# Limits at Infinity: End Behavior and Horizontal Asymptotes

> Learn how to find limits at infinity: the degree rules for rational functions, horizontal asymptotes, square roots, and worked examples with graphs.

Source: https://practicecalculus.com/blog/limits-at-infinity
Updated: 2026-09-14

## Key takeaways

- A limit at infinity describes what $f(x)$ does as $x$ grows without bound, which is the end behavior of the graph.
- For rational functions, compare degrees: bottom bigger gives $0$, equal degrees give the ratio of leading coefficients, top bigger gives no finite limit.
- If $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$, then $y = L$ is a horizontal asymptote.
- The quickest method is to divide every term by the highest power of $x$ in the denominator.
- With square roots, remember $\sqrt{x^2} = |x|$, so the limits at $+\infty$ and $-\infty$ can differ.

What happens to a function far out on the graph, past the edge of your calculator screen? Does it level off, shoot upward, or keep wobbling? **Limits at infinity** answer that question, and for rational functions there is a simple set of rules that settles almost every problem in seconds.

## What is a limit at infinity?

A **limit at infinity** describes the value a function approaches as $x$ grows without bound. We write

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

to mean that when $x$ is large enough, $f(x)$ stays as close to $L$ as you like. Similarly, $\lim_{x \to -\infty} f(x)$ describes what happens as $x$ becomes very negative.

Together these two limits describe the **end behavior** of a function: what the graph does at its far left and far right.

There are three possible outcomes:

- The outputs approach a number $L$. Then the line $y = L$ is a **horizontal asymptote**.
- The outputs grow without bound, and we write $\infty$ or $-\infty$.
- The outputs never settle, like $\sin x$, which keeps oscillating. Then the limit does not exist.

## The building block: 1 over x to a power

Everything below rests on one fact. For any positive power $p$,

$$\lim_{x \to \infty} \frac{1}{x^p} = 0$$

Dividing a fixed number by an enormous number gives something tiny. $\frac{1}{1000}$ is small, $\frac{1}{1000^2}$ is smaller still. The same holds as $x \to -\infty$ whenever $x^p$ is defined there.

More generally, $\frac{c}{x^p} \to 0$ for any constant $c$. The strategy for complicated expressions is to rewrite them so every leftover piece looks like $\frac{c}{x^p}$.

## The method: divide by the highest power in the denominator

For a rational function, divide every term in the numerator and denominator by the highest power of $x$ that appears in the **denominator**. Then let each $\frac{c}{x^p}$ term go to $0$.

### Worked example 1: equal degrees

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

**Step 1. Find the highest power in the denominator.** It is $x^2$.

**Step 2. Divide every term by $x^2$.**

$$\frac{2x^2 - 3}{x^2 + 1} = \frac{2 - \frac{3}{x^2}}{1 + \frac{1}{x^2}}$$

**Step 3. Let the small terms vanish.** As $x \to \infty$, $\frac{3}{x^2} \to 0$ and $\frac{1}{x^2} \to 0$:

$$\lim_{x \to \infty} \frac{2x^2 - 3}{x^2 + 1} = \frac{2 - 0}{1 + 0} = 2$$

The same steps give $2$ as $x \to -\infty$, so $y = 2$ is a horizontal asymptote on both ends.

```graph
title: 'Graph of y = (2x² − 3)/(x² + 1) leveling off toward the horizontal asymptote y = 2'
caption: 'As $x \to \pm\infty$ the curve approaches $y = 2$, the ratio of the leading coefficients $\frac{2}{1}$.'
x: [-10, 10]
y: [-4, 3]
functions:
  - expr: (2x^2-3)/(x^2+1)
    label: y = (2x² − 3)/(x² + 1)
  - expr: '2'
    label: asymptote y = 2
    dashed: true
points:
  - [0, -3, '(0, −3)']
```

### Worked example 2: bottom-heavy

Find $\lim_{x \to \infty} \frac{5x + 2}{x^2 + 1}$.

**Step 1.** The highest power in the denominator is $x^2$.

**Step 2. Divide every term by $x^2$.**

$$\frac{5x + 2}{x^2 + 1} = \frac{\frac{5}{x} + \frac{2}{x^2}}{1 + \frac{1}{x^2}}$$

**Step 3. Take the limit.** The top goes to $0 + 0 = 0$ and the bottom to $1$:

$$\lim_{x \to \infty} \frac{5x + 2}{x^2 + 1} = \frac{0}{1} = 0$$

So $y = 0$ is a horizontal asymptote. Notice this graph actually crosses $y = 0$ at $x = -\frac{2}{5}$. That is allowed: an asymptote only describes the far ends.

### Worked example 3: top-heavy

Find $\lim_{x \to \infty} \frac{x^3 - 1}{2x + 5}$.

**Step 1.** The highest power in the denominator is $x$.

**Step 2. Divide every term by $x$.**

$$\frac{x^3 - 1}{2x + 5} = \frac{x^2 - \frac{1}{x}}{2 + \frac{5}{x}}$$

**Step 3. Take the limit.** The bottom goes to $2$, but the top contains $x^2$, which grows without bound:

$$\lim_{x \to \infty} \frac{x^3 - 1}{2x + 5} = \infty$$

There is no horizontal asymptote. For large $x$ the function behaves like $\frac{x^3}{2x} = \frac{x^2}{2}$, a parabola.

## The degree rules for rational functions

Once you have done a few examples, you will notice that only the **leading terms** matter. For

$$f(x) = \frac{a x^n + \dots}{b x^m + \dots}$$

with $n$ the degree of the top and $m$ the degree of the bottom:

| Degrees | $\lim_{x \to \pm\infty} f(x)$ | Horizontal asymptote |
| --- | --- | --- |
| $n < m$ (bottom bigger) | $0$ | $y = 0$ |
| $n = m$ (equal) | $\frac{a}{b}$ | $y = \frac{a}{b}$ |
| $n > m$ (top bigger) | $\pm\infty$ | none |

When $n > m$, the sign of the infinity comes from the leading terms: $\frac{a x^n}{b x^m} = \frac{a}{b} x^{n-m}$. Check the sign of that for large positive or large negative $x$.

> **Tip:** if $n = m + 1$, the graph has a slant (oblique) asymptote instead. Polynomial division of the top by the bottom gives the line.

## Square roots: why the two ends can differ

Rational functions always have the same horizontal asymptote on both ends. Square roots can break that symmetry, because

$$\sqrt{x^2} = |x|$$

which equals $x$ when $x > 0$ but $-x$ when $x < 0$.

### Worked example 4: a root in the numerator

Find both $\lim_{x \to \infty}$ and $\lim_{x \to -\infty}$ of $f(x) = \frac{\sqrt{4x^2 + 1}}{x + 3}$.

**Step 1. Factor $x^2$ out of the root.**

$$\sqrt{4x^2 + 1} = \sqrt{x^2\left(4 + \frac{1}{x^2}\right)} = |x|\sqrt{4 + \frac{1}{x^2}}$$

**Step 2. As $x \to \infty$, $|x| = x$.** Divide top and bottom by $x$:

$$\frac{x\sqrt{4 + \frac{1}{x^2}}}{x + 3} = \frac{\sqrt{4 + \frac{1}{x^2}}}{1 + \frac{3}{x}} \to \frac{\sqrt{4}}{1} = 2$$

**Step 3. As $x \to -\infty$, $|x| = -x$.** Now the same division gives an extra minus sign:

$$\frac{-x\sqrt{4 + \frac{1}{x^2}}}{x + 3} = \frac{-\sqrt{4 + \frac{1}{x^2}}}{1 + \frac{3}{x}} \to -2$$

**Step 4. State the asymptotes.** $y = 2$ on the right and $y = -2$ on the left. (There is also a vertical asymptote at $x = -3$, where the denominator is $0$.)

```graph
title: 'Graph of y = √(4x² + 1)/(x + 3) with horizontal asymptotes y = 2 on the right and y = −2 on the left'
caption: 'Because $\sqrt{x^2} = |x|$, the limit is $2$ as $x \to \infty$ but $-2$ as $x \to -\infty$. The dashed vertical line is the asymptote $x = -3$.'
x: [-30, 30]
y: [-8, 8]
functions:
  - expr: sqrt(4x^2+1)/(x+3)
    label: y = √(4x² + 1)/(x + 3)
  - expr: '2'
    label: y = 2
    dashed: true
  - expr: '-2'
    label: y = −2
    dashed: true
vertical:
  - {x: -3, label: 'x = −3'}
```

## Infinity minus infinity: use the conjugate

Some limits at infinity are not fractions at all. A difference of two growing terms gives the form $\infty - \infty$, which is indeterminate: the answer depends on which term grows faster, and by how much.

### Worked example 5: a difference with a root

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

**Step 1. Identify the form.** Both $\sqrt{x^2 + x}$ and $x$ grow without bound, so this is $\infty - \infty$. You cannot conclude anything yet.

**Step 2. Multiply by the conjugate over itself.**

$$\left(\sqrt{x^2 + x} - x\right) \cdot \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x} = \frac{(x^2 + x) - x^2}{\sqrt{x^2 + x} + x} = \frac{x}{\sqrt{x^2 + x} + x}$$

**Step 3. Divide top and bottom by $x$.** For $x > 0$, $\frac{\sqrt{x^2 + x}}{x} = \sqrt{1 + \frac{1}{x}}$, so

$$\frac{x}{\sqrt{x^2 + x} + x} = \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}$$

**Step 4. Take the limit.**

$$\lim_{x \to \infty} \left(\sqrt{x^2 + x} - x\right) = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}$$

So the two terms do not cancel to $0$. The root is always about $\frac{1}{2}$ ahead of $x$.

## Limits at infinity vs infinite limits

These two phrases sound alike but describe different things, and mixing them up is common.

- A **limit at infinity** has $x \to \pm\infty$. It describes end behavior and gives horizontal asymptotes. Example: $\lim_{x \to \infty} \frac{1}{x} = 0$.
- An **infinite limit** has the *output* going to $\pm\infty$ as $x$ approaches a finite number. It gives vertical asymptotes. Example: $\lim_{x \to 0^+} \frac{1}{x} = \infty$.

A single function can have both, as worked example 4 shows: horizontal asymptotes $y = \pm 2$ and a vertical asymptote at $x = -3$.

## Other functions you should know

Not every limit at infinity involves a fraction of polynomials. These come up constantly:

| Limit | Value | Why |
| --- | --- | --- |
| $\lim_{x \to \infty} e^{-x}$ | $0$ | $e^{-x} = \frac{1}{e^x}$ and $e^x$ grows without bound |
| $\lim_{x \to -\infty} e^{x}$ | $0$ | Same fact viewed from the left |
| $\lim_{x \to \infty} \ln x$ | $\infty$ | Slowly, but without bound |
| $\lim_{x \to \infty} \arctan x$ | $\frac{\pi}{2}$ | Horizontal asymptote $y = \frac{\pi}{2}$ |
| $\lim_{x \to -\infty} \arctan x$ | $-\frac{\pi}{2}$ | Horizontal asymptote $y = -\frac{\pi}{2}$ |
| $\lim_{x \to \infty} \frac{\sin x}{x}$ | $0$ | Squeeze: $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$ |
| $\lim_{x \to \infty} \sin x$ | does not exist | Oscillates forever |

It also helps to know the **growth hierarchy** for large $x$: logarithms grow slower than any positive power of $x$, and powers grow slower than exponentials. So $\frac{\ln x}{x} \to 0$ and $\frac{x^{10}}{e^x} \to 0$. You can prove these with [L'Hôpital's rule](/blog/lhopitals-rule).

## Common mistakes with limits at infinity

1. **Dividing by the wrong power.** Divide by the highest power in the *denominator*. Dividing by something smaller leaves a term that still grows, and the limit stays hidden.
2. **Treating $\frac{\infty}{\infty}$ as $1$.** It is indeterminate. $\frac{x^2}{x}$ and $\frac{x}{x^2}$ are both $\frac{\infty}{\infty}$ but have limits $\infty$ and $0$.
3. **Treating $\infty - \infty$ as $0$.** Worked example 5 shows the answer can be any number, in that case $\frac{1}{2}$.
4. **Forgetting $\sqrt{x^2} = |x|$.** This is the classic source of a wrong sign as $x \to -\infty$.
5. **Believing graphs never cross asymptotes.** Vertical asymptotes cannot be crossed, but horizontal ones often are.

## Practice reading end behavior

The degree rules take a minute to learn and a few dozen problems to trust. In the interactive lesson below you will predict end behavior, find horizontal asymptotes, and get instant feedback that points out exactly which power or sign you missed. Once these become automatic, curve sketching and [continuity](/blog/continuity) questions get much easier too.

## Frequently asked questions

### What does a limit at infinity mean?

It describes the value $f(x)$ approaches as $x$ gets larger and larger (or more and more negative). If the outputs settle toward a number $L$, we write $\lim_{x \to \infty} f(x) = L$, and the graph flattens out toward the line $y = L$.

### How do you find the horizontal asymptote of a rational function?

Compare the degree of the numerator to the degree of the denominator. If the bottom has the higher degree, the asymptote is $y = 0$. If the degrees are equal, it is $y = \frac{a}{b}$ where $a$ and $b$ are the leading coefficients. If the top has the higher degree, there is no horizontal asymptote.

### Can a graph cross its horizontal asymptote?

Yes. A horizontal asymptote only describes behavior far to the left or right. In the middle of the graph the curve can cross it, sometimes many times. For example $\frac{5x+2}{x^2+1}$ crosses $y = 0$ at $x = -\frac{2}{5}$.

### Is infinity minus infinity equal to zero?

No. $\infty - \infty$ is an indeterminate form, meaning the answer depends on how fast each part grows. Rewrite the expression, for example by multiplying by a conjugate, before taking the limit.