Limits at Infinity and Horizontal…

Part 1 · Hook

Why this matters

A cup of hot coffee cools quickly at first, then more and more slowly, settling toward room temperature but never quite reaching it. Questions about the long run, where a quantity heads as time goes on and on, are limits at infinity. They describe where a graph levels off, and which of two growing quantities wins.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. For large $x$ , which term of $3x^2 - 5x + 7$ matters most?

$3x^2$
$-5x$
$7$
They matter equally

Show the answer

The highest power grows fastest.

Correct: $3x^2$ :
$-5x$ :
$7$ :
They matter equally:

2. What is $\lim_{x \to 2^+} \frac{1}{x - 2}$ ?

$\infty$
$-\infty$
$0$
$2$

Show the answer

One over a tiny positive number grows without bound.

Correct: $\infty$ :
$-\infty$ :
$0$ :
$2$ :

3. $e^{-x}$ equals

$\frac{1}{e^x}$
$-e^x$
$e^{1/x}$
$\frac{1}{e^{-x}}$

Show the answer

A negative exponent means a reciprocal.

Correct: $\frac{1}{e^x}$ :
$-e^x$ :
$e^{1/x}$ :
$\frac{1}{e^{-x}}$ :

Part 4 · See it

See it first

Left: the graph of f(x) = (2x² + x)/(x² + 1) for x from −10 to 10 with the dashed horizontal asymptote y = 2. The graph crosses the asymptote at x = 2 and then approaches it from above. Right: ln x, x³ and eˣ for x from 0 to 6. At first x³ is above eˣ, but eˣ passes it near x = 4.54 and stays ahead; ln x grows slowest. — A horizontal asymptote, and the race between ln x, a power and an exponential. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$x$ grows without boundterms like $\frac{1}{x}$ , $\frac{5}{x^2}$ and $e^{-x}$ shrink to 0
You divide the top and bottom of a quotient by the dominant termevery other term becomes one of those shrinking terms, and the limit can be read off
$\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$ is finite $y = L$ is a horizontal asymptote; the graph levels off there (and may cross it in the middle)
Two functions both grow without boundthe faster one dominates: as $x \to \infty$ , $\ln x$ loses to any power $x^p$ ( $p > 0$ ), and $x^p$ loses to $e^x$

Part 6 · Key ideas

Key ideas

$\lim_{x \to \infty} f(x) = L$ means $f(x)$ approaches $L$ as $x$ grows without bound. A finite limit at $\infty$ or $-\infty$ gives a horizontal asymptote $y = L$ .
For a rational function, divide by the highest power of $x$ in the denominator. Equal degrees: ratio of leading coefficients. Bottom higher: 0. Top higher: no horizontal asymptote.
$\sqrt{x^2} = |x|$ . As $x \to -\infty$ , $\sqrt{x^2} = -x$ , which can flip the sign of the answer.
Dominant terms: in the long run $\ln x$ loses to every positive power, and $e^x$ beats every power.
Vertical asymptote: $\lim_{x \to c} f(x) = \pm\infty$ . Horizontal asymptote: $\lim_{x \to \pm\infty} f(x) = L$ . Do not mix them up.

Part 7 · Misconception

A common mistake

The wrong idea: A graph never crosses its horizontal asymptote, and the limit at $-\infty$ is the same as at $\infty$ .

What actually happens: A horizontal asymptote only describes the far ends: $\frac{2x^2 + x}{x^2 + 1}$ crosses $y = 2$ at $x = 2$ . And the two ends can differ: $\frac{4x}{\sqrt{x^2 + 1}}$ approaches 4 on the right and $-4$ on the left.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Four rational functions

Four rational functions, to be studied as x grows without bound in either direction
Function	Formula
$f_1$	$f_1(x)$ $= \frac{3x^2 - x}{2x^2 + 5}$
$f_2$	$f_2(x)$ $= \frac{4x + 1}{x^2 - 3}$
$f_3$	$f_3(x)$ $= \frac{x^3 + 2}{5x^2 + x}$
$f_4$	$f_4(x)$ $= \frac{6x^2 - 1}{1 - 3x^2}$

1. Divide by the highest power of $x$ to find $\lim_{x \to \infty} f_1(x)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Divide every term by $x^2$ : $f_1(x)$ $= \frac{3 - \frac{1}{x}}{2 + \frac{5}{x^2}}$ . As $x \to \infty$ , $\frac{1}{x} \to 0$ and $\frac{5}{x^2} \to 0$ , so the limit is $\frac{3}{2}$ .

Answer: $\frac{3}{2}$ ≈ 1.500

2. Which function has the horizontal asymptote $y = 0$ ?

$f_2$
$f_1$
$f_3$
$f_4$

Show the answer

In $f_2$ the denominator has the higher degree. Dividing by $x^2$ : $\frac{\frac{4}{x} + \frac{1}{x^2}}{1 - \frac{3}{x^2}} \to \frac{0}{1}$ $= 0$ .

Correct: $f_2$ : Degree 1 over degree 2: the denominator dominates, so the quotient approaches 0.
$f_1$ : Equal degrees give $y = \frac{3}{2}$ .
$f_3$ : The numerator has the higher degree, so $f_3$ is unbounded.
$f_4$ : Equal degrees give $y = \frac{6}{-3} = -2$ .

Graph

The graph of $h(x) = \frac{4x}{\sqrt{x^2 + 1}}$

3. Why is $\lim_{x \to -\infty} h(x) = -4$ rather than $4$ ?

For $x < 0$ , $\sqrt{x^2 + 1}$ is negative, so the quotient changes sign
The constant 1 under the root becomes $-1$ when $x$ is negative
Limits at $-\infty$ are the negatives of limits at $\infty$
For $x < 0$ , $\sqrt{x^2} = -x$ , so $h(x) \approx \frac{4x}{-x} = -4$

Show the answer

$\sqrt{x^2} = |x|$ . When $x$ is negative, $|x| = -x$ , so for large negative $x$ , $\frac{4x}{\sqrt{x^2 + 1}}$ $\approx \frac{4x}{-x}$ $= -4$ .

For $x < 0$ , $\sqrt{x^2 + 1}$ is negative, so the quotient changes sign: A square root is never negative; it is the numerator $4x$ that is negative.
The constant 1 under the root becomes $-1$ when $x$ is negative: The 1 stays 1; it just becomes small next to $x^2$ .
Limits at $-\infty$ are the negatives of limits at $\infty$ : Not in general: $f_1$ above has the same limit $\frac{3}{2}$ in both directions.
Correct: For $x < 0$ , $\sqrt{x^2} = -x$ , so $h(x) \approx \frac{4x}{-x} = -4$ : The root is $|x|$ , which equals $-x$ for negative $x$ , while the numerator $4x$ is negative.

4. How many horizontal asymptotes does the graph of $h$ have?

Two: $y = 4$ and $y = -4$
One: $y = 4$
One: $y = 0$
Zero, since $h$ is defined everywhere

Show the answer

A horizontal asymptote comes from a finite limit at $\infty$ or at $-\infty$ . Here they differ, 4 and $-4$ , so there are two.

Correct: Two: $y = 4$ and $y = -4$ : Each direction gives its own finite limit.
One: $y = 4$ : The left side levels off at $-4$ , a second asymptote.
One: $y = 0$ : $h(0) = 0$ , but asymptotes describe behavior far out, not at the origin.
Zero, since $h$ is defined everywhere: Being defined everywhere does not stop a graph from leveling off.

5. Find $\lim_{x \to \infty} \frac{x^3}{e^x}$ .

$\infty$
$1$
$0$
$3$

Show the answer

An exponential with base greater than 1 eventually grows faster than any power of $x$ . So $e^x$ dominates $x^3$ and the quotient approaches 0.

$\infty$ : $x^3$ is larger only for moderate $x$ ; in the long run $e^x$ wins.
$1$ : The two do not grow at the same rate.
Correct: $0$ : $e^x$ passes $x^3$ (near $x \approx 4.5$ ) and then leaves it far behind.
$3$ : The exponent 3 does not appear in the limit.

6. Can a graph cross its own horizontal asymptote? Consider $f(x)$ $= \frac{2x^2 + x}{x^2 + 1}$ , whose horizontal asymptote is $y = 2$ .

No: a graph stays off its asymptote
Yes: $f(2) = 2$ , yet $f(x) \to 2$ as $x \to \infty$
No: $f(x) > 2$ for each $x$ , so it stays above
Yes, but then $y = 2$ is no longer an asymptote

Show the answer

$f(x) = 2$ when $2x^2 + x = 2x^2 + 2$ , so at $x = 2$ . A horizontal asymptote describes behavior far out, as $x \to \pm\infty$ , and says nothing about the middle of the graph.

No: a graph stays off its asymptote: Vertical asymptotes cannot be crossed (the function is unbounded there), but horizontal ones can.
Correct: Yes: $f(2) = 2$ , yet $f(x) \to 2$ as $x \to \infty$ : $f(2) = \frac{8 + 2}{5} = 2$ : the graph crosses $y = 2$ , and still approaches it far out.
No: $f(x) > 2$ for each $x$ , so it stays above: $f(0) = 0 < 2$ , so the graph is sometimes below.
Yes, but then $y = 2$ is no longer an asymptote: The asymptote is defined by the limit at infinity, which is still 2.

Part 9 · Summary

Summary

A limit at infinity describes where

f(x)

heads as

x \to \infty

or

x \to -\infty

; a finite one gives a horizontal asymptote. Divide by the dominant term: for rational functions compare degrees. Watch

\sqrt{x^2} = |x|

on the left. In the long run

\ln x

grows slower than any power and

e^x

faster than any power.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Four rational functions

The graph of h(x)=4xx2+1h(x) = \frac{4x}{\sqrt{x^2 + 1}}

Summary

What comes next

Connections

The graph of $h(x) = \frac{4x}{\sqrt{x^2 + 1}}$