Limits at Infinity: Rules, Asymptotes & Examples

#Calculus
TL;DR
Limits at infinity describe the end behavior of a function: the value f(x)f(x) settles toward as xx runs off to +∞+\infty or −∞-\infty. When that value is a finite number LL, the line y=Ly = L is a horizontal asymptote. The one fact that unlocks the topic is lim⁡x→∞1xp=0\lim_{x \to \infty} \frac{1}{x^p} = 0 for any p>0p > 0; for a rational function you compare the degrees of the top and bottom, and the fastest way to see the answer is to divide every term by the highest power of xx.
BT
Bhanzu TeamLast updated on September 29, 202612 min read

What Are Limits at Infinity?

Limits at infinity are the values a function approaches as its input grows without bound, written lim⁡x→∞f(x)\lim_{x \to \infty} f(x) and lim⁡x→−∞f(x)\lim_{x \to -\infty} f(x). They answer one question: once xx is enormous, where does f(x)f(x) end up? This is called the end behavior of the function, and it is a statement about the far edges of the graph rather than any single point on it.

Two outcomes are possible. The limit can be a finite number LL, or it can grow without bound and be written as +∞+\infty or −∞-\infty.

lim⁡x→∞f(x)=Lorlim⁡x→∞f(x)=±∞\lim_{x \to \infty} f(x) = L \qquad \text{or} \qquad \lim_{x \to \infty} f(x) = \pm\infty

When the limit is a finite number LL, that value has a geometric name.

y=Ly = L is a horizontal asymptote of ff if lim⁡x→∞f(x)=L\lim_{x \to \infty} f(x) = L or lim⁡x→−∞f(x)=L\lim_{x \to -\infty} f(x) = L.

The two ends can behave differently, so a graph may have one horizontal asymptote on the right and a different one on the left. The single most useful fact for computing these limits is that dividing by a growing power of xx drives a term to zero. For the everyday meaning of a limit at a finite point, the limit of a function overview is the companion to this page.

What Is The Key Fact Behind Limits at Infinity?

Everything rests on one reciprocal-power rule. As xx grows, any positive power of xx in a denominator swamps the numerator, so the fraction collapses to zero.

lim⁡x→∞1xp=0for any p>0\lim_{x \to \infty} \frac{1}{x^p} = 0 \qquad \text{for any } p > 0

The same holds as x→−∞x \to -\infty whenever xpx^p is defined there. Geometrically, y=1xpy = \tfrac{1}{x^p} flattens onto the xx-axis, so y=0y = 0 is its horizontal asymptote. Every rational-function limit in this article is really this one fact applied to each term after a tidy-up step. The standard limits and limit laws pages collect the algebra rules that let you split a limit across sums, products, and quotients before this fact does the work.

How Do You Find Limits at Infinity Of A Rational Function?

For a rational function p(x)q(x)\frac{p(x)}{q(x)}, the end behavior is decided by a race between the highest power on top and the highest power on the bottom. Compare the two degrees, and there are exactly three cases.

Table: End behavior of a rational function as x→±∞x \to \pm\infty, by degree comparison.

Case

Degree comparison

lim⁡x→±∞p(x)q(x)\lim_{x \to \pm\infty} \dfrac{p(x)}{q(x)}

Asymptote

Bottom-heavy

deg⁡(p)<deg⁡(q)\deg(p) < \deg(q)

00

Horizontal y=0y = 0

Balanced

deg⁡(p)=deg⁡(q)\deg(p) = \deg(q)

ratio of leading coefficients

Horizontal y=aby = \frac{a}{b}

Top-heavy by one

deg⁡(p)=deg⁡(q)+1\deg(p) = \deg(q) + 1

±∞\pm\infty

Slant (oblique) asymptote

In the balanced case, if p(x)p(x) has leading term axnax^n and q(x)q(x) has leading term bxnbx^n, the limit is ab\frac{a}{b}. In the top-heavy-by-one case the function has no horizontal asymptote, but polynomial long division reveals a straight slant asymptote it hugs at each end. This connects directly to rational functions and their graphs.

How Do You Divide By The Highest Power Of x?

The reliable technique is to divide every term in the numerator and denominator by the highest power of xx that appears in the denominator, then send each reciprocal-power term to zero. This turns a messy fraction into a limit you can read off.

Take the balanced example 3x2+12x2−x\dfrac{3x^2 + 1}{2x^2 - x}. The highest power on the bottom is x2x^2, so divide every term by x2x^2:

lim⁡x→∞3x2+12x2−x=lim⁡x→∞3+1x22−1x\lim_{x \to \infty} \frac{3x^2 + 1}{2x^2 - x} = \lim_{x \to \infty} \frac{3 + \dfrac{1}{x^2}}{2 - \dfrac{1}{x}}

Now apply the key fact: 1x2→0\frac{1}{x^2} \to 0 and 1x→0\frac{1}{x} \to 0, leaving

lim⁡x→∞3+02−0=32.\lim_{x \to \infty} \frac{3 + 0}{2 - 0} = \frac{3}{2}.

The leading coefficients were 33 and 22 all along, which is exactly what the balanced case promised. Dividing by the highest power is just a bookkeeping trick that makes the reciprocal-power fact do the deciding.

What Are Some Worked Examples Of Limits at Infinity?

Each example is fully stepped, and every answer is checked against the degree rule that predicts it.

Example 1: A balanced rational function.

Evaluate lim⁡x→∞3x2+12x2−x\displaystyle\lim_{x \to \infty} \frac{3x^2 + 1}{2x^2 - x}.

Divide numerator and denominator by x2x^2:

lim⁡x→∞3+1x22−1x=3+02−0=32\lim_{x \to \infty} \frac{3 + \frac{1}{x^2}}{2 - \frac{1}{x}} = \frac{3 + 0}{2 - 0} = \frac{3}{2}

The degrees are equal, so the rule predicts the ratio of leading coefficients, 32\frac{3}{2}. The two agree.

Final answer: lim⁡x→∞3x2+12x2−x=32\displaystyle\lim_{x \to \infty} \frac{3x^2 + 1}{2x^2 - x} = \frac{3}{2}, so y=32y = \frac{3}{2} is a horizontal asymptote.

Example 2: A bottom-heavy rational function.

Evaluate lim⁡x→∞xx2+1\displaystyle\lim_{x \to \infty} \frac{x}{x^2 + 1}.

Divide top and bottom by x2x^2:

lim⁡x→∞1x1+1x2=01+0=0\lim_{x \to \infty} \frac{\frac{1}{x}}{1 + \frac{1}{x^2}} = \frac{0}{1 + 0} = 0

The numerator's degree (11) is less than the denominator's (22), so the rule predicts 00.

Final answer: lim⁡x→∞xx2+1=0\displaystyle\lim_{x \to \infty} \frac{x}{x^2 + 1} = 0, so y=0y = 0 is the horizontal asymptote.

Example 3: A top-heavy rational function.

Evaluate lim⁡x→∞x2x+1\displaystyle\lim_{x \to \infty} \frac{x^2}{x + 1}.

Divide top and bottom by xx, the highest power on the bottom:

lim⁡x→∞x1+1x=∞1+0=∞\lim_{x \to \infty} \frac{x}{1 + \frac{1}{x}} = \frac{\infty}{1 + 0} = \infty

The numerator's degree is one higher, so the function grows without bound. Polynomial long division writes x2x+1=x−1+1x+1\frac{x^2}{x+1} = x - 1 + \frac{1}{x+1}, so the graph hugs the slant asymptote y=x−1y = x - 1 at each end while the leftover 1x+1\frac{1}{x+1} fades to zero.

Final answer: lim⁡x→∞x2x+1=∞\displaystyle\lim_{x \to \infty} \frac{x^2}{x + 1} = \infty; there is no horizontal asymptote, but a slant asymptote y=x−1y = x - 1.

Example 4: A square root, where the sign matters.

Evaluate lim⁡x→∞x2+1x\displaystyle\lim_{x \to \infty} \frac{\sqrt{x^2 + 1}}{x} and lim⁡x→−∞x2+1x\displaystyle\lim_{x \to -\infty} \frac{\sqrt{x^2 + 1}}{x}.

Factor x2x^2 out of the root: x2+1=x2,1+1x2=∣x∣1+1x2\sqrt{x^2 + 1} = \sqrt{x^2},\sqrt{1 + \frac{1}{x^2}} = |x|\sqrt{1 + \frac{1}{x^2}}. The absolute value is the whole story. As x→∞x \to \infty, xx is positive so ∣x∣=x|x| = x:

lim⁡x→∞∣x∣1+1x2x=lim⁡x→∞x1+1x2x=1+0=1\lim_{x \to \infty} \frac{|x|\sqrt{1 + \frac{1}{x^2}}}{x} = \lim_{x \to \infty} \frac{x\sqrt{1 + \frac{1}{x^2}}}{x} = \sqrt{1 + 0} = 1

As x→−∞x \to -\infty, xx is negative so ∣x∣=−x|x| = -x, which flips the sign:

lim⁡x→−∞∣x∣1+1x2x=lim⁡x→−∞−x1+1x2x=−1+0=−1\lim_{x \to -\infty} \frac{|x|\sqrt{1 + \frac{1}{x^2}}}{x} = \lim_{x \to -\infty} \frac{-x\sqrt{1 + \frac{1}{x^2}}}{x} = -\sqrt{1 + 0} = -1

Final answer: the limit is 11 as x→∞x \to \infty and −1-1 as x→−∞x \to -\infty. The graph has two horizontal asymptotes, y=1y = 1 on the right and y=−1y = -1 on the left.

Which Functions Grow Fastest At Infinity?

Some functions run to infinity far faster than others, and knowing the pecking order lets you read off many limits without any algebra. For large xx, with constants p>0p > 0 and a>1a > 1, the standard ranking is:

ln⁡x;≪;xp;≪;ax;≪;x factorial\ln x ;\ll; x^p ;\ll; a^x ;\ll; x \text{ factorial}

where ≪\ll means "grows much slower than," and xx factorial means the product x(x−1)(x−2)⋯(2)(1)x(x-1)(x-2)\cdots(2)(1) for whole-number inputs.

  • Logarithms crawl. ln⁡x\ln x heads to infinity, but slower than any positive power of xx, so lim⁡x→∞ln⁡xx=0\lim_{x \to \infty} \frac{\ln x}{x} = 0.

  • Powers are the middle gear. Any xpx^p beats every logarithm but loses to every exponential with base above 11.

  • Exponentials sprint. axa^x with a>1a > 1 outruns every power, so lim⁡x→∞x1002x=0\lim_{x \to \infty} \frac{x^{100}}{2^x} = 0.

  • Factorials outrun them all. For whole-number growth, xx factorial eventually overtakes any fixed exponential.

Whenever a limit pits two of these against each other in a quotient, the faster-growing one wins: it dominates the fraction, sending it to 00 or to ±∞\pm\infty depending on which side it sits. Some of these standoffs produce an indeterminate form such as ∞∞\frac{\infty}{\infty}, which the hierarchy or a tool like L'Hopital's rule then resolves.

Why Do Limits at Infinity Work?

The idea feels abstract until you picture the graph running off the page. Then it becomes concrete.

  • End behavior is a long-run trend. A limit at infinity ignores the wiggles near the origin and reports only where the curve is heading once xx is huge. That is why two functions can look completely different up close yet share a horizontal asymptote.

  • A horizontal asymptote is a target, not a wall. The line y=Ly = L is the height the curve settles toward, but the graph is allowed to touch or cross it. A curve like sin⁡xx+1\frac{\sin x}{x} + 1 crosses its asymptote y=1y = 1 infinitely often while still converging to it, because the gap keeps shrinking toward zero.

  • Dividing by the highest power exposes the dominant term. Every term except the leading ones carries a reciprocal power of xx that vanishes. What survives is the race between the fastest-growing pieces, which is precisely what "end behavior" means.

Read this way, the degree rules are not three rules to memorize. They are one idea: the highest power wins, and everything slower fades out of the picture as xx grows.

Who Shaped Limits at Infinity?

Infinity was argued over for centuries before anyone pinned it down. The symbol came first, and rigor came much later.

Two named figures anchor the rigor:

  • Augustin-Louis Cauchy (1789–1857, France) rebuilt calculus around a careful notion of limit, insisting that convergence be argued, not assumed.

  • Karl Weierstrass (1815–1897, Germany) supplied the quantifier definition of a limit that underpins the modern treatment of end behavior and asymptotes.

Where Are Limits at Infinity Used In The Real World?

End behavior answers "what happens in the long run," which is a question almost every field asks.

  • Physics: the terminal velocity of a falling object is a limit at infinity of its velocity function, the steady speed it approaches once air resistance balances gravity.

  • Biology and medicine: population and dose-response models level off at a carrying capacity or saturation value, the horizontal asymptote of a growth curve.

  • Engineering and control: the steady-state response of a circuit or a controller is the limit of its output as time runs on, telling designers where a system finally settles.

  • Economics: long-run average cost approaches a floor as output grows, and diminishing-returns models flatten toward a ceiling.

  • Computer science: the growth rate of an algorithm's running time is exactly the end behavior of a function, which is why the same degree-and-dominance thinking drives big-O analysis.

One idea, "where is this heading once the inputs are large," unifies terminal velocity, saturation, steady state, and algorithm speed. Reading a graph's far edges is reading its future.

What Are The Most Common Mistakes With Limits at Infinity?

These three errors account for most lost marks on limits at infinity, and each matches a question real students ask on r/learnmath and r/calculus and in course error handouts.

Dropping the sign of x2=∣x∣\sqrt{x^2} = |x| as x→−∞x \to -\infty.

Where it slips in:

A student simplifies x2+1\sqrt{x^2 + 1} to x1+1x2x\sqrt{1 + \frac{1}{x^2}} and gets the limit 11 at both ends, missing that the left end is −1-1.

Don't do this:

Do not replace x2\sqrt{x^2} with xx. The square root of a square is the absolute value, x2=∣x∣\sqrt{x^2} = |x|.

The correct way:

Write x2=∣x∣\sqrt{x^2} = |x|, then split by direction: ∣x∣=x|x| = x as x→∞x \to \infty and ∣x∣=−x|x| = -x as x→−∞x \to -\infty. That sign flip is what turns the left-hand limit into −1-1.

Treating ∞∞\frac{\infty}{\infty} as something you can plug in.

Where it slips in:

A student substitutes x=∞x = \infty into 3x2+12x2−x\frac{3x^2 + 1}{2x^2 - x}, writes ∞∞\frac{\infty}{\infty}, and either calls it 11 or calls it undefined and stops.

Don't do this:

Do not read ∞∞\frac{\infty}{\infty} as a number. It is an indeterminate form, and it can equal any value depending on the two rates.

The correct way:

Resolve it. Divide by the highest power of xx (or use the growth hierarchy), which for this fraction gives 32\frac{3}{2}, not 11.

Believing a graph can never cross its horizontal asymptote.

Where it slips in:

A student rules out an answer because the curve touches or crosses the line y=Ly = L, assuming an asymptote is a barrier.

Don't do this:

Do not treat a horizontal asymptote as unreachable. Only vertical asymptotes are never crossed; a horizontal one can be crossed, even infinitely often.

The correct way:

Judge by the limit alone. If lim⁡x→∞f(x)=L\lim_{x \to \infty} f(x) = L, then y=Ly = L is a horizontal asymptote no matter how many times the graph meets it on the way, as sin⁡xx+1\frac{\sin x}{x} + 1 does around y=1y = 1.

Practice Problems On Limits at Infinity

Work each one, then check against the answer. Answers are verified by the degree rule or the reciprocal-power fact.

  1. Evaluate lim⁡x→∞5x3−2x4x3+x2+7\displaystyle\lim_{x \to \infty} \frac{5x^3 - 2x}{4x^3 + x^2 + 7}.
    (Answer: equal degrees, so the ratio of leading coefficients, 54\frac{5}{4}.)

  2. Evaluate lim⁡x→∞2x+3x2−5\displaystyle\lim_{x \to \infty} \frac{2x + 3}{x^2 - 5}.
    (Answer: bottom-heavy, so 00.)

  3. Evaluate lim⁡x→∞4x2−1x+6\displaystyle\lim_{x \to \infty} \frac{4x^2 - 1}{x + 6}.
    (Answer: top-heavy by one, so +∞+\infty; slant asymptote y=4x−24y = 4x - 24.)

  4. Evaluate lim⁡x→−∞9x2+4x\displaystyle\lim_{x \to -\infty} \frac{\sqrt{9x^2 + 4}}{x}.
    (Answer: 9x2=3∣x∣=−3x\sqrt{9x^2} = 3|x| = -3x for x<0x < 0, so the limit is −3-3.)

  5. Evaluate lim⁡x→∞ln⁡xx\displaystyle\lim_{x \to \infty} \frac{\ln x}{x}.
    (Answer: ln⁡x\ln x grows slower than xx, so 00.)

  6. Evaluate lim⁡x→∞x5ex\displaystyle\lim_{x \to \infty} \frac{x^5}{e^x}.
    (Answer: the exponential outruns the power, so 00.)

Where Should You Go Next After Limits at Infinity?

Limits at infinity sit at the crossroads of limits and graphing, and several natural doors open from here.

  1. Curve sketching. End behavior and asymptotes are two of the ingredients that let you draw a full graph from its equation.

  2. Indeterminate forms. Learn how to resolve ∞∞\frac{\infty}{\infty} and 00\frac{0}{0} when the growth hierarchy alone does not settle a limit.

  3. Limit formulas. Collect the standard results that speed up limits at infinity and elsewhere.

If your child is meeting limits at infinity for the first time, a live Bhanzu trainer teaches it from the end-behavior picture up, so asymptotes and degree rules feel like one idea, in the Bhanzu math program.

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Frequently Asked Questions

What are limits at infinity in simple terms?
They describe the end behavior of a function: the value f(x)f(x) approaches as xx grows without bound toward +∞+\infty or −∞-\infty. If that value is a finite number LL, the line y=Ly = L is a horizontal asymptote of the graph.
How do limits at infinity relate to horizontal asymptotes?
They are two names for the same fact. If lim⁡x→∞f(x)=L\lim_{x \to \infty} f(x) = L or lim⁡x→−∞f(x)=L\lim_{x \to -\infty} f(x) = L with LL finite, then y=Ly = L is a horizontal asymptote. The left and right ends can give different values, so a function can have up to two horizontal asymptotes.
How do you find the limit at infinity of a rational function?
Compare the degree of the numerator with the degree of the denominator. If the bottom wins the limit is 00; if the degrees are equal the limit is the ratio of leading coefficients; if the top wins by one the function goes to ±∞\pm\infty with a slant asymptote. Dividing every term by the highest power of xx shows this quickly.
Why does the answer change sign at negative infinity for square roots?
Because x2=∣x∣\sqrt{x^2} = |x|, not xx. As x→∞x \to \infty, ∣x∣=x|x| = x; but as x→−∞x \to -\infty, ∣x∣=−x|x| = -x, which introduces a negative sign and can flip the limit, for example from 11 to −1-1.
Can a graph cross its horizontal asymptote?
Yes. A horizontal asymptote describes where the curve is heading, not a line it is forbidden to touch. Functions such as sin⁡xx+1\frac{\sin x}{x} + 1 cross y=1y = 1 infinitely often while still converging to it.
What does an infinite limit at infinity mean?
It means the function grows without bound as xx grows, written lim⁡x→∞f(x)=∞\lim_{x \to \infty} f(x) = \infty. There is no horizontal asymptote in that direction, though a top-heavy-by-one rational function still has a slant asymptote it follows.
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