How do you find a horizontal asymptote?

Find lim_x→∞f(x) and lim_x→-∞f(x). Each one that is a number L gives a horizontal asymptote y=L. For fractions of polynomials, the degree comparison gives the answer quickly.

Can a function have two horizontal asymptotes?

Yes, one on each side. Functions with square roots, like fracsqrt4x^2+1x-3, and some exponential functions approach different numbers far to the left and far to the right.

What if the degree of the top is bigger than the bottom?

Then the outputs grow without bound and there is no horizontal asymptote. If the top's degree is exactly one more, the graph approaches a slanted line instead.

Is infinity minus infinity equal to zero?

No. It's an undecided form: the answer depends on how fast each piece grows. Rewrite it, usually with the conjugate, before taking the limit.

What is the difference between a limit at infinity and an infinite limit?

A limit at infinity asks what happens as x→pm∞ and gives horizontal asymptotes. An infinite limit is when the outputs grow without bound as x approaches a number, and gives vertical asymptotes.

Limits at Infinity and Horizontal Asymptotes Explained

What does a limit at infinity mean?

Many questions are about the long run: what happens to a drug's concentration after many hours, or to the average cost per item when a factory makes millions? In maths, that's the question of what the outputs $f (x)$ do as $x$ gets larger and larger without stopping. If the outputs settle down near one number $L$ , we write

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

and read it "the limit of $f (x)$ as $x$ approaches infinity is $L$ ". Here $\infty$ is not a number you can plug in; $x \to \infty$ is shorthand for " $x$ increases without bound". In the same way, $x \to - \infty$ means $x$ decreases without bound (heads far to the left).

Definition: horizontal asymptote

The line $y = L$ is a horizontal asymptote of the graph of $f$ if $lim_{x \to \infty} f (x) =$ $L$ or $lim_{x \to - \infty} f (x) =$ $L$ : far to the right (or left), the graph gets closer and closer to that horizontal line.

y =

\frac{2 x^{2}}{x^{2} + 1}

approaches the dashed line

y = 2

far to the left and far to the right.

What is the limit of 1/x as x approaches infinity?

The limit is $0$ . As $x$ grows, $\frac{1}{x}$ becomes $\frac{1}{10}$ , $\frac{1}{1000}$ , $\frac{1}{1,000,000}$ , and so on: positive, but as close to $0$ as we like. The same happens for any number divided by any positive power of $x$ , from either direction:

The basic limit at infinity

For any number

c

and any positive power

n

lim_{x \to \infty} \frac{c}{x^{n}} =

0

and

lim_{x \to - \infty} \frac{c}{x^{n}} =

0

Why it works

As $x$ grows without bound, so does $x^{n}$ (for negative $x$ it may be negative, but its size still grows).
A fixed number $c$ divided by something whose size grows without bound gets as close to $0$ as we like.

Every other technique in this article works by rewriting an expression so that it is built from pieces like $\frac{c}{x^{n}}$ .

How do you find the limit at infinity of a fraction of polynomials?

Divide every term on the top and on the bottom by the highest power of $x$ that appears in the bottom. That doesn't change the fraction (it's the same as dividing top and bottom by the same thing), but it turns most terms into $\frac{c}{x^{n}}$ pieces that head to $0$ .

Worked example: Equal degrees

Find $lim_{x \to \infty} \frac{3 x^{2} - x}{6 x^{2} + 5}$ .

The highest power in the bottom is $x^{2}$ . Divide every term by $x^{2}$ : $\frac{\frac{3 x^{2}}{x^{2}} - \frac{x}{x^{2}}}{\frac{6 x^{2}}{x^{2}} + \frac{5}{x^{2}}} =$ $\frac{3 - \frac{1}{x}}{6 + \frac{5}{x^{2}}}$ .
As $x \to \infty$ , $\frac{1}{x} \to 0$ and $\frac{5}{x^{2}} \to 0$ .
So the limit is $\frac{3 - 0}{6 + 0} =$ $\frac{3}{6} =$ $\frac{1}{2}$ .

Answer: $\frac{1}{2}$ , so $y = \frac{1}{2}$ is a horizontal asymptote.

Worked example: Smaller degree on top

Find $lim_{x \to \infty} \frac{4 x + 1}{x^{2} - 3}$ .

Divide every term by $x^{2}$ : $\frac{\frac{4}{x} + \frac{1}{x^{2}}}{1 - \frac{3}{x^{2}}}$ .
Every piece on top heads to $0$ , and the bottom heads to $1$ .
The limit is $\frac{0 + 0}{1 - 0} = 0$ .

Answer: $0$ , so $y = 0$ is a horizontal asymptote.

Working through a few of these shows a pattern that you can use as a shortcut once you understand where it comes from. The degree of a polynomial is its highest power of $x$ , and the leading coefficient is the number multiplying that power.

Degrees	Limit as $x \to \pm \infty$	Example
top smaller than bottom	$0$	$\frac{4 x + 1}{x^{2} - 3} \to 0$
top equal to bottom	ratio of leading coefficients	$\frac{3 x^{2} - x}{6 x^{2} + 5} \to \frac{3}{6} =$ $\frac{1}{2}$
top bigger than bottom	grows without bound (no horizontal asymptote)	$\frac{x^{3}}{x + 1} \to \infty$ as $x \to \infty$

Common mistake

Dividing by the highest power on the top when it's bigger than the bottom's, then concluding the answer is $0$ . Use the bottom's highest power, or simply compare degrees with the table.

How do you handle square roots in limits at infinity?

Square roots need one extra piece of care: $\sqrt{x^{2}}$ is not always $x$ . It equals $| x |$ , the size of $x$ without its sign. When $x \to \infty$ , $x$ is positive and $| x | = x$ ; but when $x \to - \infty$ , $x$ is negative and $| x | = - x$ . That's why a function with a square root can have two different horizontal asymptotes.

Worked example: Different answers on each side

Find $lim_{x \to \infty} \frac{\sqrt{4 x^{2} + 1}}{x - 3}$ and $lim_{x \to - \infty} \frac{\sqrt{4 x^{2} + 1}}{x - 3}$ .

Take $x^{2}$ out of the root: $\sqrt{4 x^{2} + 1} =$ $\sqrt{x^{2} (4 + \frac{1}{x^{2}})} =$ $| x | \sqrt{4 + \frac{1}{x^{2}}}$ .
As $x \to \infty$ , $| x | = x$ . Divide top and bottom by $x$ : $\frac{\sqrt{4 + \frac{1}{x^{2}}}}{1 - \frac{3}{x}} \to \frac{\sqrt{4}}{1} =$ $2$ .
As $x \to - \infty$ , $| x | = - x$ , so the top is $- x \sqrt{4 + \frac{1}{x^{2}}}$ . Divide top and bottom by $x$ : $\frac{- \sqrt{4 + \frac{1}{x^{2}}}}{1 - \frac{3}{x}} \to -$ $2$ .

Answer: $2$ as $x \to \infty$ and $- 2$ as $x \to - \infty$ : two horizontal asymptotes, $y = 2$ and $y = - 2$ .

What do you do with infinity minus infinity?

Expressions like $\sqrt{x^{2} + 6 x} - x$ have two pieces that both grow without bound, giving the form $\infty - \infty$ . That form doesn't decide the answer: the difference depends on how fast each piece grows. Multiply and divide by the conjugate (the same two pieces added), which turns the difference into a fraction.

Worked example: Using the conjugate

Find $lim_{x \to \infty} (\sqrt{x^{2} + 6 x} - x)$ .

Multiply by $\frac{\sqrt{x^{2} + 6 x} + x}{\sqrt{x^{2} + 6 x} + x}$ , which equals $1$ .
The top becomes $(x^{2} + 6 x) - x^{2} = 6 x$ , using $(A - B) (A + B) = A^{2} - B^{2}$ . So the expression is $\frac{6 x}{\sqrt{x^{2} + 6 x} + x}$ .
For $x > 0$ , $\sqrt{x^{2} + 6 x} =$ $x \sqrt{1 + \frac{6}{x}}$ . Divide top and bottom by $x$ : $\frac{6}{\sqrt{1 + \frac{6}{x}} + 1}$ .
As $x \to \infty$ : $\frac{6}{\sqrt{1} + 1} =$ $\frac{6}{2} =$ $3$ .

Answer: $3$

Can a graph cross its horizontal asymptote?

Yes. A horizontal asymptote only describes the far left and far right of the graph, so in the middle the graph can cross it, even infinitely often. For example, $\frac{\sin x}{x}$ crosses the $x$ -axis every time $\sin x = 0$ , yet it still approaches $0$ as $x \to \pm \infty$ , because the top stays between $- 1$ and $1$ while the bottom grows.

y = \frac{\sin x}{x}

crosses its horizontal asymptote

y = 0

again and again, with smaller and smaller waves.

By contrast, a graph can never cross a vertical asymptote, because the function isn't defined there. Vertical asymptotes come from infinite limits, where the outputs grow without bound as $x$ approaches a number.

What about exponential functions?

Exponentials settle or explode depending on the sign of the exponent: $e^{- x} \to 0$ as $x \to \infty$ , so $y = 0$ is a horizontal asymptote of $e^{- x}$ , while $e^{x}$ grows without bound. Exponentials also grow faster than any power of $x$ , so for example $\frac{x^{5}}{e^{x}} \to 0$ as $x \to \infty$ ; L'Hôpital's rule is the tool that shows this.

Limits at infinity and horizontal asymptotes, explained