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Limits at Infinity and Horizontal Asymptotes

Reference: Stewart 3.4  •  Chapter: 3  •  Section: 4

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What Happens Far Away?

Imagine driving on a perfectly straight road toward a distant mountain. The closer you get, the larger the mountain appears; but what happens if you drive forever into an infinite desert? The flat horizon stays at the same height no matter how far you go.

This is the idea behind limits at infinity: we're asking what value (if any) a function settles toward as $x$ grows without bound. While vertical asymptotes describe what happens near "explosive" points, horizontal asymptotes describe the function's long-term, far-away behavior.

Key distinction: We already know what $\lim_{x \to a} f(x) = \infty$ means (function blows up near $a$). Now we're asking: what does $\lim_{x \to \infty} f(x) = L$ mean? Here $x$ is going to infinity, but the function values approach a finite number $L$.

Prerequisite Map

This skillLimits at Infinity and Horizontal Asymptotes

Quick Reference

Property Value
Concept Limits
Course MATH161
Section Stewart 2.6
Difficulty Intermediate
Time ~20 minutes

Key Concepts

Definition: Limit at Infinity

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

This means: as $x$ increases without bound, the values of $f(x)$ get arbitrarily close to $L$.

Intuitive version: For any tolerance level you specify, I can find a point beyond which all function values stay within that tolerance of $L$.

Precise version ($\varepsilon$-$N$): For every $\varepsilon > 0$, there exists a number $N$ such that: $$\text{if } x > N \text{ then } \vert f(x) - L\vert < \varepsilon$$

Similarly for negative infinity:

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

means $f(x) \to L$ as $x$ decreases without bound (becomes large negative).

The Fundamental Example: $\frac{1}{x}$

$$\lim_{x \to \infty} \frac{1}{x} = 0 \quad \text{and} \quad \lim_{x \to -\infty} \frac{1}{x} = 0$$

Why? As $x$ gets larger, $\frac{1}{x}$ gets smaller:

We can make $\frac{1}{x}$ as close to $0$ as we want by taking $x$ large enough.

This extends to any positive power:

$$\boxed{\lim_{x \to \infty} \frac{1}{x^r} = 0 \quad \text{for any } r > 0}$$

Visualizing Limits at Infinity

    y
    |
  L ├─────────────────────────────  ← horizontal asymptote
    |          ~~~~~~~~~~~~~~~~~~~
    |       ~~~
    |    ~~~
    |  ~~
    | ~
────+───────────────────────────→ x

As x → ∞, f(x) approaches L from below

Different functions can approach the asymptote in different ways:

Definition: Horizontal Asymptote

The line $y = L$ is a horizontal asymptote of the curve $y = f(x)$ if:

$$\lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L$$

Important observations:

Limits That Don't Exist at Infinity

Not every function has a limit at infinity.

Example: $\lim_{x \to \infty} \sin x$ does not exist.

As $x \to \infty$, $\sin x$ keeps oscillating between $-1$ and $1$ forever. It never settles down to any particular value.

Key test: Does the function eventually settle toward a single value, or does it keep oscillating/varying without bound?

Comparing Types of Limits

Notation Meaning Example
$\lim_{x \to a} f(x) = L$ Ordinary limit $\lim_{x \to 2} x^2 = 4$
$\lim_{x \to a} f(x) = \infty$ Infinite limit $\lim_{x \to 0} \frac{1}{x^2} = \infty$
$\lim_{x \to \infty} f(x) = L$ Limit at infinity $\lim_{x \to \infty} \frac{1}{x} = 0$
$\lim_{x \to \infty} f(x) = \infty$ Infinite limit at infinity $\lim_{x \to \infty} x^2 = \infty$

Practice Problems

Level 1 Interpreting Limit at Infinity Notation

What does $\lim_{x \to \infty} f(x) = 7$ mean in plain English? Does this tell us anything about $f(100)$?

Thought Process

Break this into two questions: First, what does the statement mean? Second, what does it NOT guarantee about specific function values?

Show Answer

Meaning: As $x$ becomes arbitrarily large, the values of $f(x)$ get arbitrarily close to $7$. The function values eventually stay as close to $7$ as we want if we look far enough to the right.

About $f(100)$: This tells us almost nothing specific about $f(100)$! The function might oscillate, dip, or spike before settling near $7$. We only know that eventually (for large enough $x$), the values stay close to $7$. The value $x = 100$ might not be "large enough."

For instance, $f(x) = 7 + \frac{1000}{x}$ has limit $7$ as $x \to \infty$, but $f(100) = 17$.

Level 2 Basic Limits at Infinity

Evaluate the following limits:

  1. $\lim_{x \to \infty} \frac{5}{x^3}$
  2. $\lim_{x \to -\infty} \frac{2}{x^4}$
  3. $\lim_{x \to \infty} \left(3 + \frac{1}{x}\right)$
Thought Process

Use the key result: $\lim_{x \to \pm\infty} \frac{1}{x^r} = 0$ for any $r > 0$. Then apply limit laws (constants pull out, limits of sums are sums of limits).

Show Answer

(a) $\lim_{x \to \infty} \frac{5}{x^3} = 5 \cdot \lim_{x \to \infty} \frac{1}{x^3} = 5 \cdot 0 = \boxed{0}$

(b) $\lim_{x \to -\infty} \frac{2}{x^4} = 2 \cdot \lim_{x \to -\infty} \frac{1}{x^4} = 2 \cdot 0 = \boxed{0}$

Note: $x^4 \to +\infty$ as $x \to -\infty$ because the exponent is even.

(c) $\lim_{x \to \infty} \left(3 + \frac{1}{x}\right) = \lim_{x \to \infty} 3 + \lim_{x \to \infty} \frac{1}{x} = 3 + 0 = \boxed{3}$

The horizontal asymptote is $y = 3$.

Level 3 Identifying Horizontal Asymptotes from a Graph

A function $f$ has the following behavior:

  • As $x \to \infty$, $f(x)$ approaches $4$
  • As $x \to -\infty$, $f(x)$ approaches $-1$
  • $f$ has vertical asymptotes at $x = -3$ and $x = 2$
  1. What are the horizontal asymptotes of $f$?
  2. How many asymptotes does $f$ have in total?
  3. Must the graph of $f$ lie entirely above or below its horizontal asymptotes?
Thought Process

A horizontal asymptote appears whenever a limit at $\pm\infty$ equals a finite number. Count vertical and horizontal asymptotes separately. Remember that horizontal asymptotes describe end behavior; they say nothing about what happens in between.

Show Answer

(a) The horizontal asymptotes are:

  • $y = 4$ (from the limit as $x \to \infty$)
  • $y = -1$ (from the limit as $x \to -\infty$)

(b) Total asymptotes: $2$ horizontal + $2$ vertical = $\boxed{4}$ asymptotes

(c) No! The graph can cross its horizontal asymptotes any number of times. The asymptotes only describe what happens for very large $\vert x\vert $. For moderate values of $x$, the function can be above, below, or crossing the horizontal asymptote lines.

For example, $f(x) = \frac{\sin x}{x}$ has $y = 0$ as a horizontal asymptote but crosses it infinitely many times.

Level 4 Finding N for a Given Tolerance

Consider $f(x) = \frac{1}{x}$ with limit $L = 0$ as $x \to \infty$.

  1. Find a value of $N$ such that $\vert f(x) - 0\vert < 0.01$ whenever $x > N$.
  2. Find a value of $N$ such that $\vert f(x) - 0\vert < 0.0001$ whenever $x > N$.
  3. In general, if $\varepsilon > 0$, what is the smallest $N$ that works?
Thought Process

We need $\vert f(x) - 0\vert = \vert \frac{1}{x}\vert < \varepsilon$. For $x > 0$, this becomes $\frac{1}{x} < \varepsilon$. Solve for $x$.

Show Answer

For $x > 0$, we have $\vert f(x) - 0\vert = \frac{1}{x}$.

We need $\frac{1}{x} < \varepsilon$, which means $x > \frac{1}{\varepsilon}$.

(a) For $\varepsilon = 0.01$: We need $x > \frac{1}{0.01} = 100$. So $N = 100$ works (or any larger value).

(b) For $\varepsilon = 0.0001$: We need $x > \frac{1}{0.0001} = 10{,}000$. So $N = 10{,}000$ works.

(c) In general, the smallest $N$ that works is $N = \frac{1}{\varepsilon}$.

Pattern: Smaller tolerance $\varepsilon$ requires larger $N$. This is the $\varepsilon$-$N$ definition in action: for any $\varepsilon > 0$ we specify, we can always find an $N$ (namely, $N = \frac{1}{\varepsilon}$) that makes the condition work.

Level 5 Proving a Limit at Infinity

Use the $\varepsilon$-$N$ definition to prove that $\lim_{x \to \infty} \frac{1}{x^2} = 0$.

Your proof should:

  1. Start with "Let $\varepsilon > 0$ be given"
  2. Find an appropriate $N$ in terms of $\varepsilon$
  3. Show that $x > N$ implies $\vert f(x) - 0\vert < \varepsilon$
Thought Process

Work backwards: We need $\frac{1}{x^2} < \varepsilon$. Solve for $x$: $x^2 > \frac{1}{\varepsilon}$, so $x > \frac{1}{\sqrt{\varepsilon}}$. This suggests choosing $N = \frac{1}{\sqrt{\varepsilon}}$.

Show Answer

Proof:

Let $\varepsilon > 0$ be given.

Choose $N = \frac{1}{\sqrt{\varepsilon}}$.

Now suppose $x > N$. We need to show $\left\vert \frac{1}{x^2} - 0\right\vert < \varepsilon$.

Since $x > N = \frac{1}{\sqrt{\varepsilon}} > 0$, we have: $$x > \frac{1}{\sqrt{\varepsilon}}$$

Squaring both sides (valid since both sides are positive): $$x^2 > \frac{1}{\varepsilon}$$

Taking reciprocals (which reverses the inequality since both sides are positive): $$\frac{1}{x^2} < \varepsilon$$

Therefore: $$\left\vert \frac{1}{x^2} - 0\right\vert = \frac{1}{x^2} < \varepsilon$$

This confirms that for any $\varepsilon > 0$, choosing $N = \frac{1}{\sqrt{\varepsilon}}$ ensures $\vert f(x) - 0\vert < \varepsilon$ whenever $x > N$.

By the definition of limit at infinity: $$\lim_{x \to \infty} \frac{1}{x^2} = 0 \quad \blacksquare$$

Mastery Checklist

Mental Model

The Settling Horizon: Think of limits at infinity like watching a boat sail toward the horizon. No matter how far the boat goes, it appears to approach the horizon line but never quite reaches it. The horizon line is the horizontal asymptote, a visual boundary the function approaches but doesn't necessarily reach. Unlike a vertical wall (vertical asymptote) that blocks the boat's path, the horizon stretches on forever, and the boat can sail toward it indefinitely.


Connections

Looking back:

Looking ahead:

Real-world connections:


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Last updated: 2026-01-22