The behavior of a function as [latex]x\to \pm \infty[/latex] is called the function’s end behavior. At each of the function’s ends, the function could exhibit one of the following types of behavior:

1. The function [latex]f(x)[/latex] approaches a horizontal asymptote [latex]y=L[/latex].
2. The function [latex]f(x)\to \infty[/latex] or [latex]f(x)\to −\infty[/latex].
3. The function does not approach a finite limit, nor does it approach [latex]\infty[/latex] or [latex]−\infty[/latex]. In this case, the function may have some oscillatory behavior.

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 [latex]f(x)=x^n[/latex] where [latex]n[/latex] is a positive integer. From Figure 11 and Figure 12, we see that

and

Figure 11. For power functions with an even power of [latex]n[/latex], [latex]\underset{x\to \infty }{\lim}x^n=\infty =\underset{x\to −\infty }{\lim}x^n[/latex].

 

Figure 12. For power functions with an odd power of [latex]n[/latex], [latex]\underset{x\to \infty }{\lim}x^n=\infty [/latex] and [latex]\underset{x\to −\infty}{\lim}x^n=−\infty [/latex].

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

If [latex]c<0[/latex], the graph of [latex]y=cx^n[/latex] is a vertical stretch or compression combined with a reflection about the [latex]x[/latex]-axis, and therefore

If [latex]c=0, \, y=cx^n=0[/latex], in which case [latex]\underset{x\to \infty }{\lim}cx^n=0=\underset{x\to −\infty }{\lim}cx^n[/latex].

Watch the following video to see the worked solution to Example: Limits at Infinity for Power Functions.

We now look at how the limits at infinity for power functions can be used to determine [latex]\underset{x\to \pm \infty }{\lim}f(x)[/latex] for any polynomial function [latex]f[/latex]. Consider a polynomial function

of degree [latex]n \ge 1[/latex] so that [latex]a_n \ne 0[/latex]. Factoring, we see that

As [latex]x\to \pm \infty[/latex], all the terms inside the parentheses approach zero except the first term. We conclude that

For example, the function [latex]f(x)=5x^3-3x^2+4[/latex] behaves like [latex]g(x)=5x^3[/latex] as [latex]x\to \pm \infty[/latex] as shown in Figure 13 and Table 1.

Figure 13. The end behavior of a polynomial is determined by the behavior of the term with the largest exponent.

Before proceeding, consider the graph of [latex]f(x)=\frac{(3x^2+4x)}{(x+2)}[/latex] shown in Figure 17. 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

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

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] (Figure 17).

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].

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

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

1. 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].
2. If the degree of the numerator is less than the degree of the denominator [latex](n
3. 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.
   Now let’s consider the end behavior for functions involving a radical.

Example: Determining End Behavior for a Function Involving a Radical

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].

Show Solution

Let’s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of [latex]x[/latex]. To determine the appropriate power of [latex]x[/latex], consider the expression [latex]\sqrt{4x^2+5}[/latex] in the denominator. Since

[latex]\sqrt{4x^2+5}\approx \sqrt{4x^2}=2|x|[/latex]

for large values of [latex]x[/latex] in effect [latex]x[/latex] appears just to the first power in the denominator. Therefore, we divide the numerator and denominator by [latex]|x|[/latex]. Then, using the fact that [latex]|x|=x[/latex] for [latex]x>0[/latex], [latex]|x|=−x[/latex] for [latex]x<0[/latex], and [latex]|x|=\sqrt{x^2}[/latex] for all [latex]x[/latex], we calculate the limits as follows:

[latex]\begin{array}{lll} \underset{x\to \infty }{\lim}\frac{3x-2}{\sqrt{4x^2+5}} & = & \underset{x\to \infty }{\lim}\frac{(1/|x|)(3x-2)}{(1/|x|)\sqrt{4x^2+5}} \\ & = & \underset{x\to \infty }{\lim}\frac{(1/x)(3x-2)}{\sqrt{(1/x^2)(4x^2+5)}} \\ & = & \underset{x\to \infty }{\lim}\frac{3-2/x}{\sqrt{4+5/x^2}}=\frac{3}{\sqrt{4}}=\frac{3}{2} \\ \underset{x\to −\infty }{\lim}\frac{3x-2}{\sqrt{4x^2+5}} & = & \underset{x\to −\infty }{\lim}\frac{(1/|x|)(3x-2)}{(1/|x|)\sqrt{4x^2+5}} \\ & = & \underset{x\to −\infty }{\lim}\frac{(-1/x)(3x-2)}{\sqrt{(1/x^2)(4x^2+5)}} \\ & = & \underset{x\to −\infty }{\lim}\frac{-3+2/x}{\sqrt{4+5/x^2}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}. \end{array}[/latex]

 

Therefore, [latex]f(x)[/latex] approaches the horizontal asymptote [latex]y=\frac{3}{2}[/latex] as [latex]x\to \infty[/latex] and the horizontal asymptote [latex]y=-\frac{3}{2}[/latex] as [latex]x\to −\infty[/latex] as shown in the following graph.

 

Figure 18. This function has two horizontal asymptotes and it crosses one of the asymptotes.

Figure 19. The function [latex]f(x)= \sin x[/latex] oscillates between 1 and -1 as [latex]x\to \pm \infty [/latex]

 

Figure 20. The function [latex]f(x)= \tan x[/latex] does not approach a limit and does not approach [latex]\pm \infty [/latex] as [latex]x\to \pm \infty [/latex]

Recall that for any base [latex]b>0, \, b\ne 1[/latex], the function [latex]y=b^x[/latex] is an exponential function with domain [latex](−\infty ,\infty )[/latex] and range [latex](0,\infty )[/latex]. If [latex]b>1, \, y=b^x[/latex] is increasing over [latex](−\infty ,\infty )[/latex]. If [latex]01[/latex]. Therefore, [latex]f(x)=e^x[/latex] is increasing on [latex](−\infty ,\infty )[/latex] and the range is [latex](0,\infty)[/latex]. The exponential function [latex]f(x)=e^x[/latex] approaches [latex]\infty[/latex] as [latex]x\to \infty[/latex] and approaches 0 as [latex]x\to −\infty[/latex] as shown in (Figure) and (Figure).

End behavior of the natural exponential function

[latex]x[/latex]	-5	-2	0	2	5
[latex]e^x[/latex]	0.00674	0.135	1	7.389	148.413

Figure 21. The exponential function approaches zero as [latex]x\to −\infty [/latex] and approaches [latex]\infty [/latex] as [latex]x\to \infty[/latex].

Recall that the natural logarithm function [latex]f(x)=\ln (x)[/latex] is the inverse of the natural exponential function [latex]y=e^x[/latex]. Therefore, the domain of [latex]f(x)=\ln (x)[/latex] is [latex](0,\infty )[/latex] and the range is [latex](−\infty ,\infty )[/latex]. The graph of [latex]f(x)=\ln (x)[/latex] is the reflection of the graph of [latex]y=e^x[/latex] about the line [latex]y=x[/latex]. Therefore, [latex]\ln (x)\to −\infty[/latex] as [latex]x\to 0^+[/latex] and [latex]\ln (x)\to \infty[/latex] as [latex]x\to \infty[/latex] as shown in (Figure) and (Figure).

End behavior of the natural logarithm function

[latex]x[/latex]	0.01	0.1	1	10	100
[latex]\ln (x)[/latex]	-4.605	-2.303	0	2.303	4.605

Figure 22. The natural logarithm function approaches [latex]\infty [/latex] as [latex]x\to \infty[/latex].

example: Determining End Behavior for a Transcendental Function

Find the limits as [latex]x\to \infty[/latex] and [latex]x\to −\infty[/latex] for [latex]f(x)=\frac{(2+3e^x)}{(7-5e^x)}[/latex] and describe the end behavior of [latex]f[/latex].

Show Solution

To find the limit as [latex]x\to \infty[/latex], divide the numerator and denominator by [latex]e^x[/latex]:

[latex]\begin{array}{ll} \underset{x\to \infty }{\lim}f(x) & =\underset{x\to \infty }{\lim}\frac{2+3e^x}{7-5e^x} \\ & =\underset{x\to \infty }{\lim}\frac{(2/e^x)+3}{(7/e^x)-5}. \end{array}[/latex]

As shown in Figure 21, [latex]e^x\to \infty[/latex] as [latex]x\to \infty[/latex]. Therefore,

[latex]\underset{x\to \infty }{\lim}\frac{2}{e^x}=0=\underset{x\to \infty }{\lim}\frac{7}{e^x}[/latex].

 

We conclude that [latex]\underset{x\to \infty }{\lim}f(x)=-\frac{3}{5}[/latex], and the graph of [latex]f[/latex] approaches the horizontal asymptote [latex]y=-\frac{3}{5}[/latex] as [latex]x\to \infty[/latex]. To find the limit as [latex]x\to −\infty[/latex], use the fact that [latex]e^x \to 0[/latex] as [latex]x\to −\infty[/latex] to conclude that [latex]\underset{x\to -\infty }{\lim}f(x)=\frac{2}{7}[/latex], and therefore the graph of approaches the horizontal asymptote [latex]y=\frac{2}{7}[/latex] as [latex]x\to −\infty[/latex].

Watch the following video to see the worked solution to Example: Determining End Behavior for a Transcendental Function.