<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 1.8: Infinite Limits

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

A student will be able to:

• Find infinite limits of functions.
• Analyze properties of infinite limits.
• Identify asymptotes of functions.
• Analyze end behavior of functions.

## Introduction

In this lesson we will discuss infinite limits. In our discussion the notion of infinity is discussed in two contexts. First, we can discuss infinite limits in terms of the value a function as we increase x\begin{align*}x\end{align*} without bound. In this case we speak of the limit of f(x)\begin{align*}f(x)\end{align*} as x\begin{align*}x\end{align*} approaches \begin{align*}\infty\end{align*} and write limxf(x)\begin{align*}\lim_{x \to \infty} f(x)\end{align*}. We could similarly refer to the limit of f(x)\begin{align*}f(x)\end{align*} as x\begin{align*}x\end{align*} approaches -\begin{align*} \infty\end{align*} and write limxf(x)\begin{align*}\lim_{x \to -\infty} f(x)\end{align*}.

The second context in which we speak of infinite limits involves situations where the function values increase without bound. For example, in the case of a rational function such as f(x)=(x+1)/(x2+1)\begin{align*}f(x) = (x + 1)/(x^2+1)\end{align*}, a function we discussed in previous lessons:

At x=1\begin{align*}x = 1\end{align*}, we have the situation where the graph grows without bound in both a positive and a negative direction. We say that we have a vertical asymptote at x=1\begin{align*}x = 1\end{align*}, and this is indicated by the dotted line in the graph above.

In this example we note that limx1f(x)\begin{align*}\lim_{x \to 1} f(x)\end{align*} does not exist. But we could compute both one-sided limits as follows.

limx1f(x)=\begin{align*}\lim_{x \to 1^-} f(x) = -\infty\end{align*} and limx1f(x)=+\begin{align*}\lim_{x \to 1^-} f(x) = +\infty\end{align*}.

More formally, we define these as follows:

Definition:

The right-hand limit of the function f(x)\begin{align*}f(x)\end{align*} at x=a\begin{align*}x = a\end{align*} is infinite, and we write limxa+f(x)=\begin{align*}\lim_{x \to a^+} f(x) = \infty\end{align*}, if for every positive number k\begin{align*}k\end{align*}, there exists an open interval (a,a+δ)\begin{align*}(a, a + \delta)\end{align*} contained in the domain of f(x)\begin{align*}f(x)\end{align*}, such that f(x)\begin{align*}f(x)\end{align*} is in (k,)\begin{align*}(k, \infty)\end{align*} for every x\begin{align*}x\end{align*} in (a,a+δ)\begin{align*}(a, a + \delta)\end{align*}.
The definition for negative infinite limits is similar.
Suppose we look at the function f(x)=(x+1)/(x21)\begin{align*}f(x) = (x + 1)/(x^2 - 1)\end{align*} and determine the infinite limits limxf(x)\begin{align*}\lim_{x \to \infty} f(x)\end{align*} and limxf(x)\begin{align*}\lim_{x \to -\infty} f(x)\end{align*}.
We observe that as x\begin{align*}x\end{align*} increases in the positive direction, the function values tend to get smaller. The same is true if we decrease x\begin{align*}x\end{align*} in the negative direction. Some of these extreme values are indicated in the following table.

x100200100200f(x).0101.0053.0099.005\begin{align*}x && f(x) \\ 100 && .0101 \\ 200 && .0053 \\ -100 && -.0099 \\ -200 && -.005\end{align*}

We observe that the values are getting closer to f(x)=0.\begin{align*}f(x) = 0.\end{align*} Hence limxf(x)=0\begin{align*}\lim_{x \to \infty}f(x) = 0\end{align*} and limxf(x)=0\begin{align*}\lim_{x \to -\infty}f(x) = 0\end{align*}.
Since our original function was roughly of the form f(x)=1x\begin{align*}f(x)=\frac{1}{x}\end{align*}, this enables us to determine limits for all other functions of the form \begin{align*}f(x)= \frac{1}{x^p} \end{align*} with \begin{align*}p > 0.\end{align*} Specifically, we are able to conclude that \begin{align*}\lim_{x \to \infty} \frac{1} {x^p} = 0\end{align*}. This shows how we can find infinite limits of functions by examining the end behavior of the function \begin{align*}f(x) = \frac{1}{x^p}, \; p > 0.\end{align*}
The following example shows how we can use this fact in evaluating limits of rational functions.

Example 1:

Find \begin{align*}\lim_{x \to \infty} \frac{2x^3 - x^2 + x - 1} {x^6 - x^5 + 3x^4 - 2x + 1}\end{align*}.

Solution:

Note that we have the indeterminate form, so Limit Property #5 does not hold. However, if we first divide both numerator and denominator by the quantity \begin{align*}x^6\end{align*}, we will then have a function of the form

\begin{align*}\frac{f(x)} {g(x)} = \frac{\frac{2x^3} {x^6} - \frac{x^2} {x^6} + \frac{x} {x^6} - \frac{1} {x^6}} {\frac{x^6} {x^6} - \frac{x^5} {x^6} + \frac{3x^4} {x^6} - \frac{2x} {x^6} + \frac{1} {x^6}} = \frac{\frac{2} {x^3} - \frac{1} {x^4} + \frac{1} {x^5} - \frac{1} {x^6}} {1 - \frac{1} {x} + \frac{3} {x^2} - \frac{2} {x^5} + \frac{1} {x^6}}.\end{align*}

We observe that the limits \begin{align*}\lim_{x \to \infty}\end{align*} \begin{align*}f(x)\end{align*} and \begin{align*}\lim_{x \to \infty}g(x)\end{align*} both exist. In particular, \begin{align*}\lim_{x \to \infty} f(x) = 0\end{align*} and \begin{align*}\lim_{x \to \infty} g(x) = 1\end{align*}. Hence Property #5 now applies and we have \begin{align*}\lim_{x \to \infty} \frac{2x^3 - x^2 + x - 1} {x^6 - x^5 + 3x^4 - 2x + 1} = \frac{0} {1} = 0\end{align*}.

## Lesson Summary

1. We learned to find infinite limits of functions.
2. We analyzed properties of infinite limits.
3. We identified asymptotes of functions.
4. We analyzed end behavior of functions.

## Multimedia Links

For more examples of limits at infinity (1.0), see Math Video Tutorials by James Sousa, Limits at Infinity (9:42).

## Review Questions

In problems 1 - 7, find the limits if they exist.

1. \begin{align*}\lim_{x \to 3^+} \frac{(x + 2)^2} {(x - 2)^2 - 1}\end{align*}
2. \begin{align*}\lim_{x \to \infty} \frac{(x + 2)^2} {(x - 2)^2 - 1}\end{align*}
3. \begin{align*}\lim_{x \to 1^+} \frac{(x + 2)^2} {(x - 2)^2 - 1}\end{align*}
4. \begin{align*}\lim_{x \to \infty} \frac{2x - 1} {x + 1}\end{align*}
5. \begin{align*}\lim_{x \to -\infty} \frac{x^5 + 3x^4 + 1} {x^3 - 1}\end{align*}
6. \begin{align*}\lim_{x \to \infty} \frac{3x^4 - 2x^2 + 3x + 1} {2x^4 - 2x^2 + x - 3}\end{align*}
7. \begin{align*}\lim_{x \to \infty} \frac{2x^2 - x + 3} {x^5 - 2x^3 + 2x - 3}\end{align*}

In problems 8 - 10, analyze the given function and identify all asymptotes and the end behavior of the graph.

1. \begin{align*}f(x) = \frac{(x + 4)^2} {(x - 4)^2 - 1}\end{align*}
2. \begin{align*}f(x) = -3x^3 - x^2 + 2x + 2\end{align*}
3. \begin{align*}f(x) = \frac{2x^2 - 8} {x + 2}\end{align*}
4. Consider \begin{align*}f(x) = \frac{1} {x + 1}, g(x) = x^2\end{align*}. We previously found \begin{align*}\lim_{x \to -1} (f \circ g) (x) = \frac{1} {2}\end{align*}. Find \begin{align*}\lim_{x \to -1} (g \circ f)(x).\end{align*}

## Texas Instruments Resources

In the CK-12 Texas Instruments Calculus FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/book/CK-12-Texas-Instruments-Calculus-Student-Edition/section/2.0/.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Show More

### Image Attributions

Show Hide Details
Description
Tags:
Subjects:
Grades:
Date Created:
Feb 23, 2012
Last Modified:
Jul 18, 2016
Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the section. Click Customize to make your own copy.
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original

CK.MAT.ENG.SE.1.Calculus.1.8
Here