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

Page 4 / 14

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 $f (x) = x^{n}$ where $n$ is a positive integer. From [link] and [link] , we see that

lim_{x \to \infty} x^{n} = \infty; n = 1, 2, 3,…

and

lim_{x \to − \infty} x^{n} = {\begin{matrix} \infty; n = 2, 4, 6,… \\ − \infty; n = 1, 3, 5,… \end{matrix} .

The functions x2, x4, and x6 are graphed, and it is apparent that as the exponent grows the functions increase more quickly. — For power functions with an even power of $n,$ $lim_{x \to \infty} x^{n} = \infty = lim_{x \to − \infty} x^{n} .$

The functions x, x3, and x5 are graphed, and it is apparent that as the exponent grows the functions increase more quickly. — For power functions with an odd power of $n,$ $lim_{x \to \infty} x^{n} = \infty$ and $lim_{x \to − \infty} x^{n} = − \infty .$

Using these facts, it is not difficult to evaluate $lim_{x \to \infty} c x^{n}$ and $lim_{x \to − \infty} c x^{n},$ 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

lim_{x \to \infty} c x^{n} = lim_{x \to \infty} x^{n} and lim_{x \to − \infty} c x^{n} = lim_{x \to − \infty} x^{n} if 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

lim_{x \to \infty} c x^{n} = − lim_{x \to \infty} x^{n} and lim_{x \to − \infty} c x^{n} = − lim_{x \to − \infty} x^{n} if c < 0 .

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

Limits at infinity for power functions

For each function $f,$ evaluate $lim_{x \to \infty} f (x)$ and $lim_{x \to − \infty} f (x) .$

$f (x) = −5 x^{3}$
$f (x) = 2 x^{4}$

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

Got questions? Get instant answers now!

Let $f (x) = −3 x^{4} .$ Find $lim_{x \to \infty} f (x) .$

$− \infty$

Got questions? Get instant answers now!

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

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + … + a_{1} x + a_{0}

of degree $n \geq 1$ so that $a_{n} \neq 0 .$ Factoring, we see that

f (x) = a_{n} x^{n} (1 + \frac{a_{n - 1}}{a_{n}} \frac{1}{x} + … + \frac{a_{1}}{a_{n}} \frac{1}{x^{n - 1}} + \frac{a_{0}}{a_{n}}) .

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

lim_{x \to ± \infty} f (x) = lim_{x \to ± \infty} a_{n} x^{n} .

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

Both functions f(x) = 5x3 – 3x2 + 4 and g(x) = 5x3 are plotted. Their behavior for large positive and large negative numbers converges. — The end behavior of a polynomial is determined by the behavior of the term with the largest exponent.

A polynomial’s end behavior is determined by the term with the largest exponent.
$x$	$10$	$100$	$1000$
$f (x) = 5 x^{3} - 3 x^{2} + 4$	$4704$	$4,970,004$	$4,997,000,004$
$g (x) = 5 x^{3}$	$5000$	$5,000,000$	$5,000,000,000$
$x$	$−10$	$−100$	$−1000$
$f (x) = 5 x^{3} - 3 x^{2} + 4$	$−5296$	$−5,029,996$	$−5,002,999,996$
$g (x) = 5 x^{3}$	$−5000$	$−5,000,000$	$−5,000,000,000$

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 $f (x) = \frac{p (x)}{q (x)}$ 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 $x \to − \infty .$ Then, use this information to describe the end behavior of the function.

$f (x) = \frac{3 x - 1}{2 x + 5}$ (Note: The degree of the numerator and the denominator are the same.)
$f (x) = \frac{3 x^{2} + 2 x}{4 x^{3} - 5 x + 7}$ (Note: The degree of numerator is less than the degree of the denominator.)
$f (x) = \frac{3 x^{2} + 4 x}{x + 2}$ (Note: The degree of numerator is greater than the degree of the denominator.)

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

$\begin{matrix} lim_{x \to ± \infty} \frac{3 x - 1}{2 x + 5} & = lim_{x \to ± \infty} \frac{3 - 1 / x}{2 + 5 / x} \\ = \frac{lim_{x \to ± \infty} (3 - 1 / x)}{lim_{x \to ± \infty} (2 + 5 / x)} \\ = \frac{lim_{x \to ± \infty} 3 - lim_{x \to ± \infty} 1 / x}{lim_{x \to ± \infty} 2 + lim_{x \to ± \infty} 5 / x} \\ = \frac{3 - 0}{2 + 0} = \frac{3}{2} . \end{matrix}$

Since $lim_{x \to ± \infty} f (x) = \frac{3}{2},$ we know that $y = \frac{3}{2}$ is a horizontal asymptote for this function as shown in the following graph.

The graph of this rational function approaches a horizontal asymptote as $x \to ± \infty .$
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

$lim_{x \to ± \infty} \frac{3 x^{2} + 2 x}{4 x^{3} - 5 x + 7} = lim_{x \to ± \infty} \frac{3 / x + 2 / x^{2}}{4 - 5 / x^{2} + 7 / x^{3}} = \frac{3.0 + 2.0}{4 - 5.0 + 7.0} = 0 .$

Therefore $f$ has a horizontal asymptote of $y = 0$ as shown in the following graph.

The graph of this rational function approaches the horizontal asymptote $y = 0$ as $x \to ± \infty .$
Dividing the numerator and denominator by $x,$ we have

$lim_{x \to ± \infty} \frac{3 x^{2} + 4 x}{x + 2} = lim_{x \to ± \infty} \frac{3 x + 4}{1 + 2 / x} .$

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

As $x \to \infty,$ the values $f (x) \to \infty .$ As $x \to − \infty,$ the values $f (x) \to − \infty .$

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Practice Key Terms 5

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?

Ask

©flickr: Rod	Medieval Africa By Jesenia Wofford Start Quiz
	Respiratory System Drugs By CB Biern Start Quiz
	27 AP Key Terms 27 The Reproductive System By OpenStax Start Key Terms
	17 AP 17 Endocrine System MCQ By OpenStax Start Quiz
	11 Clin Sci I - GI Twedt quiz 2 By Brooke Delaney Start Exam
	3 Arts Society: Theater 3 By Jonathan Long Start Quiz
	Chemistry Practice Test By Sandhills MLT Start Test
	English Composition 2 Final Practice By Madison Christian Start Test
©flickr: Abraham	Multicellular Organisms Test By Monty Hartfield Start Test
	16 Dr. Amberg Pharm quiz 2 By Brooke Delaney Start Exam