Connecting limits at infinity and horizontal asymptotes

AP Calculus BC· AP Calculus BC CED — Limits and Continuity· 14 min read

1. Core Definition: The Connection Between Concepts★★☆☆☆⏱ 3 min

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.

📘 Definition

Limit at Infinity

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

$lim_{x \to \infty} f (x) = L$ 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.

📘 Definition

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.

2. Evaluating Limits at Infinity for Rational Functions★★☆☆☆⏱ 4 min

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 $k > 0$ .

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

📐 Worked Example

Find $lim_{x \to \infty} \frac{5 x ^{3} - 2 x ^{2} + 7}{- 3 x ^{3} + 10 x - 4}$ 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 $x^{2} = ∣ x ∣ = - x$ for $x < 0$ ; always check the sign to avoid errors.

3. Finding All Horizontal Asymptotes★★★☆☆⏱ 3 min

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 $f (x) = \frac{2 e ^{x} + 5}{e ^{x} - 3}$ .

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.

4. Limits at Infinity for Non-Rational Functions★★★☆☆⏱ 4 min

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 $a > 0$ : $lim_{x \to \infty} e^{- a x} = 0$ , $lim_{x \to \infty} e^{a x} = \infty$
For any $p > 0$ : $lim_{x \to \infty} \frac{l n x}{x ^{p}} = 0$ (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 $f (x) = x e^{- 3 x}$ .

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 $e^{- 3 x} \to \infty$ , 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.

5. AP-Style Practice: Worked Examples★★★★☆⏱ 4 min

✓ Quick check

Test your understanding with these AP-style practice problems

Which of the following gives the sum of all distinct horizontal asymptotes of $f (x) = \frac{4 x ^{2} + 2 x - 7}{- 2 x ^{2} + 9} + \frac{e ^{x}}{e ^{x} + 1}$ ?
- -3
- -2
- -1
- 0
Reveal answer
-3 —
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 $- 1 + (- 2) = - 3$ .
Let $f (x) = \frac{x ^{2} - 4}{4 x ^{4} + 9}$ . (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 $lim_{x \to \infty} (2 f (x) - \frac{1}{x})$ .
Reveal answer
(a) Both limits equal $\frac{1}{2}$. (b) Only horizontal asymptote is $y=\frac{1}{2}$. (c) The limit equals $1$. —
Divide numerator and denominator by $x^{2}$ ; $4 x^{4} + 9 = x^{2} 4 + 9/ x^{4}$ for all non-zero $x$ , so both limits simplify to $\frac{1}{4} = \frac{1}{2}$ . Applying limit laws gives $2 (\frac{1}{2}) - 0 = 1$ .
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.
Reveal answer
Horizontal asymptote at $y=25000$. This means total cumulative sales will approach 25,000 units in the long run, representing market saturation. —
Evaluate $lim_{t \to \infty} S (t) = \frac{5000}{0.2} = 25000$ , which is the long-run maximum total sales the market will support.

6. Common Pitfalls

Wrong move:

For $f (x) = \frac{3 x}{x ^{2} + 2}$ , conclude $lim_{x \to - \infty} f (x) = 3$

Why:

Forgets that $x^{2} = ∣ x ∣ = - x$ for negative $x$ , so the sign is incorrect.

Correct move:

Always factor out $∣ x ∣$ from roots, substitute $∣ x ∣ = - x$ for $x \to - \infty$ to get the correct sign.

Wrong move:

Claim $y = 3$ cannot be a horizontal asymptote because $f (2) = 3$ (the function crosses $y = 3$ at $x = 2$ )

Why:

Believes the myth that functions cannot cross their asymptotes, which 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, conclude it has a horizontal asymptote at $y = 0$

Why:

Reverses the outcome of the degree rule for horizontal asymptotes.

Correct move:

Follow the degree rule explicitly: $n < m \to 0$ , $n = m \to$ ratio, $n > m \to$ no finite HA.

Wrong move:

For $f (x) = \frac{2 e ^{x} + 1}{e ^{x} + 5}$ , only find $y = 2$ as a horizontal asymptote and stop

Why:

Assumes exponentials only have one horizontal asymptote and forgets 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:

Conclude $lim_{x \to \infty} (x - x^{2} + 3) = 0$ because $\infty - \infty = 0$

Why:

Confuses 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:

Conclude $y = 0$ is not a horizontal asymptote of $\frac{s i n x}{x}$ 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.

7. Quick Reference 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: $n = m$	$lim_{x \to \pm \infty} f (x) = \frac{leading coefficient P}{leading coefficient Q}$	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 ( $k > 0$ )	$lim_{x \to \infty} e^{k x} = \infty$ , $lim_{x \to - \infty} e^{k x} = 0$	Reverse results for $k < 0$ ; always check both ends
Inverse Tangent Limits	$lim_{x \to \infty} arctan (x) = \frac{π}{2}$ , $lim_{x \to - \infty} arctan (x) = - \frac{π}{2}$	Two distinct HAs for $y = arctan (x)$
Logarithm Growth Rule	$lim_{x \to \infty} \frac{l n x}{x ^{p}} = 0$ 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 $\frac{s i n x}{x}$ , $\frac{c o s x}{x ^{2}}$ , etc.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Find horizontal asymptotes of exponential function
2019 · FRQ
Interpret HA in real context

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.