Unit 1Analyzing horizontal asymptotes using limits at infinity.

Concept 4.141Built for May 2027 AP Exam

Analyzing horizontal asymptotes using limits at infinity.

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

A horizontal asymptote is a claim about one end of a graph at a time. The line $y = L$ is a horizontal asymptote of $f$ when $\lim_{x \to \infty} f(x) = L$ or when $\lim_{x \to -\infty} f(x) = L$ , and the two directions are checked separately because they can settle at different heights. A function may have none, one, or two of these lines, and two is the maximum, since there are only two directions to run. A rational function with equal degrees has the same height on both sides and therefore one line; an exponential expression or a radical can produce two, because one end collapses in a way the other does not. Touching or crossing the line is irrelevant: the asymptote describes where the curve is heading, not where it has been.

On the AP Exam

The exam asks for horizontal asymptotes in three forms, and the form decides the work. The first is a direct limit request, where the answer is a number and the asymptote is that number written as a line. The second is a graph-reading request, where the curve is given and the limit statements must be matched to it. The third is the one that separates students: a function whose two ends behave differently, requiring two separate limits and a conclusion that names both lines. Multiple choice usually offers one line as a trap for a function with two, and free response pays for the second limit separately, so a response that computes only the rightward limit has answered half the question.

What you’ll see: $\frac{x}{\sqrt{x^{2}+1}}$ , $\arctan x$ , expressions built from $e^{x}$ , and rational functions whose two ends agree, offered next to curves that cross their own asymptote.
What you’ll do: Compute each directional limit separately, convert each finite value into a line, then count how many distinct lines the pair produces.
Memory hook: One end, one limit, one line. Two ends means two limits, and possibly two lines.
Use it on the exam: For MCQs: If a function is even in its degrees, expect one line; if a root or an exponential flips the sign at one end, expect two. Counting the directions before computing keeps the option list honest. For FRQs: Write both limits, then write the lines as equations. "Since $\lim_{x \to \infty} f(x) = 2$ and $\lim_{x \to -\infty} f(x) = -3$ , the graph has horizontal asymptotes $y = 2$ and $y = -3$ " is the complete answer; naming only one loses the second point.

Concept Review

The Definition, Taken One Direction at a Time

There is no such thing as "the" horizontal asymptote of a function in general. There is a right-end limit and a left-end limit, and each one either produces a line or does not.

Right-end behaviour	Left-end behaviour	Horizontal asymptotes
$\lim_{x \to \infty} f(x) = 2$	$\lim_{x \to -\infty} f(x) = 2$	One line, $y = 2$
$\lim_{x \to \infty} f(x) = 2$	$\lim_{x \to -\infty} f(x) = -3$	Two lines, $y = 2$ and $y = -3$
$\lim_{x \to \infty} f(x) = 0$	$\lim_{x \to -\infty} f(x) = +\infty$	One line, $y = 0$ , on the right only
$\lim_{x \to \infty} f(x) = +\infty$	$\lim_{x \to -\infty} f(x) = +\infty$	No horizontal asymptote

A vertical line and a horizontal line can coexist in the same graph, and they answer different questions: the vertical line is about a finite input where the output blows up, and the horizontal line is about the output as the input blows up. Neither one excludes the other.

Two Is the Ceiling

Two directions exist, so two distinct lines are the maximum. The interesting cases are the ones that reach that maximum.

Radicals with an odd effect at negative infinity. $f(x) = \frac{x}{\sqrt{x^{2}+1}}$ tends to $1$ on the right and to $-1$ on the left, because the denominator behaves like $\lvert x \rvert$ and that is $-x$ on the left.
Inverse tangent. $\arctan x \to \frac{\pi}{2}$ as $x \to \infty$ and $\to -\frac{\pi}{2}$ as $x \to -\infty$ . Two lines, and neither is reached.
Exponentials. $e^{x} \to 0$ as $x \to -\infty$ and $\to \infty$ as $x \to \infty$ , so there is one line, $y = 0$ , and it belongs to the left end only.
Rational functions with equal degrees. The leading-coefficient ratio is the same on both sides, so the two ends agree and only one line exists.

Figure 1 — One function, two horizontal asymptotes.

Caption. The curve is $f(x) = \frac{2e^{x}-3}{e^{x}+1}$ . On the right it settles toward $y = 2$ ; on the left it settles toward $y = -3$ . Both dashed lines are horizontal asymptotes, and each belongs to one direction only.

Graph of a function built from e to the x that settles toward y equals 2 on the right and toward y equals minus 3 on the left, with both levels drawn as dashed horizontal lines

Crossing the Line Does Not Disqualify It

A horizontal asymptote is not a barrier. It records where the curve is heading, and many functions cross the line on the way there.

$f(x) = \frac{2x}{x^{2}+1} \quad \text{has} \quad \lim_{x \to \pm\infty} f(x) = 0$

The graph passes through the origin, which lies on $y = 0$ , and the line is still a horizontal asymptote. The same holds for $\frac{\sin x}{x}$ , which crosses $y = 0$ infinitely often as it decays. What matters is the limiting height, not the set of points where the curve happens to meet it. Students who reject a candidate line because the graph touches it are applying a vertical-asymptote instinct to a horizontal question.

Topic-Specific Study Method

The Two-Ends Audit

Borrowed idea: A horizontal asymptote is a statement about the far end of the graph, not about the middle of it. Which end are you looking at, and does the other end agree? Nothing in the definition forbids the curve from crossing the line on its way out, so a crossing is not a counterexample and never was. Does the curve have to stay on one side of the line? It does not, because the limit records the height the curve drifts toward, and the drift can approach from either side or from both. A function can have no horizontal asymptote, one, or two, and the count depends entirely on how the two directions behave. Two is the maximum, because two directions are all there are. The work is therefore to compute each direction separately and then read the pair of results as a description of the ends.

Compute the right-end limit first, on its own line. Writing one limit per line prevents the second from being silently copied from the first.
Compute the left-end limit next, treating the sign of the input as new information. Any root encountered here needs $\lvert x \rvert = -x$ .
Convert each finite value into a line equation. A limit of $2$ produces the line $y = 2$ , and a limit that is infinite produces no line at all.
Count the distinct lines. One value for both ends is one line; two different values are two lines; an infinite limit removes one line.
State the conclusion in the language of the question. If the item asks whether a stated line is an asymptote, answer with the limit that establishes or refutes it.

Function	Right end	Left end	Lines
$\frac{2x^{2}+1}{x^{2}-3}$	$2$	$2$	$y = 2$
$\frac{x}{\sqrt{x^{2}+1}}$	$1$	$-1$	$y = 1$ and $y = -1$
$\arctan x$	$\frac{\pi}{2}$	$-\frac{\pi}{2}$	$y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$
$e^{x}$	$+\infty$	$0$	$y = 0$ , left end only
$\frac{2e^{x}-3}{e^{x}+1}$	$2$	$-3$	$y = 2$ and $y = -3$
$x^{2}$	$+\infty$	$+\infty$	None

Training & Error Diagnosis

Core Archetype (Simplest Form): Find every horizontal asymptote of $f(x) = \dfrac{2x^{2}+3x}{x^{2}-1}$ .

Standard Execution:

Right end. Divide by $x^{2}$ : $\frac{2 + 3/x}{1 - 1/x^{2}} \to 2$ , so $\lim_{x \to \infty} f(x) = 2$ .

Left end. The same division applies, and no root or exponential changes the sign, so $\lim_{x \to -\infty} f(x) = 2$ as well.

Count the lines. Both ends settle at $2$ , so the only horizontal asymptote is $y = 2$ .

Separate the vertical lines. The denominator is zero at $x = 1$ and $x = -1$ , and neither factor cancels, so $x = 1$ and $x = -1$ are vertical asymptotes and have nothing to do with the horizontal line.

Exam Variation (Context / Parameter Shift): Find every horizontal asymptote of $g(x) = \dfrac{2e^{x}-3}{e^{x}+1}$ .

Solution: For $x \to \infty$ , divide top and bottom by $e^{x}$ : $\frac{2 - 3e^{-x}}{1 + e^{-x}} \to 2$ . For $x \to -\infty$ , $e^{x} \to 0$ , so the quotient tends to $\frac{0-3}{0+1} = -3$ . The two ends disagree, which gives two horizontal asymptotes: $y = 2$ and $y = -3$ . This cannot happen for a rational function with equal degrees, because a polynomial is even-handed at both ends in a way an exponential is not.

Common Traps & Failure Signals:

The single-end answer (Signal: one limit written, one line named): the left end has not been computed, and for radicals and exponentials it frequently disagrees.
The barrier belief (Signal: "the graph crosses the line, so it is not an asymptote"): crossing is irrelevant; only the limit matters.
The vertical confusion (Signal: a denominator zero offered as a horizontal asymptote): denominator zeros produce vertical lines, and the two families are found by different operations.

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

One-Sided Effort: You stopped at the right end.
Symmetry Assumption: You assumed both ends must agree because the expression looks symmetric in its degrees, which fails the moment a root or an exponential changes sign at negative infinity.
Definition Drift: You answered the horizontal question with the vertical rule, treating a point where the denominator vanishes as if it were the height the curve approaches at infinity, so the two different kinds of line were merged into one category and neither was computed correctly.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; As the interpretation of a limit already computed
10-Year Total Count: 2 of 9 Exams · 2 FRQ Parts; One limit per direction, then the line equations
FRQ Years Appeared: 2021 (Q6b), 2022 (Q6d)

Practice Examples

MCQ 1

Which Function Has Two

Which function has two distinct horizontal asymptotes?

MCQ 2

Reading a Rational Function

Let $f(x) = \dfrac{2x^{2}+3x}{x^{2}-1}$ . Which statement about horizontal asymptotes is correct?

MCQ 3

Meeting the Line

The graph of $f(x) = \dfrac{2x}{x^{2}+1}$ passes through the origin. Which statement is correct?

FRQ 4

An Exponential With Two Lines

Let $f(x) = \frac{2e^{x}-3}{e^{x}+1}$

(a) Evaluate $\lim_{x \to \infty} f(x)$ .

(b) Evaluate $\lim_{x \to -\infty} f(x)$ .

(c) State the equation of every horizontal asymptote of $f$ .

(d) Explain why $f$ has two horizontal asymptotes while a rational function with a numerator and denominator of the same degree has only one.

Cheatsheet

Horizontal Asymptotes

	Right end	Left end	Lines
Rational, degrees equal	Ratio	Ratio	One line
Rational, denominator degree larger	$0$	$0$	One line, $y = 0$
Rational, numerator degree larger	$\pm\infty$	$\pm\infty$	None
Root of an even power over a linear term	$+k$	$-k$	Two lines
$e^{x}$ or $e^{-x}$	Depends on the exponent's sign	Opposite of the right end	One line
$\arctan x$	$\frac{\pi}{2}$	$-\frac{\pi}{2}$	Two lines
Bounded oscillation such as $\sin x$	No limit	No limit	None

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Unit 1Analyzing horizontal asymptotes using limits at infinity.

Concept 4.141Built for May 2027 AP Exam

Analyzing horizontal asymptotes using limits at infinity.

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: $\frac{x}{\sqrt{x^{2}+1}}$ , $\arctan x$ , expressions built from $e^{x}$ , and rational functions whose two ends agree, offered next to curves that cross their own asymptote.
What you’ll do: Compute each directional limit separately, convert each finite value into a line, then count how many distinct lines the pair produces.
Memory hook: One end, one limit, one line. Two ends means two limits, and possibly two lines.
Use it on the exam: For MCQs: If a function is even in its degrees, expect one line; if a root or an exponential flips the sign at one end, expect two. Counting the directions before computing keeps the option list honest. For FRQs: Write both limits, then write the lines as equations. "Since $\lim_{x \to \infty} f(x) = 2$ and $\lim_{x \to -\infty} f(x) = -3$ , the graph has horizontal asymptotes $y = 2$ and $y = -3$ " is the complete answer; naming only one loses the second point.

Concept Review

The Definition, Taken One Direction at a Time

There is no such thing as "the" horizontal asymptote of a function in general. There is a right-end limit and a left-end limit, and each one either produces a line or does not.

Right-end behaviour	Left-end behaviour	Horizontal asymptotes
$\lim_{x \to \infty} f(x) = 2$	$\lim_{x \to -\infty} f(x) = 2$	One line, $y = 2$
$\lim_{x \to \infty} f(x) = 2$	$\lim_{x \to -\infty} f(x) = -3$	Two lines, $y = 2$ and $y = -3$
$\lim_{x \to \infty} f(x) = 0$	$\lim_{x \to -\infty} f(x) = +\infty$	One line, $y = 0$ , on the right only
$\lim_{x \to \infty} f(x) = +\infty$	$\lim_{x \to -\infty} f(x) = +\infty$	No horizontal asymptote

Two Is the Ceiling

Two directions exist, so two distinct lines are the maximum. The interesting cases are the ones that reach that maximum.

Radicals with an odd effect at negative infinity. $f(x) = \frac{x}{\sqrt{x^{2}+1}}$ tends to $1$ on the right and to $-1$ on the left, because the denominator behaves like $\lvert x \rvert$ and that is $-x$ on the left.
Inverse tangent. $\arctan x \to \frac{\pi}{2}$ as $x \to \infty$ and $\to -\frac{\pi}{2}$ as $x \to -\infty$ . Two lines, and neither is reached.
Exponentials. $e^{x} \to 0$ as $x \to -\infty$ and $\to \infty$ as $x \to \infty$ , so there is one line, $y = 0$ , and it belongs to the left end only.
Rational functions with equal degrees. The leading-coefficient ratio is the same on both sides, so the two ends agree and only one line exists.

Figure 1 — One function, two horizontal asymptotes.

Graph of a function built from e to the x that settles toward y equals 2 on the right and toward y equals minus 3 on the left, with both levels drawn as dashed horizontal lines

Crossing the Line Does Not Disqualify It

A horizontal asymptote is not a barrier. It records where the curve is heading, and many functions cross the line on the way there.

$f(x) = \frac{2x}{x^{2}+1} \quad \text{has} \quad \lim_{x \to \pm\infty} f(x) = 0$

Topic-Specific Study Method

The Two-Ends Audit

Compute the right-end limit first, on its own line. Writing one limit per line prevents the second from being silently copied from the first.
Compute the left-end limit next, treating the sign of the input as new information. Any root encountered here needs $\lvert x \rvert = -x$ .
Convert each finite value into a line equation. A limit of $2$ produces the line $y = 2$ , and a limit that is infinite produces no line at all.
Count the distinct lines. One value for both ends is one line; two different values are two lines; an infinite limit removes one line.
State the conclusion in the language of the question. If the item asks whether a stated line is an asymptote, answer with the limit that establishes or refutes it.

Function	Right end	Left end	Lines
$\frac{2x^{2}+1}{x^{2}-3}$	$2$	$2$	$y = 2$
$\frac{x}{\sqrt{x^{2}+1}}$	$1$	$-1$	$y = 1$ and $y = -1$
$\arctan x$	$\frac{\pi}{2}$	$-\frac{\pi}{2}$	$y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$
$e^{x}$	$+\infty$	$0$	$y = 0$ , left end only
$\frac{2e^{x}-3}{e^{x}+1}$	$2$	$-3$	$y = 2$ and $y = -3$
$x^{2}$	$+\infty$	$+\infty$	None

Training & Error Diagnosis

Core Archetype (Simplest Form): Find every horizontal asymptote of $f(x) = \dfrac{2x^{2}+3x}{x^{2}-1}$ .

Standard Execution:

Right end. Divide by $x^{2}$ : $\frac{2 + 3/x}{1 - 1/x^{2}} \to 2$ , so $\lim_{x \to \infty} f(x) = 2$ .

Left end. The same division applies, and no root or exponential changes the sign, so $\lim_{x \to -\infty} f(x) = 2$ as well.

Count the lines. Both ends settle at $2$ , so the only horizontal asymptote is $y = 2$ .

Separate the vertical lines. The denominator is zero at $x = 1$ and $x = -1$ , and neither factor cancels, so $x = 1$ and $x = -1$ are vertical asymptotes and have nothing to do with the horizontal line.

Exam Variation (Context / Parameter Shift): Find every horizontal asymptote of $g(x) = \dfrac{2e^{x}-3}{e^{x}+1}$ .

Solution: For $x \to \infty$ , divide top and bottom by $e^{x}$ : $\frac{2 - 3e^{-x}}{1 + e^{-x}} \to 2$ . For $x \to -\infty$ , $e^{x} \to 0$ , so the quotient tends to $\frac{0-3}{0+1} = -3$ . The two ends disagree, which gives two horizontal asymptotes: $y = 2$ and $y = -3$ . This cannot happen for a rational function with equal degrees, because a polynomial is even-handed at both ends in a way an exponential is not.

Common Traps & Failure Signals:

The single-end answer (Signal: one limit written, one line named): the left end has not been computed, and for radicals and exponentials it frequently disagrees.
The barrier belief (Signal: "the graph crosses the line, so it is not an asymptote"): crossing is irrelevant; only the limit matters.
The vertical confusion (Signal: a denominator zero offered as a horizontal asymptote): denominator zeros produce vertical lines, and the two families are found by different operations.

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

One-Sided Effort: You stopped at the right end.
Symmetry Assumption: You assumed both ends must agree because the expression looks symmetric in its degrees, which fails the moment a root or an exponential changes sign at negative infinity.
Definition Drift: You answered the horizontal question with the vertical rule, treating a point where the denominator vanishes as if it were the height the curve approaches at infinity, so the two different kinds of line were merged into one category and neither was computed correctly.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; As the interpretation of a limit already computed
10-Year Total Count: 2 of 9 Exams · 2 FRQ Parts; One limit per direction, then the line equations
FRQ Years Appeared: 2021 (Q6b), 2022 (Q6d)

Practice Examples

MCQ 1

Which Function Has Two

Which function has two distinct horizontal asymptotes?

MCQ 2

Reading a Rational Function

Let $f(x) = \dfrac{2x^{2}+3x}{x^{2}-1}$ . Which statement about horizontal asymptotes is correct?

MCQ 3

Meeting the Line

The graph of $f(x) = \dfrac{2x}{x^{2}+1}$ passes through the origin. Which statement is correct?

FRQ 4

An Exponential With Two Lines

Let $f(x) = \frac{2e^{x}-3}{e^{x}+1}$

(a) Evaluate $\lim_{x \to \infty} f(x)$ .

(b) Evaluate $\lim_{x \to -\infty} f(x)$ .

(c) State the equation of every horizontal asymptote of $f$ .

(d) Explain why $f$ has two horizontal asymptotes while a rational function with a numerator and denominator of the same degree has only one.

Cheatsheet

Horizontal Asymptotes

	Right end	Left end	Lines
Rational, degrees equal	Ratio	Ratio	One line
Rational, denominator degree larger	$0$	$0$	One line, $y = 0$
Rational, numerator degree larger	$\pm\infty$	$\pm\infty$	None
Root of an even power over a linear term	$+k$	$-k$	Two lines
$e^{x}$ or $e^{-x}$	Depends on the exponent's sign	Opposite of the right end	One line
$\arctan x$	$\frac{\pi}{2}$	$-\frac{\pi}{2}$	Two lines
Bounded oscillation such as $\sin x$	No limit	No limit	None

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Start Topic Practice Next Topic: 1.20 The Intermediate Value Theorem (Conditions and Conclusion)

Unit 1Analyzing horizontal asymptotes using limits at infinity.

Concept 4.141Built for May 2027 AP Exam

Analyzing horizontal asymptotes using limits at infinity.

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: $\frac{x}{\sqrt{x^{2}+1}}$ , $\arctan x$ , expressions built from $e^{x}$ , and rational functions whose two ends agree, offered next to curves that cross their own asymptote.
What you’ll do: Compute each directional limit separately, convert each finite value into a line, then count how many distinct lines the pair produces.
Memory hook: One end, one limit, one line. Two ends means two limits, and possibly two lines.
Use it on the exam: For MCQs: If a function is even in its degrees, expect one line; if a root or an exponential flips the sign at one end, expect two. Counting the directions before computing keeps the option list honest. For FRQs: Write both limits, then write the lines as equations. "Since $\lim_{x \to \infty} f(x) = 2$ and $\lim_{x \to -\infty} f(x) = -3$ , the graph has horizontal asymptotes $y = 2$ and $y = -3$ " is the complete answer; naming only one loses the second point.

Concept Review

The Definition, Taken One Direction at a Time

There is no such thing as "the" horizontal asymptote of a function in general. There is a right-end limit and a left-end limit, and each one either produces a line or does not.

Right-end behaviour	Left-end behaviour	Horizontal asymptotes
$\lim_{x \to \infty} f(x) = 2$	$\lim_{x \to -\infty} f(x) = 2$	One line, $y = 2$
$\lim_{x \to \infty} f(x) = 2$	$\lim_{x \to -\infty} f(x) = -3$	Two lines, $y = 2$ and $y = -3$
$\lim_{x \to \infty} f(x) = 0$	$\lim_{x \to -\infty} f(x) = +\infty$	One line, $y = 0$ , on the right only
$\lim_{x \to \infty} f(x) = +\infty$	$\lim_{x \to -\infty} f(x) = +\infty$	No horizontal asymptote

Two Is the Ceiling

Two directions exist, so two distinct lines are the maximum. The interesting cases are the ones that reach that maximum.

Radicals with an odd effect at negative infinity. $f(x) = \frac{x}{\sqrt{x^{2}+1}}$ tends to $1$ on the right and to $-1$ on the left, because the denominator behaves like $\lvert x \rvert$ and that is $-x$ on the left.
Inverse tangent. $\arctan x \to \frac{\pi}{2}$ as $x \to \infty$ and $\to -\frac{\pi}{2}$ as $x \to -\infty$ . Two lines, and neither is reached.
Exponentials. $e^{x} \to 0$ as $x \to -\infty$ and $\to \infty$ as $x \to \infty$ , so there is one line, $y = 0$ , and it belongs to the left end only.
Rational functions with equal degrees. The leading-coefficient ratio is the same on both sides, so the two ends agree and only one line exists.

Figure 1 — One function, two horizontal asymptotes.

Graph of a function built from e to the x that settles toward y equals 2 on the right and toward y equals minus 3 on the left, with both levels drawn as dashed horizontal lines

Crossing the Line Does Not Disqualify It

A horizontal asymptote is not a barrier. It records where the curve is heading, and many functions cross the line on the way there.

$f(x) = \frac{2x}{x^{2}+1} \quad \text{has} \quad \lim_{x \to \pm\infty} f(x) = 0$

Topic-Specific Study Method

The Two-Ends Audit

Compute the right-end limit first, on its own line. Writing one limit per line prevents the second from being silently copied from the first.
Compute the left-end limit next, treating the sign of the input as new information. Any root encountered here needs $\lvert x \rvert = -x$ .
Convert each finite value into a line equation. A limit of $2$ produces the line $y = 2$ , and a limit that is infinite produces no line at all.
Count the distinct lines. One value for both ends is one line; two different values are two lines; an infinite limit removes one line.
State the conclusion in the language of the question. If the item asks whether a stated line is an asymptote, answer with the limit that establishes or refutes it.

Function	Right end	Left end	Lines
$\frac{2x^{2}+1}{x^{2}-3}$	$2$	$2$	$y = 2$
$\frac{x}{\sqrt{x^{2}+1}}$	$1$	$-1$	$y = 1$ and $y = -1$
$\arctan x$	$\frac{\pi}{2}$	$-\frac{\pi}{2}$	$y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$
$e^{x}$	$+\infty$	$0$	$y = 0$ , left end only
$\frac{2e^{x}-3}{e^{x}+1}$	$2$	$-3$	$y = 2$ and $y = -3$
$x^{2}$	$+\infty$	$+\infty$	None

Training & Error Diagnosis

Core Archetype (Simplest Form): Find every horizontal asymptote of $f(x) = \dfrac{2x^{2}+3x}{x^{2}-1}$ .

Standard Execution:

Right end. Divide by $x^{2}$ : $\frac{2 + 3/x}{1 - 1/x^{2}} \to 2$ , so $\lim_{x \to \infty} f(x) = 2$ .

Left end. The same division applies, and no root or exponential changes the sign, so $\lim_{x \to -\infty} f(x) = 2$ as well.

Count the lines. Both ends settle at $2$ , so the only horizontal asymptote is $y = 2$ .

Separate the vertical lines. The denominator is zero at $x = 1$ and $x = -1$ , and neither factor cancels, so $x = 1$ and $x = -1$ are vertical asymptotes and have nothing to do with the horizontal line.

Exam Variation (Context / Parameter Shift): Find every horizontal asymptote of $g(x) = \dfrac{2e^{x}-3}{e^{x}+1}$ .

Solution: For $x \to \infty$ , divide top and bottom by $e^{x}$ : $\frac{2 - 3e^{-x}}{1 + e^{-x}} \to 2$ . For $x \to -\infty$ , $e^{x} \to 0$ , so the quotient tends to $\frac{0-3}{0+1} = -3$ . The two ends disagree, which gives two horizontal asymptotes: $y = 2$ and $y = -3$ . This cannot happen for a rational function with equal degrees, because a polynomial is even-handed at both ends in a way an exponential is not.

Common Traps & Failure Signals:

The single-end answer (Signal: one limit written, one line named): the left end has not been computed, and for radicals and exponentials it frequently disagrees.
The barrier belief (Signal: "the graph crosses the line, so it is not an asymptote"): crossing is irrelevant; only the limit matters.
The vertical confusion (Signal: a denominator zero offered as a horizontal asymptote): denominator zeros produce vertical lines, and the two families are found by different operations.

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

One-Sided Effort: You stopped at the right end.
Symmetry Assumption: You assumed both ends must agree because the expression looks symmetric in its degrees, which fails the moment a root or an exponential changes sign at negative infinity.
Definition Drift: You answered the horizontal question with the vertical rule, treating a point where the denominator vanishes as if it were the height the curve approaches at infinity, so the two different kinds of line were merged into one category and neither was computed correctly.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; As the interpretation of a limit already computed
10-Year Total Count: 2 of 9 Exams · 2 FRQ Parts; One limit per direction, then the line equations
FRQ Years Appeared: 2021 (Q6b), 2022 (Q6d)

Practice Examples

MCQ 1

Which Function Has Two

Which function has two distinct horizontal asymptotes?

MCQ 2

Reading a Rational Function

Let $f(x) = \dfrac{2x^{2}+3x}{x^{2}-1}$ . Which statement about horizontal asymptotes is correct?

MCQ 3

Meeting the Line

The graph of $f(x) = \dfrac{2x}{x^{2}+1}$ passes through the origin. Which statement is correct?

FRQ 4

An Exponential With Two Lines

Let $f(x) = \frac{2e^{x}-3}{e^{x}+1}$

(a) Evaluate $\lim_{x \to \infty} f(x)$ .

(b) Evaluate $\lim_{x \to -\infty} f(x)$ .

(c) State the equation of every horizontal asymptote of $f$ .

(d) Explain why $f$ has two horizontal asymptotes while a rational function with a numerator and denominator of the same degree has only one.

Cheatsheet

Horizontal Asymptotes

	Right end	Left end	Lines
Rational, degrees equal	Ratio	Ratio	One line
Rational, denominator degree larger	$0$	$0$	One line, $y = 0$
Rational, numerator degree larger	$\pm\infty$	$\pm\infty$	None
Root of an even power over a linear term	$+k$	$-k$	Two lines
$e^{x}$ or $e^{-x}$	Depends on the exponent's sign	Opposite of the right end	One line
$\arctan x$	$\frac{\pi}{2}$	$-\frac{\pi}{2}$	Two lines
Bounded oscillation such as $\sin x$	No limit	No limit	None

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