A limit at infinity describes a function’s end behavior: what happens to \( f(x) \) as \( x \) grows without bound. If \( \lim_{x\to\infty}f(x) = L \), the line \( y = L \) is a horizontal asymptote.
This guide gives you the one fact every method rests on, the dividing technique for rational functions, the degree shortcut, radicals and the sign trap at negative infinity, exponentials, and the mistakes that cost the most points on exams.
Why it works: only the biggest term matters
Picture \( f(x) = 3x^2 + 1 \) at \( x = 1000 \). The \( 3x^2 \) part is three million, while the \( +1 \) is a rounding error. As \( x \) keeps growing, smaller terms become an ever tinier fraction of the total. So for large \( x \), a polynomial behaves like its leading term.
A ratio of two such expressions then behaves like the ratio of the leading terms. Every technique below is a careful way of saying “throw away what doesn’t matter,” while making sure what you throw away really does go to zero.
Graphically, a limit at infinity asks where the curve settles as you scroll far to the right (or left). If it flattens toward a height \( L \), you get a horizontal asymptote. If it keeps climbing, the limit is \( \infty \). If it keeps wiggling without settling, the limit does not exist.
The basic building block
$$\lim_{x\to\infty}\frac{1}{x^p} = 0 \quad\text{for any } p > 0$$
Every technique below reduces a problem to this fact. Why is it true? Pick any small tolerance, say 0.001. Once \( x^p > 1000 \), the fraction is below 0.001, and it stays below from then on. The same works for every tolerance, which is exactly what a limit of 0 means.
Rational functions: divide by the highest power
The method: find the highest power of \( x \) in the denominator, divide every term on top and bottom by it, then let each \( \frac{c}{x^p} \) go to 0.
Example 1. \( \lim_{x\to\infty}\frac{3x^2 + 1}{5x^2 - x} \). Divide top and bottom by \( x^2 \):
$$\lim_{x\to\infty}\frac{3 + \frac{1}{x^2}}{5 - \frac1x} = \frac{3}{5}$$
The horizontal asymptote is \( y = \frac35 \): the ratio of the leading coefficients.
Example 2. \( \lim_{x\to\infty}\frac{2x + 1}{x^2 + 4} = \lim\frac{\frac2x + \frac{1}{x^2}}{1 + \frac4{x^2}} = 0 \)
The denominator grows faster, so the fraction shrinks to 0.
Example 3. \( \lim_{x\to\infty}\frac{x^3 + 1}{2x^2} = \infty \): the top wins. Dividing by \( x^2 \) leaves \( \frac{x + 1/x^2}{2} \), and \( x \) alone grows without bound.
The degree shortcut
For \( \frac{P(x)}{Q(x)} \) with leading terms \( ax^n \) and \( bx^m \):
| Degrees | Limit as \( x\to\pm\infty \) | Asymptote |
|---|---|---|
| \( n < m \) | \( 0 \) | \( y = 0 \) |
| \( n = m \) | \( \frac{a}{b} \) | \( y = \frac ab \) |
| \( n > m \) | \( \pm\infty \) | none (maybe slant) |
When \( n > m \), the sign of the infinity depends on the signs of \( a \) and \( b \) and, at \( -\infty \), on whether \( n - m \) is odd or even. Look at the simplified leading-term ratio to decide.
Example 4 (slant asymptote). \( f(x) = \frac{x^2 + 1}{x - 1} \) has degree 2 over degree 1, so the limit at infinity is \( \infty \). Long division gives more detail:
$$\frac{x^2 + 1}{x - 1} = x + 1 + \frac{2}{x - 1}$$
The leftover fraction goes to 0, so the graph hugs the line \( y = x + 1 \) for large \( x \). That line is a slant (oblique) asymptote. It appears exactly when the top’s degree is one more than the bottom’s.
Radicals: watch the sign at −∞
Example 5. \( \lim_{x\to\infty}\frac{\sqrt{4x^2 + 1}}{x + 3} \). For \( x > 0 \), \( \sqrt{x^2} = x \), so dividing by \( x \) gives \( \frac{\sqrt{4 + 1/x^2}}{1 + 3/x} \to 2 \).
As \( x \to -\infty \), \( \sqrt{x^2} = |x| = -x \), and the same limit is \( -2 \). Two different horizontal asymptotes!
The reason: a square root is never negative, but \( x + 3 \) is very negative far to the left. The top stays positive and the bottom turns negative, so the ratio must be negative.
Example 6 (∞ − ∞). \( \lim_{x\to\infty}\left(\sqrt{x^2 + x} - x\right) \). Both pieces blow up, and their difference is an indeterminate form. Multiply by the conjugate over itself:
$$\begin{aligned} \sqrt{x^2 + x} - x &= \frac{(x^2 + x) - x^2}{\sqrt{x^2 + x} + x} \\ &= \frac{x}{\sqrt{x^2+x} + x} \\ &= \frac{1}{\sqrt{1 + \frac1x} + 1} \to \frac12 \end{aligned}$$
The last step divides top and bottom by \( x \). Notice the answer is not 0: the two huge quantities differ by an amount that settles near one half.
Growth rate ranking
For large \( x \), from slowest to fastest:
$$\ln x \ll x^{p} \ll e^{x} \ll x! \ll x^{x}$$
So \( \lim\frac{\ln x}{x} = 0 \) and \( \lim\frac{x^{100}}{e^x} = 0 \). These can be proved with L’Hôpital’s rule, which applies to \( \frac{\infty}{\infty} \) forms at infinity just as it does at a finite point.
Exponentials: two different ends
Example 7. \( f(x) = \frac{2e^x + 3}{e^x - 1} \).
As \( x \to \infty \), \( e^x \) is the dominant term, so divide top and bottom by it:
$$\frac{2 + 3e^{-x}}{1 - e^{-x}} \to \frac{2}{1} = 2$$
As \( x \to -\infty \), \( e^x \to 0 \) instead, so just substitute: \( \frac{0 + 3}{0 - 1} = -3 \).
The graph has horizontal asymptote \( y = 2 \) on the right and \( y = -3 \) on the left. With exponentials, always check both ends separately, because \( e^x \) behaves completely differently in each direction.
Transcendental functions
- \( \lim_{x\to\infty}e^{-x} = 0 \)
- \( \lim_{x\to\infty}\arctan x = \frac{\pi}{2} \)
- \( \lim_{x\to\infty}\sin x \) does not exist (it oscillates), but \( \lim_{x\to\infty}\frac{\sin x}{x} = 0 \) by the squeeze theorem.
Common mistakes
- Dividing by the wrong power. Divide by the highest power in the denominator. Dividing by a smaller power leaves terms that still blow up, and you are no closer to an answer.
- Forgetting \( \sqrt{x^2} = |x| \). At \( -\infty \), pulling \( x \) out of a square root introduces a minus sign. Skipping it turns \( -2 \) into \( 2 \) in Example 5.
- Treating \( \infty - \infty \) as 0. Two infinite quantities can differ by anything. Rewrite with a conjugate or a common denominator first, as in Example 6.
- Assuming both ends match. Rational functions give the same finite limit at both ends, but radicals and exponentials often don’t. Check \( +\infty \) and \( -\infty \) separately.
- Confusing the two kinds of asymptote. A vertical asymptote comes from a limit at a finite point being infinite. A horizontal asymptote comes from a limit at infinity being finite.
Where it’s used
Limits at infinity give you the end behavior you need when sketching curves, alongside critical points and concavity. They decide whether improper integrals over infinite intervals converge. Series tests such as the ratio test boil down to a limit as \( n \to \infty \). In applications, they describe long-run values: the carrying capacity of a logistic population model, the steady-state concentration in a mixing tank, or the terminal velocity of a falling object.
Limit calculator (use inf)
Type inf or -inf as the point to evaluate a limit at infinity. For a full page version, open the limit calculator, and use the graphing calculator to see the asymptote.
Practice problems
Try these before checking the answers.
- \( \lim_{x\to\infty}\frac{4x^3 - x}{2x^3 + 7} \)
- \( \lim_{x\to-\infty}\frac{x}{\sqrt{x^2 + 1}} \)
- \( \lim_{x\to\infty}\frac{\ln x}{\sqrt x} \)
- \( \lim_{x\to\infty}\frac{5 - 2x^2}{3x^2 + x} \)
- \( \lim_{x\to-\infty}\frac{x^2 - 3}{x + 1} \)
- \( \lim_{x\to\infty}\left(\sqrt{x^2 + 4x} - x\right) \)
- \( \lim_{x\to\infty}xe^{-x} \)
Answers: (1) \( 2 \); (2) \( -1 \); (3) \( 0 \); (4) \( -\frac23 \); (5) \( -\infty \), since the quotient behaves like \( x \); (6) \( 2 \), using the conjugate; (7) \( 0 \), because \( e^x \) outgrows \( x \).
FAQ
Can a graph cross its horizontal asymptote?
Yes. \( \frac{\sin x}{x} \) crosses \( y = 0 \) infinitely often. Asymptotes describe end behavior only.
Is “the limit is ∞” the same as “the limit exists”?
Not strictly. An infinite limit describes how the function behaves, but the limit is not a real number, so technically it does not exist.
How do I find the horizontal asymptotes of a function?
Compute \( \lim_{x\to\infty}f(x) \) and \( \lim_{x\to-\infty}f(x) \). Each finite answer \( L \) gives an asymptote \( y = L \). A function can have zero, one or two horizontal asymptotes.
Can I use L’Hôpital’s rule for limits at infinity?
Yes, whenever the limit has the form \( \frac{\infty}{\infty} \) or \( \frac00 \). For rational functions, dividing by the highest power is usually faster.
What is the limit of a constant at infinity?
The constant itself. A function like \( f(x) = 7 \) never changes, so its limit at infinity is 7 and its graph is its own horizontal asymptote.
Further reading
- Limits at Infinity, Part I (Paul’s Online Math Notes) — polynomials, rational functions and radicals, with many more worked examples.
- Limits at Infinity, Part II (Paul’s Online Math Notes) — exponentials, logarithms and inverse tangent at both ends.
- Limits at Infinity and Asymptotes (OpenStax Calculus Volume 1) — the textbook treatment, including oblique asymptotes and curve sketching.
Calculators for this topic
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Squeeze Theorem: Statement, How to Use It and Examples
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L’Hôpital’s Rule: When and How to Use It (with Examples)
L’Hôpital’s rule evaluates 0/0 and ∞/∞ limits by differentiating top and bottom. Conditions, 6 worked examples, other indeterminate forms and common errors.