Horizontal Asymptotes
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.6: “Limits at Infinity and Asymptotes” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-6-limits-at-infinity-and-asymptotes |
| Textbook used in class | Stewart, Calculus, Section 3.4: “Limits at Infinity; Horizontal Asymptotes” |
Opening Scenario
A cup of hot tea cools toward room temperature. As time goes on, the temperature gets closer and closer to $20^\circ$C but never falls below it. If you graphed temperature against time, the curve would approach the horizontal line $y = 20$ without ever crossing it. That horizontal line is a horizontal asymptote.
A horizontal asymptote is a level that the function’s output approaches -- either as the input grows without bound to the right, or as it grows without bound to the left. The function does not have to stay on one side of the line; it just must approach the line.
Quick Reference
Definition. The line $y = L$ is a horizontal asymptote of the curve $y = f(x)$ if $$\lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L.$$
A function can have at most two horizontal asymptotes: one as $x \to +\infty$ and possibly a different one as $x \to -\infty$.
a horizontal asymptote is a wall or boundary that the function cannot cross or touch.
This is the asymptote-as-wall error. A horizontal asymptote describes long-run behavior -- the value that the function’s output approaches as the input grows without bound -- not a physical barrier. The function can cross the asymptote, touch it, or even coincide with it at isolated points, as long as it eventually settles back toward the asymptotic value. A simple example: $f(x) = \dfrac{\sin x}{x}$ oscillates across the line $y = 0$ infinitely many times (every time $\sin x = 0$), yet $\lim_{x \to \infty} f(x) = 0$, so $y = 0$ is a horizontal asymptote. The asymptote is a trend, not a fence.
Key Concepts
1. How to Find Horizontal Asymptotes
Compute $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$.
- If either limit equals a finite number $L$, then $y = L$ is a horizontal asymptote.
- If a limit is $\pm \infty$, there is no horizontal asymptote in that direction.
- The two limits can be equal (same line approached from both sides) or different (two distinct asymptotes).
Example 1. Find the horizontal asymptotes of $f(x) = \dfrac{3x^2 - 1}{2x^2 + 5}$.
Divide numerator and denominator by the highest power $x^2$: $$\lim_{x \to \infty} \frac{3x^2 - 1}{2x^2 + 5} = \lim_{x \to \infty} \frac{3 - 1/x^2}{2 + 5/x^2}.$$
As $x \to \infty$, $1/x^2 \to 0$ and $5/x^2 \to 0$: $$= \frac{3 - 0}{2 + 0} = \frac{3}{2}.$$
The same computation holds as $x \to -\infty$ (since $x^2 \to \infty$ in both directions).
Boxed answer: One horizontal asymptote, $y = 3/2$, in both directions.
2. Two Different Horizontal Asymptotes
Some functions approach different levels as $x \to +\infty$ and as $x \to -\infty$.
Example 2. Find the horizontal asymptotes of $f(x) = \arctan x$.
The arctangent function is the inverse tangent. As $x$ increases without bound, $\arctan x$ approaches $\pi/2$ (the right-hand “edge” of the tangent’s range). As $x$ decreases without bound, $\arctan x$ approaches $-\pi/2$.
$$\lim_{x \to +\infty} \arctan x = \frac{\pi}{2}, \quad \lim_{x \to -\infty} \arctan x = -\frac{\pi}{2}.$$
Boxed answer: Two horizontal asymptotes: $y = \pi/2$ as $x \to +\infty$ and $y = -\pi/2$ as $x \to -\infty$.
Recap. The graph of $y = \arctan x$ is a curve that rises steadily from the lower-left, passes through the origin, and flattens out toward $\pi/2$ in the upper-right. Both of its ends approach a horizontal line.
3. No Horizontal Asymptote
If the limit is infinite or does not exist, there is no horizontal asymptote in that direction.
Example 3. Does $f(x) = e^x$ have a horizontal asymptote?
$\lim_{x \to +\infty} e^x = +\infty$: no asymptote as $x \to +\infty$.
$\lim_{x \to -\infty} e^x = 0$: horizontal asymptote $y = 0$ as $x \to -\infty$.
Boxed answer: One horizontal asymptote: $y = 0$ (approached only from the left, as $x \to -\infty$).
4. Horizontal Asymptotes in Curve Sketching
In a complete curve sketch, horizontal asymptotes are drawn as dashed horizontal lines and labeled. They show where the graph is heading as you move far to the right or far to the left.
For rational functions, the degree comparison rule (covered in the companion lesson on rational limits) gives the horizontal asymptote immediately:
- Equal degrees: HA is the ratio of leading coefficients.
- Numerator degree less: HA is $y = 0$.
- Numerator degree greater: no HA (the function grows without bound).
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Confusing HA with a line the function cannot cross | Saying $f$ must stay below its HA | The definition is about limits, not a boundary; $f$ can cross the asymptote |
| Missing the $x \to -\infty$ direction | Only checking $x \to +\infty$ | Always check both directions; they can give different asymptotes |
| Confusing HA with VA | Calling $x = 2$ a horizontal asymptote | Horizontal asymptotes are horizontal lines $y = L$; vertical asymptotes are vertical lines $x = a$ |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Find all horizontal asymptotes of $g(x) = \dfrac{5x}{x - 3}$.
Show answer
Divide by $x$: $\dfrac{5x}{x-3} = \dfrac{5}{1 - 3/x}$.
As $x \to \pm\infty$: $3/x \to 0$, so the limit is $5/1 = 5$.
Boxed answer: One horizontal asymptote, $y = 5$, in both directions.
Problem 2. Does $f(x) = x \sin(1/x)$ have a horizontal asymptote as $x \to \infty$?
Show answer
Let $t = 1/x$. As $x \to \infty$, $t \to 0$. Then $x \sin(1/x) = \sin(t)/t \to 1$.
Boxed answer: Yes, horizontal asymptote $y = 1$ as $x \to +\infty$.
Level 2 -- Multiple Steps
Problem 3. Find all horizontal asymptotes of $h(x) = \dfrac{x}{\sqrt{x^2 + 1}}$.
Show answer
As $x \to +\infty$: factor $x$ from the square root. Since $x > 0$, $\sqrt{x^2 + 1} = x\sqrt{1 + 1/x^2}$. $$\frac{x}{x\sqrt{1 + 1/x^2}} = \frac{1}{\sqrt{1 + 1/x^2}} \to \frac{1}{\sqrt{1}} = 1.$$
As $x \to -\infty$: now $x < 0$, so $\sqrt{x^2 + 1} = |x|\sqrt{1 + 1/x^2} = -x\sqrt{1 + 1/x^2}$ (since $|x| = -x$ when $x < 0$). $$\frac{x}{-x\sqrt{1 + 1/x^2}} = \frac{-1}{\sqrt{1 + 1/x^2}} \to -1.$$
Boxed answer: Two horizontal asymptotes: $y = 1$ as $x \to +\infty$ and $y = -1$ as $x \to -\infty$.
Note: the care with signs when $x < 0$ is essential. $\sqrt{x^2} = |x|$, not $x$.
Mastery Checklist
Mental Model
A horizontal asymptote is the long-run level: the height the output settles toward as the input grows large in magnitude. It answers the question “where is the function eventually headed?”
The function does not have to approach the asymptote monotonically; it can oscillate around it. What matters is that the distance between $f(x)$ and $L$ shrinks to zero as $|x| \to \infty$.
Two different long-run levels are possible because the right end and the left end of the graph can settle toward different values.
Back to Applications of Differentiation | Previous: Limits at Infinity for Rational Functions | Next: Complete Curve Sketch