Limits at Infinity and Asymptotes: Learn It 4

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Limits at Infinity and Asymptotes: Learn It 4

End Behavior Cont.

End Behavior for Algebraic Functions

The end behavior for rational functions and functions involving radicals is a little more complicated than for polynomials.

Note that this is not your first encounter with horizontal asymptotes. It may be helpful to recall what you already know about them.

The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.

Case 1: Degree of numerator is less than degree of denominator: horizontal asymptote at [latex]y=0[/latex]
Case 2: Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote.
- If the degree of the numerator is greater than the degree of the denominator by more than one, the end behavior of the function’s graph will mimic that of the graph of the reduced ratio of leading terms.
Case 3: Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.

In the example below, we show that the limits at infinity of a rational function [latex]f(x)=\frac{p(x)}{q(x)}[/latex] 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 [latex]x[/latex] appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of [latex]x[/latex].

For each of the following functions, determine the limits as [latex]x\to \infty[/latex] and [latex]x\to −\infty[/latex]. Then, use this information to describe the end behavior of the function.

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

The highest power of [latex]x[/latex] in the denominator is [latex]x[/latex]. Therefore, dividing the numerator and denominator by [latex]x[/latex] and applying the algebraic limit laws, we see that,
[latex]\begin{array}{ll} \underset{x\to \pm \infty }{\lim}\frac{3x-1}{2x+5} & =\underset{x\to \pm \infty }{\lim}\frac{3-1/x}{2+5/x} \\ & =\frac{\underset{x\to \pm \infty }{\lim}(3-1/x)}{\underset{x\to \pm \infty }{\lim}(2+5/x)} \\ & =\frac{\underset{x\to \pm \infty }{\lim}3-\underset{x\to \pm \infty }{\lim}1/x}{\underset{x\to \pm \infty }{\lim}2+\underset{x\to \pm \infty }{\lim}5/x} \\ & =\frac{3-0}{2+0}=\frac{3}{2}. \end{array}[/latex]

Since [latex]\underset{x\to \pm \infty }{\lim}f(x)=\frac{3}{2}[/latex], we know that [latex]y=\frac{3}{2}[/latex] is a horizontal asymptote for this function as shown in the following graph.

Figure 14. The graph of this rational function approaches a horizontal asymptote as [latex]x\to \pm \infty[/latex].
Since the largest power of [latex]x[/latex] appearing in the denominator is [latex]x^3[/latex], divide the numerator and denominator by [latex]x^3[/latex]. After doing so and applying algebraic limit laws, we obtain,
[latex]\underset{x\to \pm \infty }{\lim}\frac{3x^2+2x}{4x^3-5x+7}=\underset{x\to \pm \infty }{\lim}\frac{3/x+2/x^2}{4-5/x^2+7/x^3}=\frac{3(0)+2(0)}{4-5(0)+7(0)}=0[/latex]

Therefore [latex]f[/latex] has a horizontal asymptote of [latex]y=0[/latex] as shown in the following graph.

Figure 15. The graph of this rational function approaches the horizontal asymptote [latex]y=0[/latex] as [latex]x\to \pm \infty[/latex].
Dividing the numerator and denominator by [latex]x[/latex], we have,
[latex]\underset{x\to \pm \infty }{\lim}\frac{3x^2+4x}{x+2}=\underset{x\to \pm \infty }{\lim}\frac{3x+4}{1+2/x}[/latex].

As [latex]x\to \pm \infty[/latex], the denominator approaches 1. As [latex]x\to \infty[/latex], the numerator approaches [latex]+\infty[/latex]. As [latex]x\to −\infty[/latex], the numerator approaches [latex]−\infty[/latex]. Therefore [latex]\underset{x\to \infty }{\lim}f(x)=\infty[/latex], whereas [latex]\underset{x\to −\infty }{\lim}f(x)=−\infty[/latex] as shown in the following figure.

Figure 16. As [latex]x\to \infty[/latex], the values [latex]f(x)\to \infty[/latex]. As [latex]x\to −\infty[/latex], the values [latex]f(x)\to −\infty[/latex].

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “4.6 Limits at Infinity and Asymptotes” here (opens in new window).

Before proceeding, consider the graph of [latex]f(x)=\frac{(3x^2+4x)}{(x+2)}[/latex] shown below.

The function f(x) = (3x2 + 4x)/(x + 2) is plotted as is its diagonal asymptote y = 3x – 2. — Figure 17. The graph of the rational function [latex]f(x)=(3x^2+4x)/(x+2)[/latex] approaches the oblique asymptote [latex]y=3x-2[/latex] as [latex]x\to \pm \infty[/latex].

As [latex]x\to \infty[/latex] and [latex]x\to −\infty[/latex], the graph of [latex]f[/latex] appears almost linear. Although [latex]f[/latex] is certainly not a linear function, we now investigate why the graph of [latex]f[/latex] seems to be approaching a linear function.

First, using long division of polynomials, we can write,

[latex]f(x)=\frac{3x^2+4x}{x+2}=3x-2+\frac{4}{x+2}[/latex]

Since [latex]\frac{4}{(x+2)}\to 0[/latex] as [latex]x\to \pm \infty[/latex], we conclude that,

[latex]\underset{x\to \pm \infty }{\lim}(f(x)-(3x-2))=\underset{x\to \pm \infty }{\lim}\frac{4}{x+2}=0[/latex]

Therefore, the graph of [latex]f[/latex] approaches the line [latex]y=3x-2[/latex] as [latex]x\to \pm \infty[/latex]. This line is known as an oblique asymptote for [latex]f[/latex].

We can summarize the results of the example above to make the following conclusion regarding end behavior for rational functions.

end behavior for rational functions

Consider a rational function

[latex]f(x)=\frac{p(x)}{q(x)}=\frac{a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0}{b_m x^m + b_{m-1} x^{m-1} + \cdots + b_1 x + b_0}[/latex],

where [latex]a_n\ne 0[/latex] and [latex]b_m \ne 0[/latex].

If the degree of the numerator is the same as the degree of the denominator [latex](n=m)[/latex], then [latex]f[/latex] has a horizontal asymptote of [latex]y=a_n/b_m[/latex] as [latex]x\to \pm \infty[/latex].
If the degree of the numerator is less than the degree of the denominator [latex](n < m)[/latex], then [latex]f[/latex] has a horizontal asymptote of [latex]y=0[/latex] as [latex]x\to \pm \infty[/latex].
If the degree of the numerator is greater than the degree of the denominator [latex](n>m)[/latex], then [latex]f[/latex] does not have a horizontal asymptote. The limits at infinity are either positive or negative infinity, depending on the signs of the leading terms.

In addition, using long division, the function can be rewritten as

[latex]f(x)=\frac{p(x)}{q(x)}=g(x)+\frac{r(x)}{q(x)}[/latex],

where the degree of [latex]r(x)[/latex] is less than the degree of [latex]q(x)[/latex]. As a result, [latex]\underset{x\to \pm \infty }{\lim}r(x)/q(x)=0[/latex].

Therefore, the values of [latex][f(x)-g(x)][/latex] approach zero as [latex]x\to \pm \infty[/latex].

If the degree of [latex]p(x)[/latex] is exactly one more than the degree of [latex]q(x)[/latex] [latex](n=m+1)[/latex], the function [latex]g(x)[/latex] is a linear function. In this case, we call [latex]g(x)[/latex] an oblique asymptote.

Find the limits as [latex]x\to \infty[/latex] and [latex]x\to −\infty[/latex] for [latex]f(x)=\frac{3x-2}{\sqrt{4x^2+5}}[/latex] and describe the end behavior of [latex]f[/latex].