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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Definition of limits at infinity, degree rules for rational functions, finding horizontal asymptotes via two-sided infinite limits, non-rational limit evaluation, and interpreting horizontal asymptotes in real-world contexts.

You should already know: How to evaluate basic finite limits, algebraic manipulation of rational and exponential expressions, how to compute one-sided limits.

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 BC 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 two core ideas from Unit 1: limits at infinity (limits as $x$ grows without bound in the positive or negative direction) and horizontal asymptotes, which describe the long-run end behavior of a function’s graph. According to the AP Calculus BC Course and Exam Description (CED), Unit 1 (Limits and Continuity) accounts for 10–12% of the total AP exam score, and this topic is tested in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with curve sketching, differential equation modeling, or improper integral questions.

Formally, we write ${\lim }_{x\to \infty }f(x)=L$ to mean that as $x$ increases without bound, $f(x)$ gets arbitrarily close to the finite number $L$, and ${\lim }_{x\to -\infty }f(x)=L$ for $x$ decreasing without bound. A horizontal asymptote is a horizontal line $y=L$ that the function approaches as $x$ approaches $\infty$ or $-\infty$. The key connection this topic establishes is: a function has a horizontal asymptote at $y=L$ if and only if at least one limit at infinity of $f(x)$ equals $L$.

2. Evaluating Limits at Infinity for Rational Functions

A rational function is defined as $f(x)=\frac{P_n(x)}{Q_m(x)}$, where $P_n(x)$ is a polynomial of degree $n$ (highest power of $x$ is $x^n$) and $Q_m(x)$ is a polynomial of degree $m$. For large values of $|x|$, the highest-degree term in each polynomial dominates all lower-degree terms, since lower-degree terms become negligible as $|x|$ grows. To evaluate the limit, we divide both the numerator and denominator by the highest power of $x$ in the denominator, then use the fact that ${\lim }_{x\to \pm \infty }\frac{1}{x^k}=0$ for any $k>0$.

This leads to three simple rules for rational functions:

1. If $n<m$ (numerator lower degree than denominator): ${\lim }_{x\to \pm \infty }f(x)=0$
2. If $n=m$ (degrees equal): ${\lim }_{x\to \pm \infty }f(x)=\frac{\text{leading coefficient of }{P}_n}{\text{leading coefficient of }{Q}_m}$
3. If $n>m$ (numerator higher degree than denominator): ${\lim }_{x\to \pm \infty }f(x)=\pm \infty$, no finite limit

Worked Example

Find ${\lim }_{x\to \infty }\frac{5x^3-2x^2+7}{-3x^3+10x-4}$ and confirm the end behavior.

1. Identify degrees: numerator degree $n=3$, denominator degree $m=3$, so degrees are equal.
2. Divide numerator and denominator by $x^3$ (the highest power of $x$): $$\underset{x\to \infty }{\lim }\frac{5-\frac{2}{x}+\frac{7}{x^3}}{-3+\frac{10}{x^2}-\frac{4}{x^3}}$$
3. All terms with $1/x^k$ for $k>0$ approach 0 as $x\to \infty$, so they drop out.
4. We are left with the ratio of leading coefficients: $\frac{5}{-3}=-\frac{5}{3}$.

The limit is $\boxed{-\frac{5}{3}}$.

Exam tip: When evaluating limits of even-powered roots at negative infinity, remember that $\sqrt{x^2}=|x|=-x$ for $x<0$; always check the sign for $x\to -\infty$ to avoid sign errors.

3. Finding Horizontal Asymptotes from Limits at Infinity

By definition, a function $y=f(x)$ has a horizontal asymptote at $y=L$ if either ${\lim }_{x\to \infty }f(x)=L$ or ${\lim }_{x\to -\infty }f(x)=L$, where $L$ is a finite number. A common misconception is that a function can only have one horizontal asymptote, but this is not true: non-rational functions (like exponentials, inverse trigonometric functions) can have two different horizontal asymptotes, one for the right end ($x\to \infty$) and one for the left end ($x\to -\infty$). The only requirement for a horizontal asymptote is that at least one infinite limit is finite.

Another important correction to common myths: a function can cross its horizontal asymptote at finite values of $x$, and this does not invalidate the asymptote. Asymptotes only describe end behavior, not behavior at finite $x$.

Worked Example

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

1. First evaluate the limit as $x\to \infty$: for large positive $x$, $e^x$ grows very quickly, so divide numerator and denominator by $e^x$: $$\underset{x\to \infty }{\lim }\frac{2+5e^{-x}}{1-3e^{-x}}$$
2. As $x\to \infty$, $e^{-x}\to 0$, so the limit simplifies to $\frac{2}{1}=2$. This gives a horizontal asymptote $y=2$.
3. Next evaluate the limit as $x\to -\infty$: as $x\to -\infty$, $e^x\to 0$, so substitute directly: $$\underset{x\to -\infty }{\lim }\frac{2(0)+5}{0-3}=-\frac{5}{3}$$
4. This is a second finite limit, so we have a second horizontal asymptote $y=-\frac{5}{3}$.

The full set of horizontal asymptotes is $\boxed{y=2\text{ and }y=-\frac{5}{3}}$.

Exam tip: Never assume a function only has one horizontal asymptote; always check both $x\to \infty$ and $x\to -\infty$ for functions involving exponentials, inverse trigonometric functions, or piecewise definitions.

4. Limits at Infinity for Non-Rational Functions

Not all functions with horizontal asymptotes are rational. BC exam questions regularly ask for horizontal asymptotes of non-rational functions including products/quotients of polynomials and exponentials, logarithmic functions, and oscillating bounded functions. For these problems, you will often need to use L'Hospital's Rule (covered later in the course, but commonly applied to this topic) or the Squeeze Theorem for indeterminate forms.

Key standard results for non-rational functions that you should memorize:

• For $a>0$: ${\lim }_{x\to \infty }{e}^{-ax}=0$, ${\lim }_{x\to \infty }{e}^{ax}=\infty$
• For any $p>0$: ${\lim }_{x\to \infty }\frac{\ln x}{x^p}=0$ (logarithms grow slower than any positive power of x)
• If $|g(x)|\le M$ (bounded) and ${\lim }_{x\to \infty }h(x)=0$, then ${\lim }_{x\to \infty }g(x)h(x)=0$ (Squeeze Theorem result)

Worked Example

Find all horizontal asymptotes of $f(x)=x{e}^{-3x}$.

1. Check the limit as $x\to \infty$: this gives the indeterminate form $\infty \cdot 0$. Rewrite it as a fraction to apply L'Hospital's Rule: $$\underset{x\to \infty }{\lim }x{e}^{-3x}=\underset{x\to \infty }{\lim }\frac{x}{e^{3x}}$$
2. This is now the indeterminate form $\infty /\infty$, so we can apply L'Hospital's Rule by taking derivatives of the numerator and denominator: $$\underset{x\to \infty }{\lim }\frac{\frac{d}{dx}(x)}{\frac{d}{dx}(e^{3x})}=\underset{x\to \infty }{\lim }\frac{1}{3e^{3x}}=0$$
3. Check the limit as $x\to -\infty$: as $x\to -\infty$, $x\to -\infty$ and $e^{-3x}=e^{|3x|}\to \infty$, so $x{e}^{-3x}\to -\infty$, which is not finite.
4. Only one finite limit exists, so there is one horizontal asymptote at $y=0$.

The only horizontal asymptote is $\boxed{y=0}$.

Exam tip: For indeterminate forms at infinity ($\infty -\infty$, $0\cdot \infty$, $\infty /\infty$), always rewrite the expression to apply L'Hospital's Rule or algebraic manipulation before concluding a finite limit exists.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For $f(x)=\frac{3x}{\sqrt{x^2+2}}$, concluding ${\lim }_{x\to -\infty }f(x)=3$ because the ratio of leading coefficients is 3/1. Why: Students forget $\sqrt{x^2}=|x|=-x$ for negative $x$, so the sign is incorrect for $x\to -\infty$. Correct move: Always factor out $|x|$ from roots, substitute $|x|=-x$ for $x\to -\infty$ to get the correct sign.
• Wrong move: Claiming $y=3$ cannot be a horizontal asymptote because $f(2)=3$ (the function crosses $y=3$ at $x=2$). Why: The myth that functions cannot cross their asymptotes only applies to vertical asymptotes, not horizontal. Correct move: Only use the limit definition to confirm horizontal asymptotes; ignore crossings at finite $x$.
• Wrong move: For a rational function with numerator degree 3, denominator degree 2, concluding it has a horizontal asymptote at $y=0$. Why: Confused the degree rule: students reverse the outcome for different degrees. Correct move: Follow the degree rule explicitly: $n<m\to 0$, $n=m\to$ ratio, $n>m\to$ no finite HA.
• Wrong move: For $f(x)=\frac{2e^x+1}{e^x+5}$, only finding $y=2$ as a horizontal asymptote and stopping. Why: Students assume exponentials only have one HA and forget to check the left end. Correct move: For all non-rational functions, evaluate both ${\lim }_{x\to \infty }f(x)$ and ${\lim }_{x\to -\infty }f(x)$.
• Wrong move: Concluding ${\lim }_{x\to \infty }(x-\sqrt{x^2+3})=0$ because $\infty -\infty =0$. Why: Confused finite number subtraction with the indeterminate form $\infty -\infty$. Correct move: Multiply by the conjugate to rewrite the difference as a fraction, then evaluate the limit.
• Wrong move: Concluding $y=0$ is not a horizontal asymptote of $\frac{\sin x}{x}$ because ${\lim }_{x\to \infty }\sin x$ does not exist. Why: Forgot the Squeeze Theorem applies to products of bounded functions and functions that go to zero. Correct move: Use the Squeeze Theorem for bounded oscillating functions to check for finite limits at infinity.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following gives the sum of all distinct horizontal asymptotes of $f(x)=\frac{4x^2+2x-7}{-2x^2+9}+\frac{e^x}{e^x+1}$? A) $-3$ B) $-2$ C) $-1$ D) $0$

Worked Solution: First evaluate ${\lim }_{x\to \infty }f(x)$. For the rational term, degrees are equal, so the limit is the ratio of leading coefficients $4/-2=-2$. For the exponential term, divide numerator and denominator by $e^x$ to get ${\lim }_{x\to \infty }\frac{1}{1+e^{-x}}=1$. Adding gives ${\lim }_{x\to \infty }f(x)=-2+1=-1$, so $y=-1$ is a horizontal asymptote. Next evaluate ${\lim }_{x\to -\infty }f(x)$: the rational term still approaches $-2$, and $e^x\to 0$ as $x\to -\infty$, so the exponential term approaches $0$. This gives ${\lim }_{x\to -\infty }f(x)=-2+0=-2$, so $y=-2$ is a second distinct horizontal asymptote. The sum is $-1+(-2)=-3$. The correct answer is A.

Question 2 (Free Response)

Let $f(x)=\frac{x^2-4}{\sqrt{4x^4+9}}$. (a) Find ${\lim }_{x\to \infty }f(x)$ and ${\lim }_{x\to -\infty }f(x)$. (b) State all horizontal asymptotes of $f(x)$. (c) Use limit laws to find ${\lim }_{x\to \infty }\left(2f(x)-\frac{1}{x}\right)$.

Worked Solution: (a) Divide numerator and denominator by $x^2$, the highest power of $x$ in the denominator. Note that $\sqrt{4x^4+9}=\sqrt{x^4(4+9/x^4)}=x^2\sqrt{4+9/x^4}$, since $x^2=|x^2|$ is positive for all non-zero $x$, regardless of sign. This simplifies $f(x)$ to: $$f(x)=\frac{1-4/x^2}{\sqrt{4+9/x^4}}$$ As $x\to \pm \infty$, all terms with $1/x^k$ go to $0$, so ${\lim }_{x\to \infty }f(x)={\lim }_{x\to -\infty }f(x)=\frac{1}{\sqrt{4}}=\frac{1}{2}$.

(b) Since both limits at infinity equal $\frac{1}{2}$, the only horizontal asymptote is $y=\frac{1}{2}$.

(c) Apply limit laws to split the limit: $$\underset{x\to \infty }{\lim }\left(2f(x)-\frac{1}{x}\right)=2\underset{x\to \infty }{\lim }f(x)-\underset{x\to \infty }{\lim }\frac{1}{x}=2\left(\frac{1}{2}\right)-0=1$$

Question 3 (Application / Real-World Style)

A retail analyst is modeling the total cumulative number of units of a new product sold $t$ months after launch, given by $S(t)=\frac{5000t}{0.2t+25}$ for $t\ge 0$, where $S(t)$ is measured in units. Find the horizontal asymptote of $S(t)$, and interpret its meaning in the context of the problem.

Worked Solution: To find the horizontal asymptote, evaluate the limit as $t\to \infty$: $$\underset{t\to \infty }{\lim }\frac{5000t}{0.2t+25}=\underset{t\to \infty }{\lim }\frac{5000}{0.2+25/t}=\frac{5000}{0.2}=25000$$ The horizontal asymptote is $y=25000$. In context, this means the total cumulative number of units of the product that will ever be sold approaches 25,000; market saturation limits total sales to this long-run maximum.

7. Quick Reference Cheatsheet

8. What's Next

This topic is a core building block for analyzing function end behavior, which is required across nearly every subsequent unit of AP Calculus BC. It is a prerequisite for full curve sketching, analyzing long-term behavior of solutions to differential equations, and evaluating improper integrals. Without mastering the connection between limits at infinity and horizontal asymptotes, you will struggle to correctly answer free-response questions that ask for complete descriptions of function behavior, or to identify when an improper integral converges to a finite value. Next you will extend the ideas of end behavior to oblique (slant) asymptotes, then apply limit-at-infinity techniques to improper integral convergence.

• Evaluating limits algebraically
• Vertical and oblique asymptotes
• Analyzing function behavior
• Improper integrals and convergence