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 $lim_{x \to − \infty} 2 x^{4} = \infty .$

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

$− \infty$

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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 $f (x) \to − \infty .$

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

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