Connecting limits at infinity and horizontal asymptotes — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Defining horizontal asymptotes using limits at positive and negative infinity, end-behavior rules for rational functions, finding horizontal asymptotes for non-rational functions, and interpreting horizontal asymptotes in applied contexts for AP exam questions.

You should already know: How to evaluate limits at finite and infinite values, basic polynomial and rational function algebra, properties of exponential functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Connecting limits at infinity and horizontal asymptotes?

This topic connects the algebraic concept of a function’s end behavior to the graphical concept of horizontal asymptotes, and is a core required learning outcome per the AP Calculus AB CED in Unit 1: Limits and Continuity, which accounts for 10–12% of the total AP exam score. This topic typically appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a standalone MCQ or as a component of a larger FRQ focused on curve analysis or interpretation of a contextual model. By definition, a horizontal asymptote (HA) is a horizontal line $y=L$ that the graph of $f(x)$ approaches as $x$ increases without bound (tends to positive infinity) or decreases without bound (tends to negative infinity). Formally, if ${\lim }_{x\to \infty }f(x)=L$ or ${\lim }_{x\to -\infty }f(x)=L$, then $y=L$ is a horizontal asymptote of $f$. Unlike vertical asymptotes, which correspond to infinite limits at a finite $x$-value, horizontal asymptotes describe long-run end behavior. A function can cross its horizontal asymptote, and can have 0, 1, or 2 distinct horizontal asymptotes.

2. Finding Horizontal Asymptotes for Rational Functions

Rational functions are defined as $f(x)=\frac{P(x)}{Q(x)}$, where $P(x)$ is a polynomial of degree $n$ and $Q(x)$ is a polynomial of degree $m$. The end behavior (and any resulting horizontal asymptote) depends entirely on the relationship between $n$ and $m$. We derive the rule by factoring out the leading term from both polynomials when evaluating the limit as $x\to \pm \infty$: for $P(x)=a_n{x}^n+...+a_0$ and $Q(x)=b_m{x}^m+...+b_0$, factoring gives $$\underset{x\to \pm \infty }{\lim }\frac{a_n{x}^n\left(1+...+\frac{a_0}{a_n{x}^n}\right)}{b_m{x}^m\left(1+...+\frac{b_0}{b_m{x}^m}\right)}=\frac{a_n}{b_m}\underset{x\to \pm \infty }{\lim }{x}^{n-m}$$ All lower-degree terms vanish to 0 as $x\to \pm \infty$, leaving three simple cases:

1. If $n<m$: The limit equals 0, so there is a horizontal asymptote at $y=0$.
2. If $n=m$: The limit equals $\frac{a_n}{b_m}$, so there is a horizontal asymptote at $y=\frac{a_n}{b_m}$.
3. If $n>m$: The limit approaches $\pm \infty$, so no horizontal asymptote exists. Rational functions always have the same limit (or infinite limit) for both $x\to \infty$ and $x\to -\infty$, so they can only have 0 or 1 horizontal asymptote.

Worked Example

Find all horizontal asymptotes of $f(x)=\frac{5x^3-4x^2+10}{-2x^3+7x+28}$.

1. First, identify the degrees of the numerator and denominator: the numerator $P(x)$ has degree $n=3$, and the denominator $Q(x)$ has degree $m=3$.
2. Since $n=m$, we only need the ratio of the leading coefficients: $a_n=5$ (numerator leading term) and $b_m=-2$ (denominator leading term).
3. Confirm with the limit calculation: $$\underset{x\to \infty }{\lim }\frac{5x^3-4x^2+10}{-2x^3+7x+28}=\underset{x\to \infty }{\lim }\frac{x^3\left(5-\frac{4}{x}+\frac{10}{x^3}\right)}{x^3\left(-2+\frac{7}{x^2}+\frac{28}{x^3}\right)}=\frac{5-0+0}{-2+0+0}=-\frac{5}{2}$$
4. The limit as $x\to -\infty$ is also $-\frac{5}{2}$, so only one horizontal asymptote exists.

Conclusion: The only horizontal asymptote is $y=-\frac{5}{2}$.

Exam tip: When solving for horizontal asymptotes of a rational function, don’t waste time expanding or factoring the entire polynomial. Just pull the leading term from the numerator and denominator and apply the degree rule.

3. Finding Horizontal Asymptotes for Non-Rational Functions

Not all functions with horizontal asymptotes are rational. Many common non-rational functions (exponential, logistic, combinations of polynomials and roots, etc.) require direct evaluation of both ${\lim }_{x\to \infty }f(x)$ and ${\lim }_{x\to -\infty }f(x)$, because the limit can differ in the two directions. A key rule for exponential functions to remember: for any positive constant $k$, ${\lim }_{x\to \infty }{e}^{-kx}=0$ and ${\lim }_{x\to -\infty }{e}^{kx}=0$. Unlike rational functions, non-rational functions can have two distinct horizontal asymptotes, one for each direction of infinity. This is a common point tested on the AP exam, as many students forget to check both directions.

Worked Example

Find all horizontal asymptotes of $f(x)=\frac{3e^x+6}{5e^x-10}$.

1. We need to evaluate two separate limits: one as $x\to \infty$, and one as $x\to -\infty$, since exponential behavior changes with the sign of $x$.
2. Evaluate ${\lim }_{x\to \infty }f(x)$: For $x\to \infty$, $e^x$ grows without bound, so divide numerator and denominator by $e^x$: $$\underset{x\to \infty }{\lim }\frac{3e^x+6}{5e^x-10}=\underset{x\to \infty }{\lim }\frac{3+\frac{6}{e^x}}{5-\frac{10}{e^x}}=\frac{3+0}{5-0}=\frac{3}{5}$$
3. Evaluate ${\lim }_{x\to -\infty }f(x)$: For $x\to -\infty$, $e^x\to 0$, so substitute directly: $$\underset{x\to -\infty }{\lim }\frac{3e^x+6}{5e^x-10}=\frac{3(0)+6}{5(0)-10}=-\frac{6}{10}=-\frac{3}{5}$$
4. Both limits are finite and distinct, so both lines are horizontal asymptotes.

Conclusion: The horizontal asymptotes are $y=\frac{3}{5}$ and $y=-\frac{3}{5}$.

Exam tip: Always evaluate both limits for non-rational functions. If you only check the limit as $x\to \infty$, you will miss the second horizontal asymptote, which is often a required answer point.

4. Interpreting Horizontal Asymptotes in Context

On the AP Calculus AB exam, you will often be asked to interpret the meaning of a horizontal asymptote in a real-world context, usually in FRQ questions. This requires connecting the limit definition to the problem’s variables to earn full credit. If $y=L$ is a horizontal asymptote as $x\to \infty$, where $x$ is the independent variable (usually time, number of units, etc.) and $y$ is the dependent variable (population, temperature, cost, etc.), the interpretation must explicitly state that as the independent variable grows without bound, the dependent variable approaches $L$, with units for both variables. This is most commonly tested in growth models, cooling problems, and cost analysis, where the horizontal asymptote corresponds to a meaningful long-run quantity like carrying capacity or room temperature.

Worked Example

Newton’s Law of Cooling for a cup of hot coffee gives the temperature $T(t)$ (in degrees Celsius) of the coffee $t$ minutes after it is poured as $T(t)=22+76e^{-0.08t}$. Identify the horizontal asymptote of $T(t)$ for $t\ge 0$ and interpret it in context.

1. Since time $t$ can only increase from 0, we only evaluate the limit as $t\to \infty$.
2. Use the exponential limit rule: for $k=0.08>0$, ${\lim }_{t\to \infty }{e}^{-0.08t}=0$. Substitute into the function: $$\underset{t\to \infty }{\lim }(22+76e^{-0.08t})=22+76(0)=22$$
3. The horizontal asymptote is $T=22^{\circ }C$.
4. Write the interpretation referencing both variables and units: As the number of minutes since the coffee was poured increases without bound, the temperature of the coffee approaches 22 degrees Celsius (room temperature).

Exam tip: On FRQ interpretation questions, you will not earn full credit if you only state the asymptote. You must explicitly reference the behavior of both variables in context and include units.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming $y=L$ cannot be a horizontal asymptote because $f(5)=L$ (the function crosses the line at a finite $x$). Why: Students incorrectly extend the rule for vertical asymptotes (functions never cross vertical asymptotes) to horizontal asymptotes. Correct move: Always use the limit definition: if ${\lim }_{x\to \pm \infty }f(x)=L$, $y=L$ is a horizontal asymptote regardless of crossings at finite $x$.
• Wrong move: For the rational function $f(x)=\frac{4x^3}{2x^2+1}$, claiming a horizontal asymptote at $y=2$. Why: Compares leading coefficients without first checking that degrees are equal. Correct move: Always compare degrees first. If the numerator degree is larger than the denominator degree, state that there is no horizontal asymptote.
• Wrong move: For $f(x)=\frac{2e^x}{e^x+1}$, only finding the horizontal asymptote at $y=2$ and stopping. Why: Assumes all functions have the same limit for $x\to \pm \infty$, like rational functions. Correct move: Always evaluate both ${\lim }_{x\to \infty }f(x)$ and ${\lim }_{x\to -\infty }f(x)$ for non-rational functions before listing all horizontal asymptotes.
• Wrong move: Interpreting a horizontal asymptote of $P=1000$ (population in thousands) as "the population will eventually reach 1000". Why: Confuses the limit concept of "approaches" with "reaches" in context. Correct move: Always use the language "approaches" or "gets arbitrarily close to" and state that this describes the long-run behavior as the independent variable grows without bound.
• Wrong move: Simplifying $\sqrt{x^2+2x}=x\sqrt{1+2/x}$ for $x\to -\infty$ and keeping the positive $x$ coefficient. Why: Forgets that $\sqrt{x^2}=|x|=-x$ when $x$ is negative. Correct move: Always pull out $|x|$ from square roots of quadratic terms, and adjust the sign based on the direction of the limit.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following gives all horizontal asymptotes of the function $f(x)=\frac{5x\sqrt{x^2+9}}{3x^2+2}$? A) $y=\frac{5}{3}$ only B) $y=-\frac{5}{3}$ only C) $y=\frac{5}{3}$ and $y=-\frac{5}{3}$ D) No horizontal asymptotes

Worked Solution: We evaluate both limits separately. For $x\to \infty$, $\sqrt{x^2}=x$, so we factor $x^2$ from the square root: $$\underset{x\to \infty }{\lim }\frac{5x\sqrt{x^2(1+9/x^2)}}{x^2(3+2/x^2)}=\underset{x\to \infty }{\lim }\frac{5x\cdot x\sqrt{1+9/x^2}}{3x^2}=\frac{5}{3}$$ For $x\to -\infty$, $\sqrt{x^2}=|x|=-x$, so: $$\underset{x\to -\infty }{\lim }\frac{5x(-x)\sqrt{1+9/x^2}}{3x^2}=-\frac{5}{3}$$ Both limits are finite, so there are two distinct horizontal asymptotes. The correct answer is C.

Question 2 (Free Response)

Let $f(x)=\frac{3x^2-6x}{x^2-4}$. (a) Find all horizontal asymptotes of $f(x)$, if any exist. Justify your answer with limits. (b) Find all vertical asymptotes of $f(x)$, if any exist. Justify your answer with limits. (c) Is the statement "The function $f(x)$ never crosses its horizontal asymptote" true or false? Justify your answer.

Worked Solution: (a) The numerator and denominator are both degree 2, so we evaluate the limit: $$\underset{x\to \pm \infty }{\lim }\frac{3x^2-6x}{x^2-4}=\underset{x\to \pm \infty }{\lim }\frac{x^2(3-6/x)}{x^2(1-4/x^2)}=3$$ By definition, the only horizontal asymptote is $y=3$. (b) Factor the denominator: $x^2-4=(x-2)(x+2)$, so the function is undefined at $x=2$ and $x=-2$. Simplify the function: $f(x)=\frac{3x(x-2)}{(x-2)(x+2)}=\frac{3x}{x+2}$ for $x\ne 2$. The limit as $x\to 2$ is finite ($\frac{6}{4}=\frac{3}{2}$), so $x=2$ is a hole, not a vertical asymptote. For $x\to -2^{+}$, ${\lim }_{x\to -2^{+}}\frac{3x}{x+2}=-\infty$, so $x=-2$ is the only vertical asymptote. (c) Set $f(x)=3$ to check for a crossing: $\frac{3x^2-6x}{x^2-4}=3⟹3x^2-6x=3x^2-12⟹-6x=-12⟹x=2$. $f(2)$ is undefined, so the function never equals 3 at any finite $x$. The statement is true.

Question 3 (Application / Real-World Style)

A city planner is modeling the total cost $C(x)$, in millions of dollars, to remove $x$ percent of a pollutant from a city’s water supply, given by $C(x)=\frac{12x}{100-x}$ for $0\le x<100$. Find the horizontal asymptote of $C(x)$ as $x$ approaches 100 from the left, and interpret its meaning in context.

Worked Solution: We evaluate the limit as $x\to 100^{-}$: $$\underset{x\to 100^{-}}{\lim }\frac{12x}{100-x}=\underset{x\to 100^{-}}{\lim }\frac{12(100)}{0^{+}}=+\infty$$ The limit increases without bound, so there is no finite horizontal asymptote for this direction. In context, as the percentage of pollutant removed approaches 100%, the total cost of removing the pollutant grows without bound. This means it is not feasible for the city to remove 100% of the pollutant, as the cost would become unsustainably large.

7. Quick Reference Cheatsheet

8. What's Next

This topic is a foundational prerequisite for upcoming concepts in Unit 1 and across the rest of AP Calculus AB. Immediately after mastering limits at infinity and horizontal asymptotes, you will move on to the formal definition of continuity in Unit 1, and later to full curve sketching in Unit 5: Analytical Applications of Differentiation, where you will use horizontal asymptotes to fully describe the end behavior of a function. This topic is also critical for understanding logistic growth models in Unit 7, where the carrying capacity of a population is exactly the horizontal asymptote of the solution curve. Without mastering how to find and interpret horizontal asymptotes from limits, you will struggle to justify end behavior on FRQ questions and correctly interpret contextual models of growth and decay. The follow-on topics to study next are: Continuity over an interval Intermediate Value Theorem (IVT) Sketching graphs of functions Logistic differential equations