Learning Objectives
- Calculate the limit of a function as [latex]x[/latex] increases or decreases without bound.
- Recognize a horizontal asymptote on the graph of a function.
- Estimate the end behavior of a function as [latex]x[/latex] increases or decreases without bound.
- Recognize an oblique asymptote on the graph of a function.
- Analyze a function and its derivatives to draw its graph.
We have shown how to use the first and second derivatives of a function to describe the shape of a graph. To graph a function [latex]f[/latex] defined on an unbounded domain, we also need to know the behavior of [latex]f[/latex] as [latex]x\to \text{±}\infty .[/latex] In this section, we define limits at infinity and show how these limits affect the graph of a function. At the end of this section, we outline a strategy for graphing an arbitrary function [latex]f.[/latex]
Limits at Infinity
We begin by examining what it means for a function to have a finite limit at infinity. Then we study the idea of a function with an infinite limit at infinity. Back in Introduction to Functions and Graphs, we looked at vertical asymptotes; in this section we deal with horizontal and oblique asymptotes.
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]\text{−}\infty [/latex] of [latex]f(x)[/latex] is 2 and write [latex]\underset{x\to a}{\text{lim}}f(x)=2.[/latex]
[latex]x[/latex] | 10 | 100 | 1,000 | 10,000 |
[latex]2+\frac{1}{x}[/latex] | 2.1 | 2.01 | 2.001 | 2.0001 |
[latex]x[/latex] | -10 | -100 | -1000 | -10,000 |
[latex]2+\frac{1}{x}[/latex] | 1.9 | 1.99 | 1.999 | 1.9999 |
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 \text{−}\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 \text{−}\infty }{\text{lim}}f(x)=L.[/latex] We now look at the definition of a function having a limit at infinity.
Definition
(Informal) If the values of [latex]f(x)[/latex] become arbitrarily close to [latex]L[/latex] as [latex]x[/latex] becomes sufficiently large, we say the function [latex]f[/latex] has a limit at infinity and write
If the values of [latex]f(x)[/latex] becomes arbitrarily close to [latex]L[/latex] for [latex]x<0[/latex] as [latex]|x|[/latex] becomes sufficiently large, we say that the function [latex]f[/latex] has a limit at negative infinity and write
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 \text{−}\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]
Definition
If [latex]\underset{x\to \infty }{\text{lim}}f(x)=L[/latex] or [latex]\underset{x\to \text{−}\infty }{\text{lim}}f(x)=L,[/latex] we say the line [latex]y=L[/latex] is a horizontal asymptote of [latex]f.[/latex]
A function cannot cross a vertical asymptote because the graph must approach infinity (or [latex]\text{−}\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 \text{−}\infty }{\text{lim}}f(x).[/latex] Determine the horizontal asymptote(s) for [latex]f.[/latex]
- [latex]f(x)=5-\frac{2}{{x}^{2}}[/latex]
- [latex]f(x)=\frac{ \sin x}{x}[/latex]
- [latex]f(x)={ \tan }^{-1}(x)[/latex]
Evaluate [latex]\underset{x\to \text{−}\infty }{\text{lim}}(3+\frac{4}{x})[/latex] and [latex]\underset{x\to \infty }{\text{lim}}(3+\frac{4}{x}).[/latex] Determine the horizontal asymptotes of [latex]f(x)=3+\frac{4}{x},[/latex] if any.
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 \text{−}\infty ).[/latex] In this case, we write [latex]\underset{x\to \infty }{\text{lim}}f(x)=\infty [/latex] (or [latex]\underset{x\to \text{−}\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 \text{−}\infty ),[/latex] we write [latex]\underset{x\to \infty }{\text{lim}}f(x)=\text{−}\infty [/latex] (or [latex]\underset{x\to \text{−}\infty }{\text{lim}}f(x)=\text{−}\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 \text{−}\infty ,[/latex] the values of [latex]f(x)={x}^{3}[/latex] are negative but become arbitrarily large in magnitude. Consequently, [latex]\underset{x\to \text{−}\infty }{\text{lim}}{x}^{3}=\text{−}\infty .[/latex]
[latex]x[/latex] | 10 | 20 | 50 | 100 | 1000 |
[latex]{x}^{3}[/latex] | 1000 | 8000 | 125,000 | 1,000,000 | 1,000,000,000 |
[latex]x[/latex] | -10 | -20 | -50 | -100 | -1000 |
[latex]{x}^{3}[/latex] | -1000 | -8000 | -125,000 | -1,000,000 | -1,000,000,000 |
Definition
(Informal) We say a function [latex]f[/latex] has an infinite limit at infinity and write
if [latex]f(x)[/latex] becomes arbitrarily large for [latex]x[/latex] sufficiently large. We say a function has a negative infinite limit at infinity and write
if [latex]f(x)<0[/latex] and [latex]|f(x)|[/latex] becomes arbitrarily large for [latex]x[/latex] sufficiently large. Similarly, we can define infinite limits as [latex]x\to \text{−}\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.
Definition
(Formal) We say a function [latex]f[/latex] has a limit at infinity, if there exists a real number [latex]L[/latex] such that for all [latex]\epsilon >0,[/latex] there exists [latex]N>0[/latex] such that
for all [latex]x>N.[/latex] In that case, we write
(see (Figure)).
We say a function [latex]f[/latex] has a limit at negative infinity if there exists a real number [latex]L[/latex] such that for all [latex]\epsilon >0,[/latex] there exists [latex]N<0[/latex] such that
for all [latex]x<N.[/latex] In that case, we write
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.
A Finite Limit at Infinity Example
Use the formal definition of limit at infinity to prove that [latex]\underset{x\to \infty }{\text{lim}}(\frac{2+1}{x})=2.[/latex]
Use the formal definition of limit at infinity to prove that [latex]\underset{x\to \infty }{\text{lim}}(\frac{3-1}{{x}^{2}})=3.[/latex]
Hint
Let [latex]N=\frac{1}{\sqrt{\epsilon }}.[/latex]
We now turn our attention to a more precise definition for an infinite limit at infinity.
Definition
(Formal) We say a function [latex]f[/latex] has an infinite limit at infinity and write
if for all [latex]M>0,[/latex] there exists an [latex]N>0[/latex] such that
for all [latex]x>N[/latex] (see (Figure)).
We say a function has a negative infinite limit at infinity and write
if for all [latex]M<0,[/latex] there exists an [latex]N>0[/latex] such that
for all [latex]x>N.[/latex]
Similarly we can define limits as [latex]x\to \text{−}\infty .[/latex]
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.
An Infinite Limit at Infinity
Use the formal definition of infinite limit at infinity to prove that [latex]\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty .[/latex]
Use the formal definition of infinite limit at infinity to prove that [latex]\underset{x\to \infty }{\text{lim}}3{x}^{2}=\infty .[/latex]
Hint
Let [latex]N=\sqrt{\frac{M}{3}}.[/latex]
End Behavior
The behavior of a function as [latex]x\to \text{±}\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:
- The function [latex]f(x)[/latex] approaches a horizontal asymptote [latex]y=L.[/latex]
- The function [latex]f(x)\to \infty [/latex] or [latex]f(x)\to \text{−}\infty .[/latex]
- The function does not approach a finite limit, nor does it approach [latex]\infty [/latex] or [latex]\text{−}\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) 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 \text{−}\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 \text{−}\infty }{\text{lim}}c{x}^{n}.[/latex]
Limits at Infinity for Power Functions
For each function [latex]f,[/latex] evaluate [latex]\underset{x\to \infty }{\text{lim}}f(x)[/latex] and [latex]\underset{x\to \text{−}\infty }{\text{lim}}f(x).[/latex]
- [latex]f(x)=-5{x}^{3}[/latex]
- [latex]f(x)=2{x}^{4}[/latex]
Let [latex]f(x)=-3{x}^{4}.[/latex] Find [latex]\underset{x\to \infty }{\text{lim}}f(x).[/latex]
Hint
The coefficient -3 is negative.
We now look at how the limits at infinity for power functions can be used to determine [latex]\underset{x\to \text{±}\infty }{\text{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 \text{±}\infty ,[/latex] all the terms inside the parentheses approach zero except the first term. We conclude that
For example, the function [latex]f(x)=5{x}^{3}-3{x}^{2}+4[/latex] behaves like [latex]g(x)=5{x}^{3}[/latex] as [latex]x\to \text{±}\infty [/latex] as shown in (Figure) and (Figure).
[latex]x[/latex] | 10 | 100 | 1000 |
[latex]f(x)=5{x}^{3}-3{x}^{2}+4[/latex] | 4704 | 4,970,004 | 4,997,000,004 |
[latex]g(x)=5{x}^{3}[/latex] | 5000 | 5,000,000 | 5,000,000,000 |
[latex]x[/latex] | -10 | -100 | -1000 |
[latex]f(x)=5{x}^{3}-3{x}^{2}+4[/latex] | -5296 | -5,029,996 | -5,002,999,996 |
[latex]g(x)=5{x}^{3}[/latex] | -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 (Figure), 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]
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 \text{−}\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{3{x}^{2}+2x}{4{x}^{3}-5x+7}[/latex] (Note: The degree of numerator is less than the degree of the denominator.)
- [latex]f(x)=\frac{3{x}^{2}+4x}{x+2}[/latex] (Note: The degree of numerator is greater than the degree of the denominator.)
Evaluate [latex]\underset{x\to \text{±}\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]
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 \text{−}\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 \text{±}\infty ,[/latex] we conclude that
Therefore, the graph of [latex]f[/latex] approaches the line [latex]y=3x-2[/latex] as [latex]x\to \text{±}\infty .[/latex] This line is known as an oblique asymptote for [latex]f[/latex] ((Figure)).
We can summarize the results of (Figure) to make the following conclusion regarding end behavior for rational functions. Consider a rational function
where [latex]{a}_{n}\ne 0\text{ and }{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}\text{/}{b}_{m}[/latex] as [latex]x\to \text{±}\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 \text{±}\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 \text{±}\infty }{\text{lim}}r(x)\text{/}q(x)=0.[/latex] Therefore, the values of [latex]\left[f(x)-g(x)\right][/latex] approach zero as [latex]x\to \text{±}\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.
Determining End Behavior for a Function Involving a Radical
Find the limits as [latex]x\to \infty [/latex] and [latex]x\to \text{−}\infty [/latex] for [latex]f(x)=\frac{3x-2}{\sqrt{4{x}^{2}+5}}[/latex] and describe the end behavior of [latex]f.[/latex]
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]
Determining End Behavior for Transcendental Functions
The six basic trigonometric functions are periodic and do not approach a finite limit as [latex]x\to \text{±}\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 \text{±}\infty ;[/latex] therefore, it does not approach a finite limit nor does it approach [latex]\text{±}\infty [/latex] as [latex]x\to \text{±}\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](\text{−}\infty ,\infty )[/latex] and range [latex](0,\infty ).[/latex] If [latex]b>1,y={b}^{x}[/latex] is increasing over [latex]`(\text{−}\infty ,\infty ).[/latex] If [latex]0<b<1,[/latex] [latex]y={b}^{x}[/latex] is decreasing over [latex](\text{−}\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]`(\text{−}\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 \text{−}\infty [/latex] as shown in (Figure) and (Figure).
[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](\text{−}\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 \text{−}\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).
[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 \text{−}\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]
Find the limits as [latex]x\to \infty [/latex] and [latex]x\to \text{−}\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]
Guidelines for Drawing the Graph of a Function
We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let’s look at a general strategy to use when graphing any function.
Problem-Solving Strategy: Drawing the Graph of a Function
Given a function [latex]f,[/latex] use the following steps to sketch a graph of [latex]f\text{:}[/latex]
- Determine the domain of the function.
- Locate the [latex]x[/latex]– and [latex]y[/latex]-intercepts.
- Evaluate [latex]\underset{x\to \infty }{\text{lim}}f(x)[/latex] and [latex]\underset{x\to \text{−}\infty }{\text{lim}}f(x)[/latex] to determine the end behavior. If either of these limits is a finite number [latex]L,[/latex] then [latex]y=L[/latex] is a horizontal asymptote. If either of these limits is [latex]\infty [/latex] or [latex]\text{−}\infty ,[/latex] determine whether [latex]f[/latex] has an oblique asymptote. If [latex]f[/latex] is a rational function such that [latex]f(x)=\frac{p(x)}{q(x)},[/latex] where the degree of the numerator is greater than the degree of the denominator, then [latex]f[/latex] can be written 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] The values of [latex]f(x)[/latex] approach the values of [latex]g(x)[/latex] as [latex]x\to \text{±}\infty .[/latex] If [latex]g(x)[/latex] is a linear function, it is known as an oblique asymptote.
- Determine whether [latex]f[/latex] has any vertical asymptotes.
- Calculate [latex]{f}^{\prime }.[/latex] Find all critical points and determine the intervals where [latex]f[/latex] is increasing and where [latex]f[/latex] is decreasing. Determine whether [latex]f[/latex] has any local extrema.
- Calculate [latex]f\text{″}.[/latex] Determine the intervals where [latex]f[/latex] is concave up and where [latex]f[/latex] is concave down. Use this information to determine whether [latex]f[/latex] has any inflection points. The second derivative can also be used as an alternate means to determine or verify that [latex]f[/latex] has a local extremum at a critical point.
Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function.
Sketching a Graph of a Polynomial
Sketch a graph of [latex]f(x)={(x-1)}^{2}(x+2).[/latex]
Sketch a graph of [latex]f(x)={(x-1)}^{3}(x+2).[/latex]
Hint
[latex]f[/latex] is a fourth-degree polynomial.
Sketching a Rational Function
Sketch the graph of [latex]f(x)=\frac{{x}^{2}}{(1-{x}^{2})}\text{.}[/latex]
Sketch a graph of [latex]f(x)=\frac{(3x+5)}{(8+4x)}.[/latex]
Hint
A line [latex]y=L[/latex] is a horizontal asymptote of [latex]f[/latex] if the limit as [latex]x\to \infty [/latex] or the limit as [latex]x\to \text{−}\infty [/latex] of [latex]f(x)[/latex] is [latex]L.[/latex] A line [latex]x=a[/latex] is a vertical asymptote if at least one of the one-sided limits of [latex]f[/latex] as [latex]x\to a[/latex] is [latex]\infty [/latex] or [latex]\text{−}\infty .[/latex]
Sketching a Rational Function with an Oblique Asymptote
Sketch the graph of [latex]f(x)=\frac{{x}^{2}}{(x-1)}[/latex]
Find the oblique asymptote for [latex]f(x)=\frac{(3{x}^{3}-2x+1)}{(2{x}^{2}-4)}.[/latex]
Hint
Use long division of polynomials.
Sketching the Graph of a Function with a Cusp
Sketch a graph of [latex]f(x)={(x-1)}^{2\text{/}3}.[/latex]
Consider the function [latex]f(x)=5-{x}^{2\text{/}3}.[/latex] Determine the point on the graph where a cusp is located. Determine the end behavior of [latex]f.[/latex]
Hint
A function [latex]f[/latex] has a cusp at a point [latex]a[/latex] if [latex]f(a)[/latex] exists, [latex]f\prime (a)[/latex] is undefined, one of the one-sided limits as [latex]x\to a[/latex] of [latex]f\prime (x)[/latex] is [latex]+\infty ,[/latex] and the other one-sided limit is [latex]\text{−}\infty .[/latex]
Key Concepts
- The limit of [latex]f(x)[/latex] is [latex]L[/latex] as [latex]x\to \infty [/latex] (or as [latex]x\to \text{−}\infty )[/latex] if the values [latex]f(x)[/latex] become arbitrarily close to [latex]L[/latex] as [latex]x[/latex] becomes sufficiently large.
- The limit of [latex]f(x)[/latex] is [latex]\infty [/latex] as [latex]x\to \infty [/latex] if [latex]f(x)[/latex] becomes arbitrarily large as [latex]x[/latex] becomes sufficiently large. The limit of [latex]f(x)[/latex] is [latex]\text{−}\infty [/latex] as [latex]x\to \infty [/latex] if [latex]f(x)<0[/latex] and [latex]|f(x)|[/latex] becomes arbitrarily large as [latex]x[/latex] becomes sufficiently large. We can define the limit of [latex]f(x)[/latex] as [latex]x[/latex] approaches [latex]\text{−}\infty [/latex] similarly.
- For a polynomial function [latex]p(x)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\text{…}+{a}_{1}x+{a}_{0},[/latex] where [latex]{a}_{n}\ne 0,[/latex] the end behavior is determined by the leading term [latex]{a}_{n}{x}^{n}.[/latex] If [latex]n\ne 0,[/latex] [latex]p(x)[/latex] approaches [latex]\infty [/latex] or [latex]\text{−}\infty [/latex] at each end.
- For a rational function [latex]f(x)=\frac{p(x)}{q(x)},[/latex] the end behavior is determined by the relationship between the degree of [latex]p[/latex] and the degree of [latex]q.[/latex] If the degree of [latex]p[/latex] is less than the degree of [latex]q,[/latex] the line [latex]y=0[/latex] is a horizontal asymptote for [latex]f.[/latex] If the degree of [latex]p[/latex] is equal to the degree of [latex]q,[/latex] then the line [latex]y=\frac{{a}_{n}}{{b}_{n}}[/latex] is a horizontal asymptote, where [latex]{a}_{n}[/latex] and [latex]{b}_{n}[/latex] are the leading coefficients of [latex]p[/latex] and [latex]q,[/latex] respectively. If the degree of [latex]p[/latex] is greater than the degree of [latex]q,[/latex] then [latex]f[/latex] approaches [latex]\infty [/latex] or [latex]\text{−}\infty [/latex] at each end.
For the following exercises, examine the graphs. Identify where the vertical asymptotes are located.
For the following functions [latex]f(x),[/latex] determine whether there is an asymptote at [latex]x=a.[/latex] Justify your answer without graphing on a calculator.
[latex]f(x)=\frac{x+1}{{x}^{2}+5x+4},a=-1[/latex]
[latex]f(x)=\frac{x}{x-2},a=2[/latex]
[latex]f(x)={(x+2)}^{3\text{/}2},a=-2[/latex]
[latex]f(x)={(x-1)}^{-1\text{/}3},a=1[/latex]
[latex]f(x)=1+{x}^{-2\text{/}5},a=1[/latex]
For the following exercises, evaluate the limit.
[latex]\underset{x\to \infty }{\text{lim}}\frac{1}{3x+6}[/latex]
[latex]\underset{x\to \infty }{\text{lim}}\frac{2x-5}{4x}[/latex]
[latex]\underset{x\to \infty }{\text{lim}}\frac{{x}^{2}-2x+5}{x+2}[/latex]
[latex]\underset{x\to \text{−}\infty }{\text{lim}}\frac{3{x}^{3}-2x}{{x}^{2}+2x+8}[/latex]
[latex]\underset{x\to \text{−}\infty }{\text{lim}}\frac{{x}^{4}-4{x}^{3}+1}{2-2{x}^{2}-7{x}^{4}}[/latex]
[latex]\underset{x\to \infty }{\text{lim}}\frac{3x}{\sqrt{{x}^{2}+1}}[/latex]
[latex]\underset{x\to \text{−}\infty }{\text{lim}}\frac{\sqrt{4{x}^{2}-1}}{x+2}[/latex]
[latex]\underset{x\to \infty }{\text{lim}}\frac{4x}{\sqrt{{x}^{2}-1}}[/latex]
[latex]\underset{x\to \text{−}\infty }{\text{lim}}\frac{4x}{\sqrt{{x}^{2}-1}}[/latex]
[latex]\underset{x\to \infty }{\text{lim}}\frac{2\sqrt{x}}{x-\sqrt{x}+1}[/latex]
For the following exercises, find the horizontal and vertical asymptotes.
[latex]f(x)=x-\frac{9}{x}[/latex]
[latex]f(x)=\frac{1}{1-{x}^{2}}[/latex]
[latex]f(x)=\frac{{x}^{3}}{4-{x}^{2}}[/latex]
[latex]f(x)=\frac{{x}^{2}+3}{{x}^{2}+1}[/latex]
[latex]f(x)= \sin (x) \sin (2x)[/latex]
[latex]f(x)= \cos x+ \cos (3x)+ \cos (5x)[/latex]
[latex]f(x)=\frac{x \sin (x)}{{x}^{2}-1}[/latex]
[latex]f(x)=\frac{x}{ \sin (x)}[/latex]
[latex]f(x)=\frac{1}{{x}^{3}+{x}^{2}}[/latex]
[latex]f(x)=\frac{1}{x-1}-2x[/latex]
[latex]f(x)=\frac{{x}^{3}+1}{{x}^{3}-1}[/latex]
[latex]f(x)=\frac{ \sin x+ \cos x}{ \sin x- \cos x}[/latex]
[latex]f(x)=x- \sin x[/latex]
[latex]f(x)=\frac{1}{x}-\sqrt{x}[/latex]
For the following exercises, construct a function [latex]f(x)[/latex] that has the given asymptotes.
[latex]x=1[/latex] and [latex]y=2[/latex]
[latex]x=1[/latex] and [latex]y=0[/latex]
[latex]y=4,[/latex][latex]x=-1[/latex]
[latex]x=0[/latex]
For the following exercises, graph the function on a graphing calculator on the window [latex]x=\left[-5,5\right][/latex] and estimate the horizontal asymptote or limit. Then, calculate the actual horizontal asymptote or limit.
[T][latex]f(x)=\frac{1}{x+10}[/latex]
[T][latex]f(x)=\frac{x+1}{{x}^{2}+7x+6}[/latex]
[T][latex]\underset{x\to \text{−}\infty }{\text{lim}}{x}^{2}+10x+25[/latex]
[T][latex]\underset{x\to \text{−}\infty }{\text{lim}}\frac{x+2}{{x}^{2}+7x+6}[/latex]
[T][latex]\underset{x\to \infty }{\text{lim}}\frac{3x+2}{x+5}[/latex]
For the following exercises, draw a graph of the functions without using a calculator. Be sure to notice all important features of the graph: local maxima and minima, inflection points, and asymptotic behavior.
[latex]y=3{x}^{2}+2x+4[/latex]
[latex]y={x}^{3}-3{x}^{2}+4[/latex]
[latex]y=\frac{2x+1}{{x}^{2}+6x+5}[/latex]
[latex]y=\frac{{x}^{3}+4{x}^{2}+3x}{3x+9}[/latex]
[latex]y=\frac{{x}^{2}+x-2}{{x}^{2}-3x-4}[/latex]
[latex]y=\sqrt{{x}^{2}-5x+4}[/latex]
[latex]y=2x\sqrt{16-{x}^{2}}[/latex]
[latex]y=\frac{ \cos x}{x},[/latex] on [latex]x=\left[-2\pi ,2\pi \right][/latex]
[latex]y={e}^{x}-{x}^{3}[/latex]
[latex]y=x \tan x,x=\left[\text{−}\pi ,\pi \right][/latex]
[latex]y=x\text{ln}(x),x>0[/latex]
[latex]y={x}^{2} \sin (x),x=\left[-2\pi ,2\pi \right][/latex]
For [latex]f(x)=\frac{P(x)}{Q(x)}[/latex] to have an asymptote at [latex]y=2[/latex] then the polynomials [latex]P(x)[/latex] and [latex]Q(x)[/latex] must have what relation?
For [latex]f(x)=\frac{P(x)}{Q(x)}[/latex] to have an asymptote at [latex]x=0,[/latex] then the polynomials [latex]P(x)[/latex] and [latex]Q(x).[/latex] must have what relation?
If [latex]{f}^{\prime }(x)[/latex] has asymptotes at [latex]y=3[/latex] and [latex]x=1,[/latex] then [latex]f(x)[/latex] has what asymptotes?
Both [latex]f(x)=\frac{1}{(x-1)}[/latex] and [latex]g(x)=\frac{1}{{(x-1)}^{2}}[/latex] have asymptotes at [latex]x=1[/latex] and [latex]y=0.[/latex] What is the most obvious difference between these two functions?
True or false: Every ratio of polynomials has vertical asymptotes.
Glossary
- end behavior
- the behavior of a function as [latex]x\to \infty [/latex] and [latex]x\to \text{−}\infty [/latex]
- horizontal asymptote
- if [latex]\underset{x\to \infty }{\text{lim}}f(x)=L[/latex] or [latex]\underset{x\to \text{−}\infty }{\text{lim}}f(x)=L,[/latex] then [latex]y=L[/latex] is a horizontal asymptote of [latex]f[/latex]
- infinite limit at infinity
- a function that becomes arbitrarily large as [latex]x[/latex] becomes large
- limit at infinity
- the limiting value, if it exists, of a function as [latex]x\to \infty [/latex] or [latex]x\to \text{−}\infty [/latex]
- oblique asymptote
- the line [latex]y=mx+b[/latex] if [latex]f(x)[/latex] approaches it as [latex]x\to \infty [/latex] or [latex]x\to \text{−}\infty [/latex]
Hint
[latex]\underset{x\to \text{±}\infty }{\text{lim}}1\text{/}x=0[/latex]