appeared: 2019 (Q3d), 2026 (Q6a)

Practice Examples

MCQ 1

The Plug That Returns Zero Over Zero

Evaluate $\lim_{x \to -1}\frac{x^{2} + 3x + 2}{x + 1}$

MCQ 2

Which Limit Is Legal as Written

For which of the following limits may direct substitution be applied immediately, with no rewriting?

MCQ 3

The Piecewise One-Sided Plug

Let $f(x) = \begin{cases} x^{2} - 4, & x < 3 \cr x + 1, & x \ge 3 \end{cases}$ Evaluate $\lim_{x \to 3^{-}} f(x)$ .

FRQ 4

One Expression, Four Verdicts

Let $g(x) = \dfrac{x^{2} - 9}{x - 3}$ for $x \neq 3$ , and let $h(x) = \dfrac{4}{(x-2)^{2}}$ .

(a) Evaluate $\lim_{x \to 3} g(x)$ and state why direct substitution cannot be used as written.

(b) The function $g$ has a removable discontinuity at $x = 3$ . Find the value $k$ that makes the extended function continuous at $3$ .

(c) Evaluate $\lim_{x \to 2} h(x)$ , justifying the sign of your answer.

(d) A student writes: " $\lim_{x \to 3} g(x)$ does not exist because $g(3)$ is undefined." Identify the error and correct the sentence.

Cheatsheet

Reading the Plug

	Verdict	Next action	Sentence to write
A finite number	The limit, settled	None	"Since $f$ is continuous at $a$ , the limit equals $f(a)$ ."
Zero over a nonzero number	The limit is zero	None	"The numerator tends to zero while the denominator does not, so the quotient tends to zero."
A nonzero number over zero	No finite limit	One-sided analysis with signs	"The denominator approaches zero through positive values, so the quotient grows without bound."
Zero over zero	Undetermined	Factor, rationalise, or combine	"The form is indeterminate, so the expression is rewritten before evaluation."
A negative value under an even root	Not a real limit	Check the domain near $a$	"For $x$ near $a$ the radicand is negative, so the function is not defined there."

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Unit 1Direct Substitution

Concept 4.90Built for May 2027 AP Exam

Direct Substitution

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: Polynomial, rational, radical, trigonometric, exponential, and piecewise expressions at a named input, often with one value chosen so the plug does not go through cleanly.
What you’ll do: Substitute, read the result, and either stop with a justification clause or switch methods on the strength of that result.
Memory hook: Plug, read, then decide. Three words, in that order. The plug is not the answer; it is the reading that tells you which method the item is actually about.
Use it on the exam: For MCQs: Substitute before you look at the choices. If the plug returns a number, the answer is on the page and no algebra is needed. If it returns zero over zero, you now know the item is a factoring or rationalising item, and you have saved the thirty seconds you would have spent guessing. For FRQs: Write the substitution and its verdict as a sentence, not as a bare number. "Since $f$ is a polynomial, $f$ is continuous at $x = 2$ , so $\lim_{x \to 2} f(x) = f(2) = 14$ " earns the justification. A bare $14$ does not.

Concept Review

What the Method Actually Claims

Direct substitution is a one-line application of a continuity theorem. It is not a computational trick that happens to work.

Term	What it means	Where it breaks
Direct substitution	Replacing the approaching value with the value itself and evaluating the function there.	It has no meaning unless $f(a)$ exists.
Continuity eligibility	For $f$ continuous at $a$ , $\lim_{x \to a} f(x) = f(a)$ by definition of continuity.	Discontinuous points, holes, jumps, and asymptotes all void it.
Domain eligibility	$a$ must be a real input of $f$ : no zero denominator, no negative under an even root, no logarithm of a non-positive number.	A value outside the domain has no $f(a)$ to report.

The Three-Question Gate

Ask these in order, and stop at the first no.

Is the expression built from continuous pieces? Polynomials, sine, cosine, exponentials, and logarithms on their domains are continuous everywhere they are defined. Sums, products, quotients, and composites of continuous functions stay continuous wherever they are defined.
Is the target value in the domain? Check the denominator, the radicand, and the logarithm argument before you write anything. This is a reading step, not an arithmetic step.
Does the plug return a finite number? If it does, that number is the limit and the work is over. If it does not, the form of the failure selects the next method.

Figure 1 — Four things a single substitution can return.

Caption. Only the left column ends the question. The right column means the expression, not the limit, has to change before any number can be reported.

Three Corrections Worth Memorising

“Substitution failed, so the limit does not exist.”
- Correct: substitution failed as a method. The limit may well exist, and for a rational function at a hole it usually does.
- Common error: reading a broken method as a broken limit.
“The function is undefined at $a$ , so the function is not continuous there.”
- Correct: undefined at $a$ does mean not continuous at $a$ , but the limit at $a$ does not care about $f(a)$ .
- Common error: importing the continuity verdict into the limit question.
“Zero over zero means the limit is zero.”
- Correct: zero over zero carries no information at all. It is the form that says the numerator and denominator both vanish, which is a fact about the expression, not about the limit.

Topic-Specific Study Method

The Plug-First Gate

Write the substitution as a line of work, not as a thought. Put $x = a$ under the expression and record the arithmetic. A plug you did mentally cannot be reviewed, and a wrong mental plug sends you into the wrong method.
Classify the result into one of four forms. Finite, zero, nonzero over zero, or zero over zero. Say the form out loud in your own words, because the classification is what selects the next move.
Switch methods only on evidence. Zero over zero sends you to factoring, rationalising, or a common denominator. A nonzero number over zero sends you to one-sided analysis. Neither sends you to "does not exist" as a first answer.
Close the item with a clause, not a number. One sentence naming continuity or naming the rewrite. That clause is the difference between a scored answer and a checked answer.

What the page shows	What your brain must notice	What you write	Where the marks go
A polynomial at an integer	Nothing blocks the plug	$f(a)$ and "continuous, so limit equals $f(a)$ "	Answer point only; the clause protects the justification point
A rational function that returns zero over zero	Numerator and denominator share the factor $(x-a)$	The rewrite, then the second plug	The rewrite line earns the method point
A radical that returns zero over zero	The conjugate is the only route out	Multiply by the conjugate over itself	The conjugate line must be written, not implied
A piecewise expression at its break	The side decides which branch applies	Two one-sided plugs, then a verdict	Each side is a separate scored step
A nonzero number over zero	The sign of each side, not the size	One-sided limits with signs, then "no two-sided limit"	The verdict without signs earns nothing

Training & Error Diagnosis

Core Archetype (Simplest Form): Evaluate $\lim_{x \to 2}\left(3x^{2} - x + 4\right)$ .

Standard Execution:

Check continuity. The expression is a polynomial, so it is continuous at every real input.

Substitute. $3(2)^{2} - 2 + 4 = 12 - 2 + 4$ .

Report. The value is $14$ , and the limit equals $14$ by continuity.

Solution: The left branch applies to inputs below $2$ , so $\lim_{x \to 2^{-}} f(x) = 2(2) + 1 = 5$ . The right branch applies above $2$ , so $\lim_{x \to 2^{+}} f(x) = 2^{2} + 2 = 6$ . The two one-sided limits disagree, so no value of $k$ can repair continuity at $2$ ; a jump is not a hole.

Common Traps & Failure Signals:

The undefined answer (Signal: the plug produces a denominator of zero and a choice offers "undefined"): An undefined output is not a limit. Rewrite the expression and evaluate again.
The premature verdict (Signal: you are about to write "does not exist" on the first line): Zero over zero never licenses that verdict. The form has to be resolved first.
The wrong branch (Signal: a piecewise expression with a side condition in the limit): Substitute into the branch that owns the side, not the branch that looks easiest.

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

Value Confusion: You plug in and stop checking.
Method Lock: Because years of polynomial drills made substitution feel like a complete method, the moment a rational expression returns zero over zero the page keeps going in the same handwriting instead of switching to factoring, and the item that would have earned the answer point ends with an undefined value the rubric cannot credit.
Branch Blindness: You substitute into whichever piece looks familiar, and the side condition in the limit notation never gets read.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; The opening move of almost every limit item
10-Year Total Count: 2 of 9 Exams · 2 FRQ Parts; Loading a value, then handing off to algebra or one-sided analysis
FRQ Years Appeared: 2019 (Q3d), 2026 (Q6a)

Practice Examples

MCQ 1

The Plug That Returns Zero Over Zero

Evaluate $\lim_{x \to -1}\frac{x^{2} + 3x + 2}{x + 1}$

MCQ 2

Which Limit Is Legal as Written

For which of the following limits may direct substitution be applied immediately, with no rewriting?

MCQ 3

The Piecewise One-Sided Plug

Let $f(x) = \begin{cases} x^{2} - 4, & x < 3 \cr x + 1, & x \ge 3 \end{cases}$ Evaluate $\lim_{x \to 3^{-}} f(x)$ .

FRQ 4

One Expression, Four Verdicts

Let $g(x) = \dfrac{x^{2} - 9}{x - 3}$ for $x \neq 3$ , and let $h(x) = \dfrac{4}{(x-2)^{2}}$ .

(a) Evaluate $\lim_{x \to 3} g(x)$ and state why direct substitution cannot be used as written.

(b) The function $g$ has a removable discontinuity at $x = 3$ . Find the value $k$ that makes the extended function continuous at $3$ .

(c) Evaluate $\lim_{x \to 2} h(x)$ , justifying the sign of your answer.

(d) A student writes: " $\lim_{x \to 3} g(x)$ does not exist because $g(3)$ is undefined." Identify the error and correct the sentence.

Cheatsheet

Reading the Plug

	Verdict	Next action	Sentence to write
A finite number	The limit, settled	None	"Since $f$ is continuous at $a$ , the limit equals $f(a)$ ."
Zero over a nonzero number	The limit is zero	None	"The numerator tends to zero while the denominator does not, so the quotient tends to zero."
A nonzero number over zero	No finite limit	One-sided analysis with signs	"The denominator approaches zero through positive values, so the quotient grows without bound."
Zero over zero	Undetermined	Factor, rationalise, or combine	"The form is indeterminate, so the expression is rewritten before evaluation."
A negative value under an even root	Not a real limit	Check the domain near $a$	"For $x$ near $a$ the radicand is negative, so the function is not defined there."

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