Limits at Infinity

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Textbook Reference

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Key idea

A limit at infinity asks one question: when you zoom out far enough, does the graph settle down toward a horizontal line?

You have studied limits where $x$ approaches a finite number. A limit at infinity changes only what $x$ does: instead of approaching a point, $x$ runs off to the right forever (toward $\infty$) or to the left forever (toward $-\infty$). The question is whether the height $f(x)$ approaches a fixed value $L$ as that happens.

Here is the picture. Take the function $f(x) = \dfrac{x^2 - 1}{x^2 + 1}$ and plug in larger and larger inputs: at $x = 10$ it is about $0.98$, at $x = 100$ about $0.9998$, at $x = 1000$ about $0.999998$. The values are crowding up against $1$ but never quite reaching it. We write \[ \lim_{x \to \infty} \frac{x^2 - 1}{x^2 + 1} = 1, \] and we say the line $y = 1$ is a horizontal asymptote: the graph hugs it on the far right.

The one idea to hold firmly is that $\infty$ is not a number. The notation $x \to \infty$ is shorthand for “$x$ grows without bound.” You never substitute infinity. Instead you ask what the output drifts toward as the input grows, and you have a clean technique, coming below, to answer it without any guessing.

Prerequisite Check

Before this lesson, make sure you can do all of the following:

• State what a limit means: the value $f(x)$ approaches as $x$ approaches something (review The Limit of a Function)
• Apply the limit laws for sums, products, and quotients (review Limit Laws)
• Recognize that dividing a constant by a growing number drives the quotient toward zero: $\frac{5}{x} \to 0$ as $x$ grows
• Identify the degree (highest power) of a polynomial
• Simplify a complex fraction by dividing numerator and denominator by the same expression

If “the limit value is what the output approaches, not a value you plug in” feels shaky, review the meaning of a limit first. Everything here rests on it.

Quick Reference

Limit at infinity (intuitive). $\displaystyle\lim_{x \to \infty} f(x) = L$ means $f(x)$ can be made as close to $L$ as you like by taking $x$ large enough. The line $y = L$ is then a horizontal asymptote.

The key fact. For any rational power $r > 0$, \[ \lim_{x \to \infty} \frac{1}{x^r} = 0 \qquad \text{and} \qquad \lim_{x \to -\infty} \frac{1}{x^r} = 0 \] (the second holds whenever $x^r$ is defined for negative $x$). A fixed number over a growing power goes to zero.

The technique for a rational function. To evaluate $\displaystyle\lim_{x\to\pm\infty}\frac{P(x)}{Q(x)}$, divide the numerator and denominator by the highest power of $x$ in the denominator, then send each $\frac{1}{x^k}$ term to zero.

The shortcut for a rational function (compare degrees). Let $P$ have degree $m$ and $Q$ have degree $n$:

Sign rule for radicals at $-\infty$. For $x < 0$, $\sqrt{x^2} = |x| = -x$. This sign flip is the most-missed step in limits as $x \to -\infty$.

Key Concepts

1. What a Limit at Infinity Means

Investigate $f(x) = \dfrac{x^2 - 1}{x^2 + 1}$ as $x$ grows. The outputs climb toward $1$ without reaching it, so \[ \lim_{x \to \infty} \frac{x^2 - 1}{x^2 + 1} = 1. \] (This is the opening example of Stewart 3.4.) The same happens as $x$ runs to the left: \[ \lim_{x \to -\infty} \frac{x^2 - 1}{x^2 + 1} = 1. \]

Definition (intuitive limit at infinity). Let $f$ be defined on an interval $(a, \infty)$. Then $\displaystyle\lim_{x\to\infty} f(x) = L$ means the values of $f(x)$ can be made arbitrarily close to $L$ by requiring $x$ to be sufficiently large. (Plain gloss: push $x$ far enough to the right and the height stays as near $L$ as you please.)

Definition (horizontal asymptote). The line $y = L$ is a horizontal asymptote of $y = f(x)$ if either $\displaystyle\lim_{x\to\infty} f(x) = L$ or $\displaystyle\lim_{x\to-\infty} f(x) = L$.

A graph can have at most two horizontal asymptotes, one for each direction, and the two need not be the same. Unlike a vertical asymptote, a horizontal asymptote may be crossed by the graph; it only governs the far-out behavior.

Important: $\infty$ is not a number, so you do not substitute it. Writing $\frac{\infty^2}{\infty^2} = 1$ is meaningless, because $\frac{\infty}{\infty}$ is not defined. The arrow $x \to \infty$ describes a process (the input growing), and the limit is the value the output approaches during that process. Use the technique below, not substitution.

2. The One Fact Everything Rests On

The engine behind every limit at infinity is simple: a fixed number divided by a growing quantity shrinks to zero.

Example 1. Find $\displaystyle\lim_{x\to\infty}\frac{1}{x}$ and $\displaystyle\lim_{x\to-\infty}\frac{1}{x}$.

When $x$ is large, $\frac{1}{x}$ is small: $\frac{1}{100} = 0.01$, $\frac{1}{10{,}000} = 0.0001$. By taking $x$ large enough, $\frac{1}{x}$ gets as close to $0$ as we please. So $\displaystyle\lim_{x\to\infty}\frac{1}{x} = 0$. The same reasoning with large negative $x$ gives $\displaystyle\lim_{x\to-\infty}\frac{1}{x} = 0$. (This is Stewart 3.4, Example 2.)

This generalizes to every positive power.

Theorem. If $r > 0$ is rational, then $\displaystyle\lim_{x\to\infty}\frac{1}{x^r} = 0$, and $\displaystyle\lim_{x\to-\infty}\frac{1}{x^r} = 0$ whenever $x^r$ is defined for $x < 0$.

So $\frac{1}{x}$, $\frac{1}{x^2}$, $\frac{1}{\sqrt{x}}$, and $\frac{1}{x^{3/2}}$ all go to zero as $x$ grows. Every technique below works by manufacturing these $\frac{1}{x^k}$ terms and then setting them to zero.

3. The Technique for Rational Functions

When both the numerator and denominator grow without bound, the expression has the indeterminate form $\frac{\infty}{\infty}$ and you cannot read off the answer. The fix is algebra: divide top and bottom by the highest power of $x$ in the denominator.

Example 2. Evaluate $\displaystyle\lim_{x\to\infty}\frac{3x^2 - x - 2}{5x^2 + 4x + 1}$.

Goal. Divide numerator and denominator by $x^2$ (the highest power in the denominator), then send each $\frac{1}{x^k}$ to zero. (This is Stewart 3.4, Example 3.)

Divide every term by $x^2$: \[ \frac{3x^2 - x - 2}{5x^2 + 4x + 1} = \frac{3 - \dfrac{1}{x} - \dfrac{2}{x^2}}{5 + \dfrac{4}{x} + \dfrac{1}{x^2}}. \]

As $x \to \infty$, every term with an $x$ in the denominator goes to zero: \[ \lim_{x\to\infty}\frac{3 - \dfrac{1}{x} - \dfrac{2}{x^2}}{5 + \dfrac{4}{x} + \dfrac{1}{x^2}} = \frac{3 - 0 - 0}{5 + 0 + 0} = \frac{3}{5}. \]

Boxed answer: $\displaystyle\lim_{x\to\infty}\frac{3x^2 - x - 2}{5x^2 + 4x + 1} = \frac{3}{5}$. The line $y = \frac{3}{5}$ is a horizontal asymptote.

Recap. After dividing, the surviving terms are the leading coefficients, $3$ on top and $5$ on the bottom. That is the source of the shortcut: when the top and bottom have the same degree, the limit is the ratio of the leading coefficients. The honest algebra above is why the shortcut works.

4. The Degree Shortcut

Once you have seen the technique, you can read off the limit of a rational function by comparing the degrees of the numerator and denominator.

• Denominator degree larger (bottom wins): the limit is $0$. Dividing by the larger power makes the numerator terms vanish.
• Equal degrees (tie): the limit is the ratio of the leading coefficients, as in Example 2.
• Numerator degree larger (top wins): the function grows without bound, so the limit is $+\infty$ or $-\infty$ and there is no horizontal asymptote.

Example 3. Evaluate $\displaystyle\lim_{x\to\infty}\frac{x^2 + x}{3 - x}$.

Goal. The numerator has degree $2$, the denominator degree $1$, so the top wins. Confirm with the technique: divide by $x$, the highest power in the denominator. (This is Stewart 3.4, Example 10.)

\[ \frac{x^2 + x}{3 - x} = \frac{x + 1}{\dfrac{3}{x} - 1} \longrightarrow \frac{\infty}{0 - 1} = -\infty \] because the numerator $x + 1 \to \infty$ while the denominator $\frac{3}{x} - 1 \to -1$.

Boxed answer: $\displaystyle\lim_{x\to\infty}\frac{x^2 + x}{3 - x} = -\infty$. There is no horizontal asymptote.

Recap. When the top wins, the function blows up; the sign ($-\infty$ here) comes from the leading behavior, a large positive numerator over a quantity approaching $-1$.

5. The Sign Trap: Radicals as $x \to -\infty$

Limits involving square roots as $x \to -\infty$ hide a sign flip. The rule $\sqrt{x^2} = x$ is only correct when $x \geq 0$. In general $\sqrt{x^2} = |x|$, and for $x < 0$ that equals $-x$.

Example 4. Evaluate $\displaystyle\lim_{x\to-\infty}\frac{\sqrt{2x^2 + 1}}{3x - 5}$.

Goal. Divide by $x$, but because $x < 0$, pull the $x$ inside the root as $-\sqrt{x^2}$. (This is Stewart 3.4, the $x \to -\infty$ companion to Example 4.)

For $x < 0$, dividing the numerator by $x$ means dividing inside the root by $-\sqrt{x^2}$: \[ \frac{\sqrt{2x^2 + 1}}{x} = \frac{\sqrt{2x^2 + 1}}{-\sqrt{x^2}} = -\sqrt{\frac{2x^2 + 1}{x^2}} = -\sqrt{2 + \frac{1}{x^2}}. \]

Dividing the denominator by $x$ as usual, the limit becomes \[ \lim_{x\to-\infty}\frac{-\sqrt{2 + \dfrac{1}{x^2}}}{3 - \dfrac{5}{x}} = \frac{-\sqrt{2 + 0}}{3 - 0} = -\frac{\sqrt{2}}{3}. \]

Boxed answer: $\displaystyle\lim_{x\to-\infty}\frac{\sqrt{2x^2 + 1}}{3x - 5} = -\frac{\sqrt{2}}{3}$, so $y = -\frac{\sqrt{2}}{3}$ is a horizontal asymptote.

Recap. The minus sign is the key result here. As $x \to +\infty$ the same function approaches $+\frac{\sqrt 2}{3}$; the two horizontal asymptotes differ in sign precisely because $\sqrt{x^2} = -x$ for negative $x$. Whenever a square root meets $x \to -\infty$, stop and write $\sqrt{x^2} = -x$ explicitly.

Important: $\sqrt{x^2} = |x|$, which is $-x$ when $x$ is negative. Treating $\sqrt{x^2}$ as $x$ for negative $x$ flips the sign of the answer and is the most common error in this section. Write the absolute value first, then replace it with $-x$ for the $x \to -\infty$ branch.

6. The Conjugate Trick for Differences

A difference like $\sqrt{x^2 + 1} - x$ has the indeterminate form $\infty - \infty$, which the limit laws cannot touch. Multiplying by the conjugate turns it into a fraction you can handle.

Example 5. Evaluate $\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2 + 1} - x\right)$.

Goal. Multiply and divide by the conjugate $\sqrt{x^2+1} + x$ to rationalize. (This is Stewart 3.4, Example 5.)

\[ \sqrt{x^2 + 1} - x = \frac{\left(\sqrt{x^2 + 1} - x\right)\left(\sqrt{x^2 + 1} + x\right)}{\sqrt{x^2 + 1} + x} = \frac{(x^2 + 1) - x^2}{\sqrt{x^2 + 1} + x} = \frac{1}{\sqrt{x^2 + 1} + x}. \]

As $x \to \infty$, the denominator grows without bound, so the fraction goes to zero: \[ \lim_{x\to\infty}\left(\sqrt{x^2 + 1} - x\right) = \lim_{x\to\infty}\frac{1}{\sqrt{x^2 + 1} + x} = 0. \]

Boxed answer: $\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2 + 1} - x\right) = 0$.

Recap. The conjugate clears the radical from the numerator, leaving a constant over a growing denominator. The same trick handles $\infty - \infty$ differences whenever a conjugate is available.

7. End Behavior and the Precise Definition

The end behavior of a function is how it acts as $x \to \infty$ and as $x \to -\infty$. For a polynomial, the leading term decides it: $\displaystyle\lim_{x\to\infty}(x^2 - x) = \lim_{x\to\infty} x(x-1) = \infty$, since both factors grow. (This is Stewart 3.4, Example 9.) Knowing the two end behaviors, plus the intercepts, sketches a rough graph of any polynomial without calculus.

The intuitive definition can be made fully precise.

Definition (precise limit at infinity). $\displaystyle\lim_{x\to\infty} f(x) = L$ means that for every $\varepsilon > 0$ there is a number $N$ such that if $x > N$, then $|f(x) - L| < \varepsilon$. (Plain gloss: name any error tolerance $\varepsilon$, no matter how small; there is a cutoff $N$ so that past it the graph stays within $\varepsilon$ of the line $y = L$.)

This is the same logic as the $\varepsilon$-$\delta$ definition for finite limits, with “$x$ close to $a$” replaced by “$x$ larger than $N$.” It is what makes the word “approaches” mathematically exact, and it is the rigorous foundation under every shortcut in this lesson.

Common Errors Summary

Common Misconceptions

Leveled Practice

Level 1 -- Direct Application

Problem 1. Evaluate $\displaystyle\lim_{x\to\infty}\frac{2x^2 - 7}{5x^2 + x - 3}$.

Problem 2. Evaluate $\displaystyle\lim_{x\to\infty}\frac{x}{x^2 + 1}$.

Problem 3. Find the horizontal asymptote of $y = \dfrac{5 + 4x}{x + 3}$.

Level 2 -- Multiple Steps

Problem 4. Evaluate $\displaystyle\lim_{x\to-\infty}\frac{4 - \sqrt{x}}{2 + \sqrt{x}}$, or explain why it does not exist.

Problem 5. A tank holds $5000$ liters of pure water. Brine containing $30$ grams of salt per liter is pumped in at $25$ liters per minute. The salt concentration after $t$ minutes is $C(t) = \dfrac{30t}{200 + t}$ grams per liter. What happens to the concentration as $t \to \infty$?

Problem 6. Evaluate $\displaystyle\lim_{x\to-\infty}\frac{\sqrt{9x^2 + 1}}{2x + 3}$.

Level 3 -- Deeper Problems

Problem 7. Evaluate $\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2 + x} - x\right)$.

Problem 8. Find both horizontal asymptotes of $y = \dfrac{x}{\sqrt{x^2 + 1}}$.

Problem 9. Two functions have the same end behavior if the limit of their ratio is $1$ as $x \to \infty$. Show that $P(x) = 3x^5 - 5x^3 + 2x$ and $Q(x) = 3x^5$ have the same end behavior.

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:

• State, in plain words, what $\lim_{x\to\infty} f(x) = L$ means, without “plugging in infinity”
• Use $\lim_{x\to\infty} \frac{1}{x^r} = 0$ to evaluate limits at infinity
• Evaluate the limit at infinity of a rational function by dividing by the highest power in the denominator
• Apply the degree shortcut: bottom wins gives $0$, a tie gives the ratio of leading coefficients, top wins gives $\pm\infty$
• Identify horizontal asymptotes from limits at infinity, including the case of two different ones
• Handle $\sqrt{x^2} = -x$ correctly when $x \to -\infty$
• Use the conjugate trick on an $\infty - \infty$ difference of radicals
• Describe the end behavior of a polynomial from its leading term

Mental Model

Think of a limit at infinity as zooming out on a graph until it stops moving.

Up close, near the origin, the graph of a rational function can wiggle, cross axes, and shoot up at vertical asymptotes. Zoom out far enough, though, and the local drama becomes invisible. What you see at the far right and far left is whether the curve flattens toward a horizontal line.

Three things follow from the picture:

• The limit is a destination, not a value you reach. The graph of $\frac{x^2-1}{x^2+1}$ never equals $1$, yet $1$ is its limit, because it gets and stays arbitrarily close. A limit at infinity names where the curve is heading, not a point it lands on.
• Only the biggest terms survive the zoom-out. In a rational function, the lower-degree terms become rounding error once $x$ is huge. That is why the answer depends only on the leading terms, and why comparing degrees works as a shortcut. The honest algebra (dividing by the highest power) is just making “the small terms vanish” precise.
• Left and right can have different destinations. Because squaring erases sign but a square root reintroduces it as $-x$ for negative inputs, a function can level off at one height on the right and a different height on the left. Always ask the question twice, once for $+\infty$ and once for $-\infty$.

Connections

Within Applications of Differentiation (Chapter 3)

• Limit laws: Limits at infinity obey the same sum, product, and quotient laws as finite limits (with the same caution that the laws fail for $\frac{\infty}{\infty}$ and $\infty - \infty$). This lesson rests on the laws plus the single new fact $\frac{1}{x^r} \to 0$.
• Infinite limits and vertical asymptotes: A vertical asymptote describes the graph shooting to $\pm\infty$ as $x$ approaches a finite point; a horizontal asymptote describes the graph leveling off as $x$ runs to $\pm\infty$. The two are mirror questions, and Example 1 of Stewart 3.4 reads both off one graph.
• Horizontal asymptotes: A horizontal asymptote is exactly a finite limit at infinity, so every asymptote computation is a limit-at-infinity computation. The methods here are how you find them.

Toward later calculus (MATH161 and beyond)

• Curve sketching: The end behavior you compute here, together with intercepts and the first and second derivatives, produces a complete, accurate graph. End behavior is the outer frame the rest of the sketch fills in.
• L’Hopital’s rule: For indeterminate forms like $\frac{\infty}{\infty}$ that resist algebra (for example $\frac{\ln x}{x}$ or $\frac{x^2}{e^x}$), L’Hopital’s rule evaluates the limit by differentiating top and bottom. It extends the reach of this lesson to functions that are not rational.
• Improper integrals: In Calculus 2, integrals over infinite intervals, $\int_a^\infty f(x)\,dx$, are defined as limits as an upper bound runs to infinity. Whether such an integral converges depends on the limit-at-infinity behavior of the function, the very thing you are learning here.

Audience Notes

For students who find math intimidating: You do not need the hard definition to get full marks. Memorize one rule and one habit. The rule: compare the top and bottom degrees (bottom bigger gives zero, equal gives the coefficient ratio, top bigger gives infinity). The habit: when you see a square root and $x \to -\infty$, stop and write $\sqrt{x^2} = -x$. Those two carry almost every exam problem in this section.

For career-focused students: A limit at infinity is the steady-state behavior of a system, what a quantity settles to after a long time. The salt-tank problem is a model of any mixing or buffering process; the long-run concentration is the limit. Control engineers, chemical engineers, and data scientists read these limits to know where a system ends up, not just where it starts.

For gifted and curious students: Investigate why $\sqrt{x^2 + 1} - x \to 0$ but $\sqrt{x^2 + x} - x \to \frac12$ (Problem 7). The difference is the linear term inside the root. More generally, $\sqrt{x^2 + bx + c} - x \to \frac{b}{2}$ as $x \to \infty$, a fact the conjugate trick proves in one line and that previews the idea of an oblique asymptote.

For PhD-track students: The precise definition replaces “$x$ near $a$” with “$x > N$,” but the deeper unification comes from the extended real line and the substitution $t = 1/x$, which turns a limit at infinity into a one-sided limit at $0$. That change of variable, $\lim_{x\to\infty} f(x) = \lim_{t\to 0^+} f(1/t)$, is the formal bridge between the two kinds of limit and is what lets the $\varepsilon$-$N$ and $\varepsilon$-$\delta$ definitions share one theory.

Back to Applications of Differentiation | Related: Limit Laws | Next: Horizontal Asymptotes