# Connecting limits at infinity and horizontal asymptotes

> AP Calculus BC · Unit 1: Limits and Continuity
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u1-connecting-limits-at-infinity-and/

This sub-topic connects limits at infinity (end behavior as $x$ grows without bound) to horizontal asymptotes, covering degree rules for rational functions, non-rational evaluation, and real-world interpretation for AP Calculus BC.

**Prerequisites:** [Evaluating basic finite limits](https://www.owlsprep.com/study/ap-calculus-bc-u1-evaluating-limits-algebraically/); Algebraic manipulation of rational and exponential expressions; Computing one-sided limits

## Learning objectives

- Define limits at infinity and connect them to horizontal asymptotes
- Evaluate limits at infinity for rational functions using degree rules
- Find all horizontal asymptotes for rational and non-rational functions
- Interpret horizontal asymptotes in real-world contexts

## Core Definition: The Connection Between Concepts

This topic connects two core Unit 1 ideas: 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. Unit 1 accounts for 10–12% of the total AP exam score, and this topic is tested in both multiple-choice and free-response sections.

**Limit at Infinity** — $\lim_{x \to \infty} f(x) = L$ means $f(x)$ gets arbitrarily close to the finite number $L$ as $x$ increases without bound; $\lim_{x \to -\infty} f(x) = L$ is the same for $x$ decreasing without bound.

*Notation:* $\lim_{x \to \pm \infty} f(x)$

**Horizontal Asymptote** — A horizontal line $y=L$ that the function approaches as $x$ approaches $\infty$ or $-\infty$. A function has a horizontal asymptote at $y=L$ if and only if at least one limit at infinity of $f(x)$ equals $L$.

*Example:* A rational function with equal leading degrees has a horizontal asymptote at the ratio of leading coefficients.

> **Exam tip:** Asymptotes only describe end behavior, not behavior at finite $x$ — functions can cross horizontal asymptotes without violating the definition.

## 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 $n$ and $Q_m(x)$ is a polynomial of degree $m$. For large values of $|x|$, the highest-degree term dominates all lower terms, which become negligible. To evaluate the limit, divide both the numerator and denominator by the highest power of $x$ in the denominator, using the fact that $\lim_{x \to \pm \infty} \frac{1}{x^k} = 0$ for any $k>0$.

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 $x$):
3. $$\lim_{x \to \infty} \frac{5 - \frac{2}{x} + \frac{7}{x^3}}{-3 + \frac{10}{x^2} - \frac{4}{x^3}}$$
4. All terms with $1/x^k$ for $k>0$ approach 0 as $x \to \infty$, so they drop out, leaving the ratio of leading coefficients:
5. $$\frac{5}{-3} = -\frac{5}{3}$$

> **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 to avoid errors.

## Finding All Horizontal Asymptotes

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. Common misconceptions: (1) non-rational functions can have two different horizontal asymptotes, one for each end, and (2) crossing a horizontal asymptote at finite $x$ does not invalidate it.

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

1. Evaluate the limit as $x \to \infty$: divide numerator and denominator by $e^x$:
2. $$\lim_{x \to \infty} \frac{2 + 5e^{-x}}{1 - 3e^{-x}}$$
3. As $x \to \infty$, $e^{-x} \to 0$, so the limit simplifies to $\frac{2}{1} = 2$, giving a horizontal asymptote $y=2$.
4. Next evaluate the limit as $x \to -\infty$: as $x \to -\infty$, $e^x \to 0$, so substitute directly:
5. $$\lim_{x \to -\infty} \frac{2(0) + 5}{0 - 3} = -\frac{5}{3}$$
6. This is a second finite limit, so we have a second horizontal asymptote.

> **Exam tip:** Never assume a function only has one horizontal asymptote; always check both $x \to \infty$ and $x \to -\infty$ for non-rational functions.

## Limits at Infinity for Non-Rational Functions

AP 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 indeterminate forms, use L'Hospital's Rule or the Squeeze Theorem. Key standard results to memorize:

- For $a>0$: $\lim_{x \to \infty} e^{-ax} = 0$, $\lim_{x \to \infty} e^{ax} = \infty$
- For any $p>0$: $\lim_{x \to \infty} \frac{\ln x}{x^p} = 0$ (logs grow slower than any positive power of $x$)
- If $|g(x)| \leq M$ (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:
2. $$\lim_{x \to \infty} x e^{-3x} = \lim_{x \to \infty} \frac{x}{e^{3x}}$$
3. This is now the indeterminate form $\infty/\infty$, so apply L'Hospital's Rule by differentiating numerator and denominator:
4. $$\lim_{x \to \infty} \frac{\frac{d}{dx}(x)}{\frac{d}{dx}(e^{3x})} = \lim_{x \to \infty} \frac{1}{3 e^{3x}} = 0$$
5. Check the limit as $x \to -\infty$: as $x \to -\infty$, $x \to -\infty$ and $e^{-3x} \to \infty$, so the product tends to $-\infty$, which is not finite. Only one finite limit exists.

> **Exam tip:** For indeterminate forms at infinity ($\infty - \infty$, $0 \cdot \infty$, $\infty/\infty$), always rewrite the expression before concluding a finite limit exists.

## AP-Style Practice: Worked Examples

**Check your understanding**

Test your understanding with these AP-style practice problems

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

   - -3
   - -2
   - -1
   - 0

   *Why:* For $x \to \infty$, the rational term approaches $-2$ and the exponential term approaches $1$, for a total of $-1$. For $x \to -\infty$, the rational term approaches $-2$ and the exponential term approaches $0$, for a total of $-2$. Sum is $-1 + (-2) = -3$.

2. 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 $\lim_{x \to \infty} \left(2 f(x) - \frac{1}{x}\right)$.

   *Why:* Divide numerator and denominator by $x^2$; $\sqrt{4x^4 + 9} = x^2 \sqrt{4 + 9/x^4}$ for all non-zero $x$, so both limits simplify to $\frac{1}{\sqrt{4}} = \frac{1}{2}$. Applying limit laws gives $2\left(\frac{1}{2}\right) - 0 = 1$.

3. A retail analyst models cumulative product sales $t$ months after launch as $S(t) = \frac{5000 t}{0.2 t + 25}$ for $t \geq 0$. Find the horizontal asymptote of $S(t)$, and interpret its meaning in context.

   *Why:* Evaluate $\lim_{t \to \infty} S(t) = \frac{5000}{0.2} = 25000$, which is the long-run maximum total sales the market will support.

## Common pitfalls

- **Wrong:** For $f(x) = \frac{3x}{\sqrt{x^2 + 2}}$, conclude $\lim_{x \to -\infty} f(x) = 3$
  - Why it fails: Forgets that $\sqrt{x^2} = |x| = -x$ for negative $x$, so the sign is incorrect.
  - Correct: Always factor out $|x|$ from roots, substitute $|x|=-x$ for $x \to -\infty$ to get the correct sign.
- **Wrong:** Claim $y=3$ cannot be a horizontal asymptote because $f(2) = 3$ (the function crosses $y=3$ at $x=2$)
  - Why it fails: Believes the myth that functions cannot cross their asymptotes, which only applies to vertical asymptotes, not horizontal.
  - Correct: Only use the limit definition to confirm horizontal asymptotes; ignore crossings at finite $x$.
- **Wrong:** For a rational function with numerator degree 3, denominator degree 2, conclude it has a horizontal asymptote at $y=0$
  - Why it fails: Reverses the outcome of the degree rule for horizontal asymptotes.
  - Correct: Follow the degree rule explicitly: $n<m \to 0$, $n=m \to$ ratio, $n>m \to$ no finite HA.
- **Wrong:** For $f(x) = \frac{2e^x + 1}{e^x + 5}$, only find $y=2$ as a horizontal asymptote and stop
  - Why it fails: Assumes exponentials only have one horizontal asymptote and forgets to check the left end.
  - Correct: For all non-rational functions, evaluate both $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$.
- **Wrong:** Conclude $\lim_{x \to \infty} (x - \sqrt{x^2 + 3}) = 0$ because $\infty - \infty = 0$
  - Why it fails: Confuses finite number subtraction with the indeterminate form $\infty - \infty$.
  - Correct: Multiply by the conjugate to rewrite the difference as a fraction, then evaluate the limit.
- **Wrong:** Conclude $y=0$ is not a horizontal asymptote of $\frac{\sin x}{x}$ because $\lim_{x \to \infty} \sin x$ does not exist
  - Why it fails: Forgot the Squeeze Theorem applies to products of bounded functions and functions that go to zero.
  - Correct: Use the Squeeze Theorem for bounded oscillating functions to check for finite limits at infinity.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Horizontal Asymptote Definition | $y=L$ is HA if $\lim_{x \to \infty} f(x)=L$ *or* $\lim_{x \to -\infty} f(x)=L$ (L finite) | A function can have 0, 1, or 2 HAs; function can cross HA at finite $x$ |
| Rational Function: $n < m$ | $\lim_{x \to \pm \infty} f(x) = 0$ | HA at $y=0$ |
| Rational Function: $n = m$ | $\lim_{x \to \pm \infty} f(x) = \frac{\text{leading coefficient } P}{\text{leading coefficient } Q}$ | HA at $y=$ ratio of leading coefficients |
| Rational Function: $n > m$ | $\lim_{x \to \pm \infty} f(x) = \pm \infty$ | No finite horizontal asymptote |
| Exponential Limits ($k>0$) | $\lim_{x \to \infty} e^{kx} = \infty$, $\lim_{x \to -\infty} e^{kx} = 0$ | Reverse results for $k<0$; always check both ends |
| Inverse Tangent Limits | $\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}$, $\lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}$ | Two distinct HAs for $y=\arctan(x)$ |
| Logarithm Growth Rule | $\lim_{x \to \infty} \frac{\ln x}{x^p} = 0$ for all $p>0$ | Logs grow slower than any positive power of $x$ |
| Squeeze Theorem for Oscillation | If $\|g(x)\| \leq M$ (bounded) and $\lim_{x \to \infty} h(x) = 0$, then $\lim_{x \to \infty} g(x)h(x) = 0$ | Applies to $\frac{\sin x}{x}$, $\frac{\cos x}{x^2}$, etc. |

## What's next

This topic is a core building block for analyzing function end behavior, 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 these ideas to oblique asymptotes and improper integral convergence.

- [Working with the Intermediate Value Theorem (IVT)](https://www.owlsprep.com/study/ap-calculus-bc-u1-working-with-the-intermediate-value/)
- [Differentiation: Definition and Fundamental Properties](https://www.owlsprep.com/study/ap-calculus-bc-u2-overview/)
- [Defining average and instantaneous rates of change at a point](https://www.owlsprep.com/study/ap-calculus-bc-u2-defining-average-and-instantaneous-rates/)

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