# 4.6 Limits at infinity and asymptotes  (Page 4/14)

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Let’s consider several classes of functions here and look at the different types of end behaviors for these functions.

## End behavior for polynomial functions

Consider the power function $f\left(x\right)={x}^{n}$ where $n$ is a positive integer. From [link] and [link] , we see that

$\underset{x\to \infty }{\text{lim}}{x}^{n}=\infty ;n=1,2,3\text{,…}$

and

$\underset{x\to \text{−}\infty }{\text{lim}}{x}^{n}=\left\{\begin{array}{c}\infty ;n=2,4,6\text{,…}\hfill \\ \text{−}\infty ;n=1,3,5\text{,…}\hfill \end{array}.$

Using these facts, it is not difficult to evaluate $\underset{x\to \infty }{\text{lim}}c{x}^{n}$ and $\underset{x\to \text{−}\infty }{\text{lim}}c{x}^{n},$ where $c$ is any constant and $n$ is a positive integer. If $c>0,$ the graph of $y=c{x}^{n}$ is a vertical stretch or compression of $y={x}^{n},$ and therefore

$\underset{x\to \infty }{\text{lim}}c{x}^{n}=\underset{x\to \infty }{\text{lim}}{x}^{n}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to \text{−}\infty }{\text{lim}}c{x}^{n}=\underset{x\to \text{−}\infty }{\text{lim}}{x}^{n}\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}c>0.$

If $c<0,$ the graph of $y=c{x}^{n}$ is a vertical stretch or compression combined with a reflection about the $x$ -axis, and therefore

$\underset{x\to \infty }{\text{lim}}c{x}^{n}=\text{−}\underset{x\to \infty }{\text{lim}}{x}^{n}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to \text{−}\infty }{\text{lim}}c{x}^{n}=\text{−}\underset{x\to \text{−}\infty }{\text{lim}}{x}^{n}\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}c<0.$

If $c=0,y=c{x}^{n}=0,$ in which case $\underset{x\to \infty }{\text{lim}}c{x}^{n}=0=\underset{x\to \text{−}\infty }{\text{lim}}c{x}^{n}.$

## Limits at infinity for power functions

For each function $f,$ evaluate $\underset{x\to \infty }{\text{lim}}f\left(x\right)$ and $\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right).$

1. $f\left(x\right)=-5{x}^{3}$
2. $f\left(x\right)=2{x}^{4}$
1. Since the coefficient of ${x}^{3}$ is $-5,$ the graph of $f\left(x\right)=-5{x}^{3}$ involves a vertical stretch and reflection of the graph of $y={x}^{3}$ about the $x$ -axis. Therefore, $\underset{x\to \infty }{\text{lim}}\left(-5{x}^{3}\right)=\text{−}\infty$ and $\underset{x\to \text{−}\infty }{\text{lim}}\left(-5{x}^{3}\right)=\infty .$
2. Since the coefficient of ${x}^{4}$ is $2,$ the graph of $f\left(x\right)=2{x}^{4}$ is a vertical stretch of the graph of $y={x}^{4}.$ Therefore, $\underset{x\to \infty }{\text{lim}}2{x}^{4}=\infty$ and $\underset{x\to \text{−}\infty }{\text{lim}}2{x}^{4}=\infty .$

Let $f\left(x\right)=-3{x}^{4}.$ Find $\underset{x\to \infty }{\text{lim}}f\left(x\right).$

$\text{−}\infty$

We now look at how the limits at infinity for power functions can be used to determine $\underset{x\to \text{±}\infty }{\text{lim}}f\left(x\right)$ for any polynomial function $f.$ Consider a polynomial function

$f\left(x\right)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\text{…}+{a}_{1}x+{a}_{0}$

of degree $n\ge 1$ so that ${a}_{n}\ne 0.$ Factoring, we see that

$f\left(x\right)={a}_{n}{x}^{n}\left(1+\frac{{a}_{n-1}}{{a}_{n}}\phantom{\rule{0.2em}{0ex}}\frac{1}{x}+\text{…}+\frac{{a}_{1}}{{a}_{n}}\phantom{\rule{0.2em}{0ex}}\frac{1}{{x}^{n-1}}+\frac{{a}_{0}}{{a}_{n}}\right).$

As $x\to \text{±}\infty ,$ all the terms inside the parentheses approach zero except the first term. We conclude that

$\underset{x\to \text{±}\infty }{\text{lim}}f\left(x\right)=\underset{x\to \text{±}\infty }{\text{lim}}{a}_{n}{x}^{n}.$

For example, the function $f\left(x\right)=5{x}^{3}-3{x}^{2}+4$ behaves like $g\left(x\right)=5{x}^{3}$ as $x\to \text{±}\infty$ as shown in [link] and [link] .

 $x$ $10$ $100$ $1000$ $f\left(x\right)=5{x}^{3}-3{x}^{2}+4$ $4704$ $4,970,004$ $4,997,000,004$ $g\left(x\right)=5{x}^{3}$ $5000$ $5,000,000$ $5,000,000,000$ $x$ $-10$ $-100$ $-1000$ $f\left(x\right)=5{x}^{3}-3{x}^{2}+4$ $-5296$ $-5,029,996$ $-5,002,999,996$ $g\left(x\right)=5{x}^{3}$ $-5000$ $-5,000,000$ $-5,000,000,000$

## End behavior for algebraic functions

The end behavior for rational functions and functions involving radicals is a little more complicated than for polynomials. In [link] , we show that the limits at infinity of a rational function $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ depend on the relationship between the degree of the numerator and the degree of the denominator. To evaluate the limits at infinity for a rational function, we divide the numerator and denominator by the highest power of $x$ appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of $x.$

## Determining end behavior for rational functions

For each of the following functions, determine the limits as $x\to \infty$ and $x\to \text{−}\infty .$ Then, use this information to describe the end behavior of the function.

1. $f\left(x\right)=\frac{3x-1}{2x+5}$ (Note: The degree of the numerator and the denominator are the same.)
2. $f\left(x\right)=\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}$ (Note: The degree of numerator is less than the degree of the denominator.)
3. $f\left(x\right)=\frac{3{x}^{2}+4x}{x+2}$ (Note: The degree of numerator is greater than the degree of the denominator.)
1. The highest power of $x$ in the denominator is $x.$ Therefore, dividing the numerator and denominator by $x$ and applying the algebraic limit laws, we see that
$\begin{array}{cc}\hfill \underset{x\to \text{±}\infty }{\text{lim}}\frac{3x-1}{2x+5}& =\underset{x\to \text{±}\infty }{\text{lim}}\frac{3-1\text{/}x}{2+5\text{/}x}\hfill \\ & =\frac{\underset{x\to \text{±}\infty }{\text{lim}}\left(3-1\text{/}x\right)}{\underset{x\to \text{±}\infty }{\text{lim}}\left(2+5\text{/}x\right)}\hfill \\ & =\frac{\underset{x\to \text{±}\infty }{\text{lim}}3-\underset{x\to \text{±}\infty }{\text{lim}}1\text{/}x}{\underset{x\to \text{±}\infty }{\text{lim}}2+\underset{x\to \text{±}\infty }{\text{lim}}5\text{/}x}\hfill \\ & =\frac{3-0}{2+0}=\frac{3}{2}.\hfill \end{array}$

Since $\underset{x\to \text{±}\infty }{\text{lim}}f\left(x\right)=\frac{3}{2},$ we know that $y=\frac{3}{2}$ is a horizontal asymptote for this function as shown in the following graph.
2. Since the largest power of $x$ appearing in the denominator is ${x}^{3},$ divide the numerator and denominator by ${x}^{3}.$ After doing so and applying algebraic limit laws, we obtain
$\underset{x\to \text{±}\infty }{\text{lim}}\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}=\underset{x\to \text{±}\infty }{\text{lim}}\frac{3\text{/}x+2\text{/}{x}^{2}}{4-5\text{/}{x}^{2}+7\text{/}{x}^{3}}=\frac{3.0+2.0}{4-5.0+7.0}=0.$

Therefore $f$ has a horizontal asymptote of $y=0$ as shown in the following graph.
3. Dividing the numerator and denominator by $x,$ we have
$\underset{x\to \text{±}\infty }{\text{lim}}\frac{3{x}^{2}+4x}{x+2}=\underset{x\to \text{±}\infty }{\text{lim}}\frac{3x+4}{1+2\text{/}x}.$

As $x\to \text{±}\infty ,$ the denominator approaches $1.$ As $x\to \infty ,$ the numerator approaches $+\infty .$ As $x\to \text{−}\infty ,$ the numerator approaches $\text{−}\infty .$ Therefore $\underset{x\to \infty }{\text{lim}}f\left(x\right)=\infty ,$ whereas $\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=\text{−}\infty$ as shown in the following figure.

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