Limits at Infinity and Horizontal Asymptotes

Recall that [latex]\underset{x\to a}{\text{lim}}f(x)=L[/latex] means [latex]f(x)[/latex] becomes arbitrarily close to [latex]L[/latex] as long as [latex]x[/latex] is sufficiently close to [latex]a.[/latex] We can extend this idea to limits at infinity. For example, consider the function [latex]f(x)=2+\frac{1}{x}.[/latex] As can be seen graphically in (Figure) and numerically in (Figure) , as the values of [latex]x[/latex] get larger, the values of [latex]f(x)[/latex] approach 2. We say the limit as [latex]x[/latex] approaches [latex]\infty[/latex] of [latex]f(x)[/latex] is 2 and write [latex]\underset{x\to \infty }{\text{lim}}f(x)=2.[/latex] Similarly, for [latex]x < 0,[/latex] as the values [latex]|x|[/latex] get larger, the values of [latex]f(x)[/latex] approaches 2. We say the limit as [latex]x[/latex] approaches [latex]-\infty[/latex] of [latex]f(x)[/latex] is 2 and write [latex]\underset{x\to a}{\text{lim}}f(x)=2.[/latex]

More generally, for any function [latex]f,[/latex] we say the limit as [latex]x\to \infty[/latex] of [latex]f(x)[/latex] is [latex]L[/latex] if [latex]f(x)[/latex] becomes arbitrarily close to [latex]L[/latex] as long as [latex]x[/latex] is sufficiently large. In that case, we write [latex]\underset{x\to a}{\text{lim}}f(x)=L.[/latex] Similarly, we say the limit as [latex]x\to -\infty[/latex] of [latex]f(x)[/latex] is [latex]L[/latex] if [latex]f(x)[/latex] becomes arbitrarily close to [latex]L[/latex] as long as [latex]x < 0[/latex] and [latex]|x|[/latex] is sufficiently large. In that case, we write [latex]\underset{x\to -\infty }{\text{lim}}f(x)=L.[/latex] We now look at the definition of a function having a limit at infinity.

If the values [latex]f(x)[/latex] are getting arbitrarily close to some finite value [latex]L[/latex] as [latex]x\to \infty[/latex] or [latex]x\to -\infty ,[/latex] the graph of [latex]f[/latex] approaches the line [latex]y=L.[/latex] In that case, the line [latex]y=L[/latex] is a horizontal asymptote of [latex]f[/latex] ( (Figure) ). For example, for the function [latex]f(x)=\frac{1}{x},[/latex] since [latex]\underset{x\to \infty }{\text{lim}}f(x)=0,[/latex] the line [latex]y=0[/latex] is a horizontal asymptote of [latex]f(x)=\frac{1}{x}.[/latex]

A function cannot cross a vertical asymptote because the graph must approach infinity (or [latex]-\infty )[/latex] from at least one direction as [latex]x[/latex] approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function [latex]f(x)=\frac{( \cos x)}{x}+1[/latex] shown in (Figure) intersects the horizontal asymptote [latex]y=1[/latex] an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.

The algebraic limit laws and squeeze theorem we introduced in Introduction to Limits also apply to limits at infinity. We illustrate how to use these laws to compute several limits at infinity.

Computing Limits at Infinity

For each of the following functions [latex]f,[/latex] evaluate [latex]\underset{x\to \infty }{\text{lim}}f(x)[/latex] and [latex]\underset{x\to -\infty }{\text{lim}}f(x).[/latex] Determine the horizontal asymptote(s) for [latex]f.[/latex]

1. [latex]f(x)=5-\frac{2}{{x}^{2}}[/latex]
2. [latex]f(x)=\frac{ \sin x}{x}[/latex]
3. [latex]f(x)={ \tan }^{-1}(x)[/latex]

Solution

1. Using the algebraic limit laws, we have [latex]\underset{x\to \infty }{\text{lim}}(5-\frac{2}{{x}^{2}})=\underset{x\to \infty }{\text{lim}}5-2(\underset{x\to \infty }{\text{lim}}\frac{1}{x}).(\underset{x\to \infty }{\text{lim}}\frac{1}{x})=5-2\cdot 0=5.[/latex]Similarly, [latex]\underset{x\to \infty }{\text{lim}}f(x)=5.[/latex]Therefore, [latex]f(x)=5-\frac{2}{{x}^{2}}[/latex] has a horizontal asymptote of [latex]y=5[/latex] and [latex]f[/latex] approaches this horizontal asymptote as [latex]x\to \pm \infty[/latex] as shown in the following graph.

   

2. Since [latex]-1\le \sin x\le 1[/latex] for all [latex]x,[/latex] we have
   [latex]\frac{-1}{x}\le \frac{ \sin x}{x}\le \frac{1}{x}[/latex]

   for all [latex]x\ne 0.[/latex] Also, since

   [latex]\underset{x\to \infty }{\text{lim}}\frac{-1}{x}=0=\underset{x\to \infty }{\text{lim}}\frac{1}{x},[/latex]

   we can apply the squeeze theorem to conclude that

   [latex]\underset{x\to \infty }{\text{lim}}\frac{ \sin x}{x}=0.[/latex]

   Similarly,

   [latex]\underset{x\to -\infty }{\text{lim}}\frac{ \sin x}{x}=0.[/latex]

    

   Thus, [latex]f(x)=\frac{ \sin x}{x}[/latex] has a horizontal asymptote of [latex]y=0[/latex] and [latex]f(x)[/latex] approaches this horizontal asymptote as [latex]x\to \pm \infty[/latex] as shown in the following graph.

   

3. To evaluate [latex]\underset{x\to \infty }{\text{lim}}{ \tan }^{-1}(x)[/latex] and [latex]\underset{x\to -\infty }{\text{lim}}{ \tan }^{-1}(x),[/latex] we first consider the graph of [latex]y= \tan (x)[/latex] over the interval [latex](-\pi \text{/}2,\pi \text{/}2)[/latex] as shown in the following graph.
   
   The graph of [latex]\tan x[/latex] has vertical asymptotes at [latex]x=\pm \frac{\pi }{2}[/latex]

Since

[latex]\underset{x\to {(\pi \text{/}2)}^{-}}{\text{lim}} \tan x=\infty ,[/latex]

it follows that

[latex]\underset{x\to \infty }{\text{lim}}{ \tan }^{-1}(x)=\frac{\pi }{2}.[/latex]

Similarly, since

[latex]\underset{x\to {(\pi \text{/}2)}^{+}}{\text{lim}} \tan x=-\infty ,[/latex]

it follows that

[latex]\underset{x\to -\infty }{\text{lim}}{ \tan }^{-1}(x)=-\frac{\pi }{2}.[/latex]

As a result, [latex]y=\frac{\pi }{2}[/latex] and [latex]y=-\frac{\pi }{2}[/latex] are horizontal asymptotes of [latex]f(x)={ \tan }^{-1}(x)[/latex] as shown in the following graph.

Infinite Limits at Infinity

Sometimes the values of a function [latex]f[/latex] become arbitrarily large as [latex]x\to \infty[/latex] (or as [latex]x\to -\infty ).[/latex] In this case, we write [latex]\underset{x\to \infty }{\text{lim}}f(x)=\infty[/latex] (or [latex]\underset{x\to -\infty }{\text{lim}}f(x)=\infty ).[/latex] On the other hand, if the values of [latex]f[/latex] are negative but become arbitrarily large in magnitude as [latex]x\to \infty[/latex] (or as [latex]x\to -\infty ),[/latex] we write [latex]\underset{x\to \infty }{\text{lim}}f(x)=-\infty[/latex] (or [latex]\underset{x\to -\infty }{\text{lim}}f(x)=-\infty ).[/latex]

For example, consider the function [latex]f(x)={x}^{3}.[/latex] As seen in (Figure) and (Figure) , as [latex]x\to \infty[/latex] the values [latex]f(x)[/latex] become arbitrarily large. Therefore, [latex]\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty .[/latex] On the other hand, as [latex]x\to -\infty ,[/latex] the values of [latex]f(x)={x}^{3}[/latex] are negative but become arbitrarily large in magnitude. Consequently, [latex]\underset{x\to -\infty }{\text{lim}}{x}^{3}=-\infty .[/latex]

Formal Definitions

Earlier, we used the terms arbitrarily close , arbitrarily large , and sufficiently large to define limits at infinity informally. Although these terms provide accurate descriptions of limits at infinity, they are not precise mathematically. Here are more formal definitions of limits at infinity. We then look at how to use these definitions to prove results involving limits at infinity.

Earlier in this section, we used graphical evidence in (Figure) and numerical evidence in (Figure) to conclude that [latex]\underset{x\to \infty }{\text{lim}}(\frac{2+1}{x})=2.[/latex] Here we use the formal definition of limit at infinity to prove this result rigorously.

We now turn our attention to a more precise definition for an infinite limit at infinity.

Earlier, we used graphical evidence ( (Figure) ) and numerical evidence ( (Figure) ) to conclude that [latex]\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty .[/latex] Here we use the formal definition of infinite limit at infinity to prove that result.

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) and (Figure) , we see that

and

Using these facts, it is not difficult to evaluate [latex]\underset{x\to \infty }{\text{lim}}c{x}^{n}[/latex] and [latex]\underset{x\to -\infty }{\text{lim}}c{x}^{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=c{x}^{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=c{x}^{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=c{x}^{n}=0,[/latex] in which case [latex]\underset{x\to \infty }{\text{lim}}c{x}^{n}=0=\underset{x\to -\infty }{\text{lim}}c{x}^{n}.[/latex]

End Behavior for Algebraic Functions

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

Determining End Behavior for Rational Functions

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.

1. [latex]f(x)=\frac{3x-1}{2x+5}[/latex]
2. [latex]f(x)=\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}[/latex]
3. [latex]f(x)=\frac{3{x}^{2}+4x}{x+2}[/latex]

Solution

1. 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}{cc}\hfill \underset{x\to \pm \infty }{\text{lim}}\frac{3x-1}{2x+5}& =\underset{x\to \pm \infty }{\text{lim}}\frac{3-1\text{/}x}{2+5\text{/}x}\hfill \\ & =\frac{\underset{x\to \pm \infty }{\text{lim}}(3-1\text{/}x)}{\underset{x\to \pm \infty }{\text{lim}}(2+5\text{/}x)}\hfill \\ & =\frac{\underset{x\to \pm \infty }{\text{lim}}3-\underset{x\to \pm \infty }{\text{lim}}1\text{/}x}{\underset{x\to \pm \infty }{\text{lim}}2+\underset{x\to \pm \infty }{\text{lim}}5\text{/}x}\hfill \\ & =\frac{3-0}{2+0}=\frac{3}{2}.\hfill \end{array}[/latex]

   Since [latex]\underset{x\to \pm \infty }{\text{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.

   

2. 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 }{\text{lim}}\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}=\underset{x\to \pm \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.[/latex]

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

   

3. Dividing the numerator and denominator by [latex]x,[/latex] we have
   [latex]\underset{x\to \pm \infty }{\text{lim}}\frac{3{x}^{2}+4x}{x+2}=\underset{x\to \pm \infty }{\text{lim}}\frac{3x+4}{1+2\text{/}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 }{\text{lim}}f(x)=\infty ,[/latex] whereas [latex]\underset{x\to -\infty }{\text{lim}}f(x)=-\infty[/latex] as shown in the following figure.

   

Evaluate [latex]\underset{x\to \pm \infty }{\text{lim}}\frac{3{x}^{2}+2x-1}{5{x}^{2}-4x+7}[/latex] and use these limits to determine the end behavior of [latex]f(x)=\frac{3{x}^{2}+2x-2}{5{x}^{2}-4x+7}.[/latex]

Hint

Divide the numerator and denominator by [latex]{x}^{2}.[/latex]

Solution

[latex]\frac{3}{5}[/latex]

Before proceeding, consider the graph of [latex]f(x)=\frac{(3{x}^{2}+4x)}{(x+2)}[/latex] shown in (Figure) . 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{3{x}^{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 }{\text{lim}}(f(x)-(3x-2))=\underset{x\to \pm \infty }{\text{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] ( (Figure) ).

Now let’s consider the end behavior for functions involving a radical.

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{4{x}^{2}+5}}[/latex] and describe the end behavior of [latex]f.[/latex]

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{4{x}^{2}+5}[/latex] in the denominator. Since

[latex]\sqrt{4{x}^{2}+5}\approx \sqrt{4{x}^{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}{ccc}\hfill \underset{x\to \infty }{\text{lim}}\frac{3x-2}{\sqrt{4{x}^{2}+5}}& =\hfill & \underset{x\to \infty }{\text{lim}}\frac{(1\text{/}|x|)(3x-2)}{(1\text{/}|x|)\sqrt{4{x}^{2}+5}}\hfill \\ & =\hfill & \underset{x\to \infty }{\text{lim}}\frac{(1\text{/}x)(3x-2)}{\sqrt{(1\text{/}{x}^{2})(4{x}^{2}+5)}}\hfill \\ & =\hfill & \underset{x\to \infty }{\text{lim}}\frac{3-2\text{/}x}{\sqrt{4+5\text{/}{x}^{2}}}=\frac{3}{\sqrt{4}}=\frac{3}{2}\hfill \\ \hfill \underset{x\to -\infty }{\text{lim}}\frac{3x-2}{\sqrt{4{x}^{2}+5}}& =\hfill & \underset{x\to -\infty }{\text{lim}}\frac{(1\text{/}|x|)(3x-2)}{(1\text{/}|x|)\sqrt{4{x}^{2}+5}}\hfill \\ & =\hfill & \underset{x\to -\infty }{\text{lim}}\frac{(-1\text{/}x)(3x-2)}{\sqrt{(1\text{/}{x}^{2})(4{x}^{2}+5)}}\hfill \\ & =\hfill & \underset{x\to -\infty }{\text{lim}}\frac{-3+2\text{/}x}{\sqrt{4+5\text{/}{x}^{2}}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.\hfill \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.

Evaluate [latex]\underset{x\to \infty }{\text{lim}}\frac{\sqrt{3{x}^{2}+4}}{x+6}.[/latex]

Hint

Divide the numerator and denominator by [latex]|x|.[/latex]

Solution

[latex]\pm \sqrt{3}[/latex]

Determining End Behavior for Transcendental Functions

The six basic trigonometric functions are periodic and do not approach a finite limit as [latex]x\to \pm \infty .[/latex] For example, [latex]\sin x[/latex] oscillates between [latex]1\text{ and }-1[/latex] ( (Figure) ). The tangent function [latex]x[/latex] has an infinite number of vertical asymptotes as [latex]x\to \pm \infty ;[/latex] therefore, it does not approach a finite limit nor does it approach [latex]\pm \infty[/latex] as [latex]x\to \pm \infty[/latex] as shown in (Figure) .

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]0 < b < 1,[/latex] [latex]y={b}^{x}[/latex] is decreasing over [latex](-\infty ,\infty ).[/latex] For the natural exponential function [latex]f(x)={e}^{x},[/latex] [latex]e\approx 2.718 < 1.[/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

Recall that the natural logarithm function [latex]f(x)=\text{ln}(x)[/latex] is the inverse of the natural exponential function [latex]y={e}^{x}.[/latex] Therefore, the domain of [latex]f(x)=\text{ln}(x)[/latex] is [latex](0,\infty )[/latex] and the range is [latex](-\infty ,\infty ).[/latex] The graph of [latex]f(x)=\text{ln}(x)[/latex] is the reflection of the graph of [latex]y={e}^{x}[/latex] about the line [latex]y=x.[/latex] Therefore, [latex]\text{ln}(x)\to -\infty[/latex] as [latex]x\to {0}^{+}[/latex] and [latex]\text{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]\text{ln}(x)[/latex]	-4.605	-2.303	0	2.303	4.605

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+3{e}^{x})}{(7-5{e}^{x})}[/latex] and describe the end behavior of [latex]f.[/latex]

Solution

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

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

As shown in (Figure) , [latex]{e}^{x}\to \infty[/latex] as [latex]x\to \infty .[/latex] Therefore,

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

We conclude that [latex]\underset{x\to \infty }{\text{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 }{\text{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]

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

Hint

[latex]\underset{x\to \infty }{\text{lim}}{e}^{x}=\infty[/latex] and [latex]\underset{x\to -\infty }{\text{lim}}{e}^{x}=0.[/latex]

Solution

[latex]\underset{x\to \infty }{\text{lim}}f(x)=\frac{3}{5}[/latex], [latex]\underset{x\to -\infty }{\text{lim}}f(x)=-2[/latex]