We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > Calculus Volume 1

Limits at Infinity and Asymptotes

Learning Objectives

  • Calculate the limit of a function as xx increases or decreases without bound.
  • Recognize a horizontal asymptote on the graph of a function.
  • Estimate the end behavior of a function as xx 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 ff defined on an unbounded domain, we also need to know the behavior of ff as x±.x\to \text{±}\infty . 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 f.f.

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 limxaf(x)=L\underset{x\to a}{\text{lim}}f\left(x\right)=L means f(x)f\left(x\right) becomes arbitrarily close to LL as long as xx is sufficiently close to a.a. We can extend this idea to limits at infinity. For example, consider the function f(x)=2+1x.f\left(x\right)=2+\frac{1}{x}. As can be seen graphically in [link] and numerically in [link], as the values of xx get larger, the values of f(x)f\left(x\right) approach 2.2. We say the limit as xx approaches \infty of f(x)f\left(x\right) is 22 and write limxf(x)=2.\underset{x\to \infty }{\text{lim}}f\left(x\right)=2. Similarly, for x<0,x<0, as the values x|x| get larger, the values of f(x)f\left(x\right) approaches 2.2. We say the limit as xx approaches \text{−}\infty of f(x)f\left(x\right) is 22 and write limxaf(x)=2.\underset{x\to a}{\text{lim}}f\left(x\right)=2.

The function approaches the asymptote y=2y=2 as xx approaches ±.\text{±}\infty .
The function f(x) 2 + 1/x is graphed. The function starts negative near y = 2 but then decreases to −∞ near x = 0. The function then decreases from ∞ near x = 0 and gets nearer to y = 2 as x increases. There is a horizontal line denoting the asymptote y = 2.
Values of a function ff as x±x\to \text{±}\infty
xx10101001001,0001,00010,00010,000
2+1x2+\frac{1}{x}2.12.12.012.012.0012.0012.00012.0001
xx10-10100-1001000-100010,000-10,000
2+1x2+\frac{1}{x}1.91.91.991.991.9991.9991.99991.9999

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

Definition

(Informal) If the values of f(x)f\left(x\right) become arbitrarily close to LL as xx becomes sufficiently large, we say the function ff has a limit at infinity and write

limxf(x)=L.\underset{x\to \infty }{\text{lim}}f\left(x\right)=L.

If the values of f(x)f\left(x\right) becomes arbitrarily close to LL for x<0x<0 as x|x| becomes sufficiently large, we say that the function ff has a limit at negative infinity and write

limxf(x)=L.\underset{x\to \infty }{\text{lim}}f\left(x\right)=L.

If the values f(x)f\left(x\right) are getting arbitrarily close to some finite value LL as xx\to \infty or x,x\to \text{−}\infty , the graph of ff approaches the line y=L.y=L. In that case, the line y=Ly=L is a horizontal asymptote of ff ([link]). For example, for the function f(x)=1x,f\left(x\right)=\frac{1}{x}, since limxf(x)=0,\underset{x\to \infty }{\text{lim}}f\left(x\right)=0, the line y=0y=0 is a horizontal asymptote of f(x)=1x.f\left(x\right)=\frac{1}{x}.

Definition

If limxf(x)=L\underset{x\to \infty }{\text{lim}}f\left(x\right)=L or limxf(x)=L,\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=L, we say the line y=Ly=L is a horizontal asymptote of f.f.

(a) As x,x\to \infty , the values of ff are getting arbitrarily close to L.L. The line y=Ly=L is a horizontal asymptote of f.f. (b) As x,x\to \text{−}\infty , the values of ff are getting arbitrarily close to M.M. The line y=My=M is a horizontal asymptote of f.f.
The figure is broken up into two figures labeled a and b. Figure a shows a function f(x) approaching but never touching a horizontal dashed line labeled L from above. Figure b shows a function f(x) approaching but never a horizontal dashed line labeled M from below.

A function cannot cross a vertical asymptote because the graph must approach infinity (or \text{−}\infty \right) from at least one direction as xx 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 f(x)=(cosx)x+1f\left(x\right)=\frac{\left(\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)}{x}+1 shown in [link] intersects the horizontal asymptote y=1y=1 an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.

The graph of f(x)=(cosx)/x+1f\left(x\right)=\left(\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)\text{/}x+1 crosses its horizontal asymptote y=1y=1 an infinite number of times.
The function f(x) = (cos x)/x + 1 is shown. It decreases from (0, ∞) and then proceeds to oscillate around y = 1 with 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 f,f, evaluate limxf(x)\underset{x\to \infty }{\text{lim}}f\left(x\right) and limxf(x).\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right). Determine the horizontal asymptote(s) for f.f.

  1. f(x)=52x2f\left(x\right)=5-\frac{2}{{x}^{2}}
  2. f(x)=sinxxf\left(x\right)=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}
  3. f(x)=tan1(x)f\left(x\right)={\text{tan}}^{-1}\left(x\right)
  1. Using the algebraic limit laws, we have limx(52x2)=limx52(limx1x).(limx1x)=520=5.\underset{x\to \infty }{\text{lim}}\left(5-\frac{2}{{x}^{2}}\right)=\underset{x\to \infty }{\text{lim}}5-2\left(\underset{x\to \infty }{\text{lim}}\frac{1}{x}\right).\left(\underset{x\to \infty }{\text{lim}}\frac{1}{x}\right)=5-2·0=5.

    Similarly, limxf(x)=5.\underset{x\to \infty }{\text{lim}}f\left(x\right)=5. Therefore, f(x)=52x2f\left(x\right)=5-\frac{2}{{x}^{2}} has a horizontal asymptote of y=5y=5 and ff approaches this horizontal asymptote as x±x\to \text{±}\infty as shown in the following graph.

    This function approaches a horizontal asymptote as x±.x\to \text{±}\infty .
    The function f(x) = 5 – 2/x2 is graphed. The function approaches the horizontal asymptote y = 5 as x approaches ±∞.
  2. Since 1sinx1-1\le \text{sin}\phantom{\rule{0.1em}{0ex}}x\le 1 for all x,x, we have

    1xsinxx1x\frac{-1}{x}\le \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}\le \frac{1}{x}

    for all x0.x\ne 0. Also, since

    limx1x=0=limx1x,\underset{x\to \infty }{\text{lim}}\frac{-1}{x}=0=\underset{x\to \infty }{\text{lim}}\frac{1}{x},

    we can apply the squeeze theorem to conclude that

    limxsinxx=0.\underset{x\to \infty }{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}=0.

    Similarly,

    limxsinxx=0.\underset{x\to \text{−}\infty }{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}=0.

    Thus, f(x)=sinxxf\left(x\right)=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x} has a horizontal asymptote of y=0y=0 and f(x)f\left(x\right) approaches this horizontal asymptote as x±x\to \text{±}\infty as shown in the following graph.

    This function crosses its horizontal asymptote multiple times.
    The function f(x) = (sin x)/x is shown. It has a global maximum at (0, 1) and then proceeds to oscillate around y = 0 with decreasing amplitude.
  3. To evaluate limxtan1(x)\underset{x\to \infty }{\text{lim}}{\text{tan}}^{-1}\left(x\right) and limxtan1(x),\underset{x\to \text{−}\infty }{\text{lim}}{\text{tan}}^{-1}\left(x\right), we first consider the graph of y=tan(x)y=\text{tan}\left(x\right) over the interval (π/2,π/2)\left(\text{−}\pi \text{/}2,\pi \text{/}2\right) as shown in the following graph.

    The graph of tanx\text{tan}\phantom{\rule{0.1em}{0ex}}x has vertical asymptotes at x=±π2x=\text{±}\frac{\pi }{2}
    The function f(x) = tan x is shown. It increases from (−π/2, −∞), passes through the origin, and then increases toward (π/2, ∞). There are vertical dashed lines marking x = ±π/2.

Since

limx(π/2)tanx=,\underset{x\to {\left(\pi \text{/}2\right)}^{-}}{\text{lim}}\text{tan}\phantom{\rule{0.1em}{0ex}}x=\infty ,

it follows that

limxtan1(x)=π2.\underset{x\to \infty }{\text{lim}}{\text{tan}}^{-1}\left(x\right)=\frac{\pi }{2}.

Similarly, since

limx(π/2)+tanx=,\underset{x\to {\left(\pi \text{/}2\right)}^{+}}{\text{lim}}\text{tan}\phantom{\rule{0.1em}{0ex}}x=\text{−}\infty ,

it follows that

limxtan1(x)=π2.\underset{x\to \text{−}\infty }{\text{lim}}{\text{tan}}^{-1}\left(x\right)=-\frac{\pi }{2}.

As a result, y=π2y=\frac{\pi }{2} and y=π2y=-\frac{\pi }{2} are horizontal asymptotes of f(x)=tan1(x)f\left(x\right)={\text{tan}}^{-1}\left(x\right) as shown in the following graph.

This function has two horizontal asymptotes.
The function f(x) = tan−1 x is shown. It increases from (−∞, −π/2), passes through the origin, and then increases toward (∞, π/2). There are horizontal dashed lines marking y = ±π/2.

Evaluate limx(3+4x)\underset{x\to \text{−}\infty }{\text{lim}}\left(3+\frac{4}{x}\right) and limx(3+4x).\underset{x\to \infty }{\text{lim}}\left(3+\frac{4}{x}\right). Determine the horizontal asymptotes of f(x)=3+4x,f\left(x\right)=3+\frac{4}{x}, if any.

Both limits are 3.3. The line y=3y=3 is a horizontal asymptote.

Hint

limx±1/x=0\underset{x\to \text{±}\infty }{\text{lim}}1\text{/}x=0

Infinite Limits at Infinity

Sometimes the values of a function ff become arbitrarily large as xx\to \infty (or as x\to \text{−}\infty \right). In this case, we write limxf(x)=\underset{x\to \infty }{\text{lim}}f\left(x\right)=\infty (or \underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=\infty \right). On the other hand, if the values of ff are negative but become arbitrarily large in magnitude as xx\to \infty (or as x\to \text{−}\infty \right), we write limxf(x)=\underset{x\to \infty }{\text{lim}}f\left(x\right)=\text{−}\infty (or \underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=\text{−}\infty \right).

For example, consider the function f(x)=x3.f\left(x\right)={x}^{3}. As seen in [link] and [link], as xx\to \infty the values f(x)f\left(x\right) become arbitrarily large. Therefore, limxx3=.\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty . On the other hand, as x,x\to \text{−}\infty , the values of f(x)=x3f\left(x\right)={x}^{3} are negative but become arbitrarily large in magnitude. Consequently, limxx3=.\underset{x\to \text{−}\infty }{\text{lim}}{x}^{3}=\text{−}\infty .

Values of a power function as x±x\to \text{±}\infty
xx10102020505010010010001000
x3{x}^{3}1000100080008000125,000125,0001,000,0001,000,0001,000,000,0001,000,000,000
xx10-1020-2050-50100-1001000-1000
x3{x}^{3}1000-10008000-8000125,000-125,0001,000,000-1,000,0001,000,000,000-1,000,000,000
For this function, the functional values approach infinity as x±.x\to \text{±}\infty .
The function f(x) = x3 is graphed. It is apparent that this function rapidly approaches infinity as x approaches infinity.
Definition

(Informal) We say a function ff has an infinite limit at infinity and write

limxf(x)=.\underset{x\to \infty }{\text{lim}}f\left(x\right)=\infty .

if f(x)f\left(x\right) becomes arbitrarily large for xx sufficiently large. We say a function has a negative infinite limit at infinity and write

limxf(x)=.\underset{x\to \infty }{\text{lim}}f\left(x\right)=\text{−}\infty .

if f(x)<0f\left(x\right)<0 and f(x)|f\left(x\right)| becomes arbitrarily large for xx sufficiently large. Similarly, we can define infinite limits as x.x\to \text{−}\infty .

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 ff has a limit at infinity, if there exists a real number LL such that for all ϵ>0,\epsilon >0, there exists N>0N>0 such that

f(x)L<ϵ|f\left(x\right)-L|<\epsilon

for all x>N.x>N. In that case, we write

limxf(x)=L\underset{x\to \infty }{\text{lim}}f\left(x\right)=L

(see [link]).

We say a function ff has a limit at negative infinity if there exists a real number LL such that for all ϵ>0,\epsilon >0, there exists N<0N<0 such that

f(x)L<ϵ|f\left(x\right)-L|<\epsilon

for all x<N.x<N. In that case, we write

limxf(x)=L.\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=L.
For a function with a limit at infinity, for all x>N,x>N, f(x)L<ϵ.|f\left(x\right)-L|<\epsilon .
The function f(x) is graphed, and it has a horizontal asymptote at L. L is marked on the y axis, as is L + ॉ and L – ॉ. On the x axis, N is marked as the value of x such that f(x) = L + ॉ.

Earlier in this section, we used graphical evidence in [link] and numerical evidence in [link] to conclude that limx(2+1x)=2.\underset{x\to \infty }{\text{lim}}\left(\frac{2+1}{x}\right)=2. 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 limx(2+1x)=2.\underset{x\to \infty }{\text{lim}}\left(\frac{2+1}{x}\right)=2.

Let ϵ>0.\epsilon >0. Let N=1ϵ.N=\frac{1}{\epsilon }. Therefore, for all x>N,x>N, we have

2+1x2=1x=1x<1N=ϵ.|2+\frac{1}{x}-2|=|\frac{1}{x}|=\frac{1}{x}<\frac{1}{N}=\epsilon \text{.}

Use the formal definition of limit at infinity to prove that limx(31x2)=3.\underset{x\to \infty }{\text{lim}}\left(\frac{3-1}{{x}^{2}}\right)=3.

Let ϵ>0.\epsilon >0. Let N=1ϵ.N=\frac{1}{\sqrt{\epsilon }}. Therefore, for all x>N,x>N, we have

31x23=1x2<1N2=ϵ|3-\frac{1}{{x}^{2}}-3|=\frac{1}{{x}^{2}}<\frac{1}{{N}^{2}}=\epsilon

Therefore, limx(31/x2)=3.\underset{x\to \infty }{\text{lim}}\left(3-1\text{/}{x}^{2}\right)=3.

Hint

Let N=1ϵ.N=\frac{1}{\sqrt{\epsilon }}.

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

Definition

(Formal) We say a function ff has an infinite limit at infinity and write

limxf(x)=\underset{x\to \infty }{\text{lim}}f\left(x\right)=\infty

if for all M>0,M>0, there exists an N>0N>0 such that

f(x)>Mf\left(x\right)>M

for all x>Nx>N (see [link]).

We say a function has a negative infinite limit at infinity and write

limxf(x)=\underset{x\to \infty }{\text{lim}}f\left(x\right)=\text{−}\infty

if for all M<0,M<0, there exists an N>0N>0 such that

f(x)<Mf\left(x\right)<M

for all x>N.x>N.

Similarly we can define limits as x.x\to \text{−}\infty .

For a function with an infinite limit at infinity, for all x>N,x>N, f(x)>M.f\left(x\right)>M.
The function f(x) is graphed. It continues to increase rapidly after x = N, and f(N) = M.

Earlier, we used graphical evidence ([link]) and numerical evidence ([link]) to conclude that limxx3=.\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty . 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 limxx3=.\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty .

Let M>0.M>0. Let N=M3.N=\sqrt[3]{M}. Then, for all x>N,x>N, we have

x3>N3=(M3)3=M.{x}^{3}>{N}^{3}={\left(\sqrt[3]{M}\right)}^{3}=M.

Therefore, limxx3=.\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty .

Use the formal definition of infinite limit at infinity to prove that limx3x2=.\underset{x\to \infty }{\text{lim}}3{x}^{2}=\infty .

Let M>0.M>0. Let N=M3.N=\sqrt{\frac{M}{3}}. Then, for all x>N,x>N, we have

3x2>3N2=3(M3)22=3M3=M3{x}^{2}>3{N}^{2}=3{\left(\sqrt{\frac{M}{3}}\right)}^{2}{2}^{}=\frac{3M}{3}=M

Hint

Let N=M3.N=\sqrt{\frac{M}{3}}.

End Behavior

The behavior of a function as x±x\to \text{±}\infty 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 f(x)f\left(x\right) approaches a horizontal asymptote y=L.y=L.
  2. The function f(x)f\left(x\right)\to \infty or f(x).f\left(x\right)\to \text{−}\infty .
  3. The function does not approach a finite limit, nor does it approach \infty or .\text{−}\infty . 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 f(x)=xnf\left(x\right)={x}^{n} where nn is a positive integer. From [link] and [link], we see that

limxxn=;n=1,2,3,\underset{x\to \infty }{\text{lim}}{x}^{n}=\infty ;n=1,2,3\text{,…}

and

\underset{x\to \text{−}\infty }{\text{lim}}{x}^{n}=\left\{\begin{array}{c}\infty ;n=2,4,6\text{,…}\hfill \\ \text{−}\infty ;n=1,3,5\text{,…}\hfill \end{array}.
For power functions with an even power of n,n, limxxn==limxxn.\underset{x\to \infty }{\text{lim}}{x}^{n}=\infty =\underset{x\to \text{−}\infty }{\text{lim}}{x}^{n}.
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 odd power of n,n, limxxn=\underset{x\to \infty }{\text{lim}}{x}^{n}=\infty and limxxn=.\underset{x\to \text{−}\infty }{\text{lim}}{x}^{n}=\text{−}\infty .
The functions x, x3, and x5 are graphed, and it is apparent that as the exponent grows the functions increase more quickly.

Using these facts, it is not difficult to evaluate limxcxn\underset{x\to \infty }{\text{lim}}c{x}^{n} and limxcxn,\underset{x\to \text{−}\infty }{\text{lim}}c{x}^{n}, where cc is any constant and nn is a positive integer. If c>0,c>0, the graph of y=cxny=c{x}^{n} is a vertical stretch or compression of y=xn,y={x}^{n}, and therefore

limxcxn=limxxnandlimxcxn=limxxnifc>0.\underset{x\to \infty }{\text{lim}}c{x}^{n}=\underset{x\to \infty }{\text{lim}}{x}^{n}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to \text{−}\infty }{\text{lim}}c{x}^{n}=\underset{x\to \text{−}\infty }{\text{lim}}{x}^{n}\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}c>0.

If c<0,c<0, the graph of y=cxny=c{x}^{n} is a vertical stretch or compression combined with a reflection about the xx-axis, and therefore

limxcxn=limxxnandlimxcxn=limxxnifc<0.\underset{x\to \infty }{\text{lim}}c{x}^{n}=\text{−}\underset{x\to \infty }{\text{lim}}{x}^{n}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to \text{−}\infty }{\text{lim}}c{x}^{n}=\text{−}\underset{x\to \text{−}\infty }{\text{lim}}{x}^{n}\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}c<0.

If c=0,y=cxn=0,c=0,y=c{x}^{n}=0, in which case limxcxn=0=limxcxn.\underset{x\to \infty }{\text{lim}}c{x}^{n}=0=\underset{x\to \text{−}\infty }{\text{lim}}c{x}^{n}.

Limits at Infinity for Power Functions

For each function f,f, evaluate limxf(x)\underset{x\to \infty }{\text{lim}}f\left(x\right) and limxf(x).\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right).

  1. f(x)=5x3f\left(x\right)=-5{x}^{3}
  2. f(x)=2x4f\left(x\right)=2{x}^{4}
  1. Since the coefficient of x3{x}^{3} is 5,-5, the graph of f(x)=5x3f\left(x\right)=-5{x}^{3} involves a vertical stretch and reflection of the graph of y=x3y={x}^{3} about the xx-axis. Therefore, limx(5x3)=\underset{x\to \infty }{\text{lim}}\left(-5{x}^{3}\right)=\text{−}\infty and limx(5x3)=.\underset{x\to \text{−}\infty }{\text{lim}}\left(-5{x}^{3}\right)=\infty .
  2. Since the coefficient of x4{x}^{4} is 2,2, the graph of f(x)=2x4f\left(x\right)=2{x}^{4} is a vertical stretch of the graph of y=x4.y={x}^{4}. Therefore, limx2x4=\underset{x\to \infty }{\text{lim}}2{x}^{4}=\infty and limx2x4=.\underset{x\to \text{−}\infty }{\text{lim}}2{x}^{4}=\infty .

Let f(x)=3x4.f\left(x\right)=-3{x}^{4}. Find limxf(x).\underset{x\to \infty }{\text{lim}}f\left(x\right).

\text{−}\infty

Hint

The coefficient 3-3 is negative.

We now look at how the limits at infinity for power functions can be used to determine limx±f(x)\underset{x\to \text{±}\infty }{\text{lim}}f\left(x\right) for any polynomial function f.f. Consider a polynomial function

f(x)=anxn+an1xn1++a1x+a0f\left(x\right)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\text{…}+{a}_{1}x+{a}_{0}

of degree n1n\ge 1 so that an0.{a}_{n}\ne 0. Factoring, we see that

f(x)=anxn(1+an1an1x++a1an1xn1+a0an).f\left(x\right)={a}_{n}{x}^{n}\left(1+\frac{{a}_{n-1}}{{a}_{n}}\phantom{\rule{0.2em}{0ex}}\frac{1}{x}+\text{…}+\frac{{a}_{1}}{{a}_{n}}\phantom{\rule{0.2em}{0ex}}\frac{1}{{x}^{n-1}}+\frac{{a}_{0}}{{a}_{n}}\right).

As x±,x\to \text{±}\infty , all the terms inside the parentheses approach zero except the first term. We conclude that

limx±f(x)=limx±anxn.\underset{x\to \text{±}\infty }{\text{lim}}f\left(x\right)=\underset{x\to \text{±}\infty }{\text{lim}}{a}_{n}{x}^{n}.

For example, the function f(x)=5x33x2+4f\left(x\right)=5{x}^{3}-3{x}^{2}+4 behaves like g(x)=5x3g\left(x\right)=5{x}^{3} as x±x\to \text{±}\infty as shown in [link] and [link].

The end behavior of a polynomial is determined by the behavior of the term with the largest exponent.
Both functions f(x) = 5x3 – 3x2 + 4 and g(x) = 5x3 are plotted. Their behavior for large positive and large negative numbers converges.
A polynomial’s end behavior is determined by the term with the largest exponent.
xx101010010010001000
f(x)=5x33x2+4f\left(x\right)=5{x}^{3}-3{x}^{2}+4470447044,970,0044,970,0044,997,000,0044,997,000,004
g(x)=5x3g\left(x\right)=5{x}^{3}500050005,000,0005,000,0005,000,000,0005,000,000,000
xx10-10100-1001000-1000
f(x)=5x33x2+4f\left(x\right)=5{x}^{3}-3{x}^{2}+45296-52965,029,996-5,029,9965,002,999,996-5,002,999,996
g(x)=5x3g\left(x\right)=5{x}^{3}5000-50005,000,000-5,000,0005,000,000,000-5,000,000,000

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)=p(x)q(x)f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)} 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 xx appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of x.x.

Determining End Behavior for Rational Functions

For each of the following functions, determine the limits as xx\to \infty and x.x\to \text{−}\infty . Then, use this information to describe the end behavior of the function.

  1. f(x)=3x12x+5f\left(x\right)=\frac{3x-1}{2x+5} (Note: The degree of the numerator and the denominator are the same.)
  2. f(x)=3x2+2x4x35x+7f\left(x\right)=\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7} (Note: The degree of numerator is less than the degree of the denominator.)
  3. f(x)=3x2+4xx+2f\left(x\right)=\frac{3{x}^{2}+4x}{x+2} (Note: The degree of numerator is greater than the degree of the denominator.)
  1. The highest power of xx in the denominator is x.x. Therefore, dividing the numerator and denominator by xx and applying the algebraic limit laws, we see that

    limx±3x12x+5=limx±31/x2+5/x=limx±(31/x)limx±(2+5/x)=limx±3limx±1/xlimx±2+limx±5/x=302+0=32.\begin{array}{cc}\hfill \underset{x\to \text{±}\infty }{\text{lim}}\frac{3x-1}{2x+5}& =\underset{x\to \text{±}\infty }{\text{lim}}\frac{3-1\text{/}x}{2+5\text{/}x}\hfill \\ & =\frac{\underset{x\to \text{±}\infty }{\text{lim}}\left(3-1\text{/}x\right)}{\underset{x\to \text{±}\infty }{\text{lim}}\left(2+5\text{/}x\right)}\hfill \\ & =\frac{\underset{x\to \text{±}\infty }{\text{lim}}3-\underset{x\to \text{±}\infty }{\text{lim}}1\text{/}x}{\underset{x\to \text{±}\infty }{\text{lim}}2+\underset{x\to \text{±}\infty }{\text{lim}}5\text{/}x}\hfill \\ & =\frac{3-0}{2+0}=\frac{3}{2}.\hfill \end{array}

    Since limx±f(x)=32,\underset{x\to \text{±}\infty }{\text{lim}}f\left(x\right)=\frac{3}{2}, we know that y=32y=\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±.x\to \text{±}\infty .
    The function f(x) = (3x + 1)/(2x + 5) is plotted as is its horizontal asymptote at y = 3/2.
  2. Since the largest power of xx appearing in the denominator is x3,{x}^{3}, divide the numerator and denominator by x3.{x}^{3}. After doing so and applying algebraic limit laws, we obtain

    limx±3x2+2x4x35x+7=limx±3/x+2/x245/x2+7/x3=3.0+2.045.0+7.0=0.\underset{x\to \text{±}\infty }{\text{lim}}\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}=\underset{x\to \text{±}\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.

    Therefore ff has a horizontal asymptote of y=0y=0 as shown in the following graph.

    The graph of this rational function approaches the horizontal asymptote y=0y=0 as x±.x\to \text{±}\infty .
    The function f(x) = (3x2 + 2x)/(4x2 – 5x + 7) is plotted as is its horizontal asymptote at y = 0.
  3. Dividing the numerator and denominator by x,x, we have

    limx±3x2+4xx+2=limx±3x+41+2/x.\underset{x\to \text{±}\infty }{\text{lim}}\frac{3{x}^{2}+4x}{x+2}=\underset{x\to \text{±}\infty }{\text{lim}}\frac{3x+4}{1+2\text{/}x}.

    As x±,x\to \text{±}\infty , the denominator approaches 1.1. As x,x\to \infty , the numerator approaches +.+\infty . As x,x\to \text{−}\infty , the numerator approaches .\text{−}\infty . Therefore limxf(x)=,\underset{x\to \infty }{\text{lim}}f\left(x\right)=\infty , whereas limxf(x)=\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=\text{−}\infty as shown in the following figure.

    As x,x\to \infty , the values f(x).f\left(x\right)\to \infty . As x,x\to \text{−}\infty , the values f(x).f\left(x\right)\to \text{−}\infty .
    The function f(x) = (3x2 + 4x)/(x + 2) is plotted. It appears to have a diagonal asymptote as well as a vertical asymptote at x = −2.

Evaluate limx±3x2+2x15x24x+7\underset{x\to \text{±}\infty }{\text{lim}}\frac{3{x}^{2}+2x-1}{5{x}^{2}-4x+7} and use these limits to determine the end behavior of f(x)=3x2+2x25x24x+7.f\left(x\right)=\frac{3{x}^{2}+2x-2}{5{x}^{2}-4x+7}.

35\frac{3}{5}

Hint

Divide the numerator and denominator by x2.{x}^{2}.

Before proceeding, consider the graph of f(x)=(3x2+4x)(x+2)f\left(x\right)=\frac{\left(3{x}^{2}+4x\right)}{\left(x+2\right)} shown in [link]. As xx\to \infty and x,x\to \text{−}\infty , the graph of ff appears almost linear. Although ff is certainly not a linear function, we now investigate why the graph of ff seems to be approaching a linear function. First, using long division of polynomials, we can write

f(x)=3x2+4xx+2=3x2+4x+2.f\left(x\right)=\frac{3{x}^{2}+4x}{x+2}=3x-2+\frac{4}{x+2}.

Since 4(x+2)0\frac{4}{\left(x+2\right)}\to 0 as x±,x\to \text{±}\infty , we conclude that

limx±(f(x)(3x2))=limx±4x+2=0.\underset{x\to \text{±}\infty }{\text{lim}}\left(f\left(x\right)-\left(3x-2\right)\right)=\underset{x\to \text{±}\infty }{\text{lim}}\frac{4}{x+2}=0.

Therefore, the graph of ff approaches the line y=3x2y=3x-2 as x±.x\to \text{±}\infty . This line is known as an oblique asymptote for ff ([link]).

The graph of the rational function f(x)=(3x2+4x)/(x+2)f\left(x\right)=\left(3{x}^{2}+4x\right)\text{/}\left(x+2\right) approaches the oblique asymptote y=3x2asx±.y=3x-2\phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}x\to \text{±}\infty .
The function f(x) = (3x2 + 4x)/(x + 2) is plotted as is its diagonal asymptote y = 3x – 2.

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

f(x)=p(x)q(x)=anxn+an1xn1++a1x+a0bmxm+bm1xm1++b1x+b0,f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}=\frac{{a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\text{…}+{a}_{1}x+{a}_{0}}{{b}_{m}{x}^{m}+{b}_{m-1}{x}^{m-1}+\text{…}+{b}_{1}x+{b}_{0}},

where an0andbm0.{a}_{n}\ne 0\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{b}_{m}\ne 0.

  1. If the degree of the numerator is the same as the degree of the denominator (n=m),\left(n=m\right), then ff has a horizontal asymptote of y=an/bmy={a}_{n}\text{/}{b}_{m} as x±.x\to \text{±}\infty .
  2. If the degree of the numerator is less than the degree of the denominator (n<m),\left(n<m\right), then ff has a horizontal asymptote of y=0y=0 as x±.x\to \text{±}\infty .
  3. If the degree of the numerator is greater than the degree of the denominator (n>m),\left(n>m\right), then ff 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

    f(x)=p(x)q(x)=g(x)+r(x)q(x),f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}=g\left(x\right)+\frac{r\left(x\right)}{q\left(x\right)},

    where the degree of r(x)r\left(x\right) is less than the degree of q(x).q\left(x\right). As a result, limx±r(x)/q(x)=0.\underset{x\to \text{±}\infty }{\text{lim}}r\left(x\right)\text{/}q\left(x\right)=0. Therefore, the values of [f(x)g(x)]\left[f\left(x\right)-g\left(x\right)\right] approach zero as x±.x\to \text{±}\infty . If the degree of p(x)p\left(x\right) is exactly one more than the degree of q(x)q\left(x\right) (n=m+1),\left(n=m+1\right), the function g(x)g\left(x\right) is a linear function. In this case, we call g(x)g\left(x\right) 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 xx\to \infty and xx\to \text{−}\infty for f(x)=3x24x2+5f\left(x\right)=\frac{3x-2}{\sqrt{4{x}^{2}+5}} and describe the end behavior of f.f.

Let’s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of x.x. To determine the appropriate power of x,x, consider the expression 4x2+5\sqrt{4{x}^{2}+5} in the denominator. Since

4x2+54x2=2x\sqrt{4{x}^{2}+5}\approx \sqrt{4{x}^{2}}=2|x|

for large values of xx in effect xx appears just to the first power in the denominator. Therefore, we divide the numerator and denominator by x.|x|. Then, using the fact that x=x|x|=x for x>0,x>0, x=x|x|=\text{−}x for x<0,x<0, and x=x2|x|=\sqrt{{x}^{2}} for all x,x, we calculate the limits as follows:

limx3x24x2+5=limx(1/x)(3x2)(1/x)4x2+5=limx(1/x)(3x2)(1/x2)(4x2+5)=limx32/x4+5/x2=34=32limx3x24x2+5=limx(1/x)(3x2)(1/x)4x2+5=limx(1/x)(3x2)(1/x2)(4x2+5)=limx3+2/x4+5/x2=34=32.\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{\left(1\text{/}|x|\right)\left(3x-2\right)}{\left(1\text{/}|x|\right)\sqrt{4{x}^{2}+5}}\hfill \\ & =\hfill & \underset{x\to \infty }{\text{lim}}\frac{\left(1\text{/}x\right)\left(3x-2\right)}{\sqrt{\left(1\text{/}{x}^{2}\right)\left(4{x}^{2}+5\right)}}\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 \text{−}\infty }{\text{lim}}\frac{3x-2}{\sqrt{4{x}^{2}+5}}& =\hfill & \underset{x\to \text{−}\infty }{\text{lim}}\frac{\left(1\text{/}|x|\right)\left(3x-2\right)}{\left(1\text{/}|x|\right)\sqrt{4{x}^{2}+5}}\hfill \\ & =\hfill & \underset{x\to \text{−}\infty }{\text{lim}}\frac{\left(-1\text{/}x\right)\left(3x-2\right)}{\sqrt{\left(1\text{/}{x}^{2}\right)\left(4{x}^{2}+5\right)}}\hfill \\ & =\hfill & \underset{x\to \text{−}\infty }{\text{lim}}\frac{-3+2\text{/}x}{\sqrt{4+5\text{/}{x}^{2}}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.\hfill \end{array}

Therefore, f(x)f\left(x\right) approaches the horizontal asymptote y=32y=\frac{3}{2} as xx\to \infty and the horizontal asymptote y=32y=-\frac{3}{2} as xx\to \text{−}\infty as shown in the following graph.

This function has two horizontal asymptotes and it crosses one of the asymptotes.
The function f(x) = (3x − 2)/(the square root of the quantity (4x2 + 5)) is plotted. It has two horizontal asymptotes at y = ±3/2, and it crosses y = −3/2 before converging toward it from below.

Evaluate limx3x2+4x+6.\underset{x\to \infty }{\text{lim}}\frac{\sqrt{3{x}^{2}+4}}{x+6}.

±3\text{±}\sqrt{3}

Hint

Divide the numerator and denominator by x.|x|.

Determining End Behavior for Transcendental Functions

The six basic trigonometric functions are periodic and do not approach a finite limit as x±.x\to \text{±}\infty . For example, sinx\text{sin}\phantom{\rule{0.1em}{0ex}}x oscillates between 1and11\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-1 ([link]). The tangent function xx has an infinite number of vertical asymptotes as x±;x\to \text{±}\infty ; therefore, it does not approach a finite limit nor does it approach ±\text{±}\infty as x±x\to \text{±}\infty as shown in [link].

The function f(x)=sinxf\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x oscillates between 1and11\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-1 as x±x\to \text{±}\infty
The function f(x) = sin x is graphed.
The function f(x)=tanxf\left(x\right)=\text{tan}\phantom{\rule{0.1em}{0ex}}x does not approach a limit and does not approach ±\text{±}\infty as x±x\to \text{±}\infty
The function f(x) = tan x is graphed.

Recall that for any base b>0,b1,b>0,b\ne 1, the function y=bxy={b}^{x} is an exponential function with domain (,)\left(\text{−}\infty ,\infty \right) and range (0,).\left(0,\infty \right). If b>1,y=bxb>1,y={b}^{x} is increasing over (,).`\left(\text{−}\infty ,\infty \right). If 0<b<1,0<b<1, y=bxy={b}^{x} is decreasing over (,).\left(\text{−}\infty ,\infty \right). For the natural exponential function f(x)=ex,f\left(x\right)={e}^{x}, e2.718>1.e\approx 2.718>1. Therefore, f(x)=exf\left(x\right)={e}^{x} is increasing on (,)`\left(\text{−}\infty ,\infty \right) and the range is (0,).`\left(0,\infty \right). The exponential function f(x)=exf\left(x\right)={e}^{x} approaches \infty as xx\to \infty and approaches 00 as xx\to \text{−}\infty as shown in [link] and [link].

End behavior of the natural exponential function
xx5-52-2002255
ex{e}^{x}0.006740.006740.1350.135117.3897.389148.413148.413
The exponential function approaches zero as xx\to \text{−}\infty and approaches \infty as x.x\to \infty .
The function f(x) = ex is graphed.

Recall that the natural logarithm function f(x)=ln(x)f\left(x\right)=\text{ln}\left(x\right) is the inverse of the natural exponential function y=ex.y={e}^{x}. Therefore, the domain of f(x)=ln(x)f\left(x\right)=\text{ln}\left(x\right) is (0,)\left(0,\infty \right) and the range is (,).\left(\text{−}\infty ,\infty \right). The graph of f(x)=ln(x)f\left(x\right)=\text{ln}\left(x\right) is the reflection of the graph of y=exy={e}^{x} about the line y=x.y=x. Therefore, ln(x)\text{ln}\left(x\right)\to \text{−}\infty as x0+x\to {0}^{+} and ln(x)\text{ln}\left(x\right)\to \infty as xx\to \infty as shown in [link] and [link].

End behavior of the natural logarithm function
xx0.010.010.10.1111010100100
ln(x)\text{ln}\left(x\right)4.605-4.6052.303-2.303002.3032.3034.6054.605
The natural logarithm function approaches \infty as x.x\to \infty .
The function f(x) = ln(x) is graphed.
Determining End Behavior for a Transcendental Function

Find the limits as xx\to \infty and xx\to \text{−}\infty for f(x)=(2+3ex)(75ex)f\left(x\right)=\frac{\left(2+3{e}^{x}\right)}{\left(7-5{e}^{x}\right)} and describe the end behavior of f.f.

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

limxf(x)=limx2+3ex75ex=limx(2/ex)+3(7/ex)5.\begin{array}{cc}\hfill \underset{x\to \infty }{\text{lim}}f\left(x\right)& =\underset{x\to \infty }{\text{lim}}\frac{2+3{e}^{x}}{7-5{e}^{x}}\hfill \\ & =\underset{x\to \infty }{\text{lim}}\frac{\left(2\text{/}{e}^{x}\right)+3}{\left(7\text{/}{e}^{x}\right)-5}.\hfill \end{array}

As shown in [link], ex{e}^{x}\to \infty as x.x\to \infty . Therefore,

limx2ex=0=limx7ex.\underset{x\to \infty }{\text{lim}}\frac{2}{{e}^{x}}=0=\underset{x\to \infty }{\text{lim}}\frac{7}{{e}^{x}}.

We conclude that limxf(x)=35,\underset{x\to \infty }{\text{lim}}f\left(x\right)=-\frac{3}{5}, and the graph of ff approaches the horizontal asymptote y=35y=-\frac{3}{5} as x.x\to \infty . To find the limit as x,x\to \text{−}\infty , use the fact that ex0{e}^{x}\to 0 as xx\to \text{−}\infty to conclude that limxf(x)=27,\underset{x\to \infty }{\text{lim}}f\left(x\right)=\frac{2}{7}, and therefore the graph of approaches the horizontal asymptote y=27y=\frac{2}{7} as x.x\to \text{−}\infty .

Find the limits as xx\to \infty and xx\to \text{−}\infty for f(x)=(3ex4)(5ex+2).f\left(x\right)=\frac{\left(3{e}^{x}-4\right)}{\left(5{e}^{x}+2\right)}.

limxf(x)=35,\underset{x\to \infty }{\text{lim}}f\left(x\right)=\frac{3}{5},limxf(x)=2\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=-2

Hint

limxex=\underset{x\to \infty }{\text{lim}}{e}^{x}=\infty and limxex=0.\underset{x\to \infty }{\text{lim}}{e}^{x}=0.

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 f,f, use the following steps to sketch a graph of f:f\text{:}

  1. Determine the domain of the function.
  2. Locate the xx- and yy-intercepts.
  3. Evaluate limxf(x)\underset{x\to \infty }{\text{lim}}f\left(x\right) and limxf(x)\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right) to determine the end behavior. If either of these limits is a finite number L,L, then y=Ly=L is a horizontal asymptote. If either of these limits is \infty or ,\text{−}\infty , determine whether ff has an oblique asymptote. If ff is a rational function such that f(x)=p(x)q(x),f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}, where the degree of the numerator is greater than the degree of the denominator, then ff can be written as

    f(x)=p(x)q(x)=g(x)+r(x)q(x),f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}=g\left(x\right)+\frac{r\left(x\right)}{q\left(x\right)},

    where the degree of r(x)r\left(x\right) is less than the degree of q(x).q\left(x\right). The values of f(x)f\left(x\right) approach the values of g(x)g\left(x\right) as x±.x\to \text{±}\infty . If g(x)g\left(x\right) is a linear function, it is known as an oblique asymptote.
  4. Determine whether ff has any vertical asymptotes.
  5. Calculate f.{f}^{\prime }. Find all critical points and determine the intervals where ff is increasing and where ff is decreasing. Determine whether ff has any local extrema.
  6. Calculate f.f\text{″}. Determine the intervals where ff is concave up and where ff is concave down. Use this information to determine whether ff has any inflection points. The second derivative can also be used as an alternate means to determine or verify that ff 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 f(x)=(x1)2(x+2).f\left(x\right)={\left(x-1\right)}^{2}\left(x+2\right).

Step 1. Since ff is a polynomial, the domain is the set of all real numbers.

Step 2. When x=0,f(x)=2.x=0,f\left(x\right)=2. Therefore, the yy-intercept is (0,2).\left(0,2\right). To find the xx-intercepts, we need to solve the equation (x1)2(x+2)=0,{\left(x-1\right)}^{2}\left(x+2\right)=0, gives us the xx-intercepts (1,0)\left(1,0\right) and (2,0)\left(-2,0\right)

Step 3. We need to evaluate the end behavior of f.f. As x,x\to \infty , (x1)2{\left(x-1\right)}^{2}\to \infty and (x+2).\left(x+2\right)\to \infty . Therefore, limxf(x)=.\underset{x\to \infty }{\text{lim}}f\left(x\right)=\infty . As x,x\to \text{−}\infty , (x1)2{\left(x-1\right)}^{2}\to \infty and (x+2).\left(x+2\right)\to \text{−}\infty . Therefore, limxf(x)=.\underset{x\to \infty }{\text{lim}}f\left(x\right)=\text{−}\infty . To get even more information about the end behavior of f,f, we can multiply the factors of f.f. When doing so, we see that

f(x)=(x1)2(x+2)=x33x+2.f\left(x\right)={\left(x-1\right)}^{2}\left(x+2\right)={x}^{3}-3x+2.

Since the leading term of ff is x3,{x}^{3}, we conclude that ff behaves like y=x3y={x}^{3} as x±.x\to \text{±}\infty .

Step 4. Since ff is a polynomial function, it does not have any vertical asymptotes.

Step 5. The first derivative of ff is

f(x)=3x23.{f}^{\prime }\left(x\right)=3{x}^{2}-3.

Therefore, ff has two critical points: x=1,1.x=1,-1. Divide the interval (,)\left(\text{−}\infty ,\infty \right) into the three smaller intervals: (,1),\left(\text{−}\infty ,-1\right), (1,1),\left(-1,1\right), and (1,).\left(1,\infty \right). Then, choose test points x=2,x=-2, x=0,x=0, and x=2x=2 from these intervals and evaluate the sign of f(x){f}^{\prime }\left(x\right) at each of these test points, as shown in the following table.

IntervalTest PointSign of Derivative f(x)=3x23=3(x1)(x+1)f\prime \left(x\right)=3{x}^{2}-3=3\left(x-1\right)\left(x+1\right)Conclusion
(,1)\left(\text{−}\infty ,-1\right)x=2x=-2(+)()()=+\left(\text{+}\right)\left(\text{−}\right)\left(\text{−}\right)=+ff is increasing.
(1,1)\left(-1,1\right)x=0x=0(+)()(+)=\left(\text{+}\right)\left(\text{−}\right)\left(\text{+}\right)=\text{−}ff is decreasing.
(1,)\left(1,\infty \right)x=2x=2(+)(+)(+)=+\left(\text{+}\right)\left(\text{+}\right)\left(\text{+}\right)=+ff is increasing.

From the table, we see that ff has a local maximum at x=1x=-1 and a local minimum at x=1.x=1. Evaluating f(x)f\left(x\right) at those two points, we find that the local maximum value is f(1)=4f\left(-1\right)=4 and the local minimum value is f(1)=0.f\left(1\right)=0.

Step 6. The second derivative of ff is

f(x)=6x.f\text{″}\left(x\right)=6x.

The second derivative is zero at x=0.x=0. Therefore, to determine the concavity of f,f, divide the interval (,)\left(\text{−}\infty ,\infty \right) into the smaller intervals (,0)\left(\text{−}\infty ,0\right) and (0,),\left(0,\infty \right), and choose test points x=1x=-1 and x=1x=1 to determine the concavity of ff on each of these smaller intervals as shown in the following table.

IntervalTest PointSign of f(x)=6xf\text{″}\left(x\right)=6xConclusion
(,0)\left(\text{−}\infty ,0\right)x=1x=-1-ff is concave down.
(0,)\left(0,\infty \right)x=1x=1++ff is concave up.

We note that the information in the preceding table confirms the fact, found in step 5,5, that ff has a local maximum at x=1x=-1 and a local minimum at x=1.x=1. In addition, the information found in step 55—namely, ff has a local maximum at x=1x=-1 and a local minimum at x=1,x=1, and f(x)=0{f}^{\prime }\left(x\right)=0 at those points—combined with the fact that ff\text{″} changes sign only at x=0x=0 confirms the results found in step 66 on the concavity of f.f.

Combining this information, we arrive at the graph of f(x)=(x1)2(x+2)f\left(x\right)={\left(x-1\right)}^{2}\left(x+2\right) shown in the following graph.

The function f(x) = (x −1)2 (x + 2) is graphed. It crosses the x axis at x = −2 and touches the x axis at x = 1.

Sketch a graph of f(x)=(x1)3(x+2).f\left(x\right)={\left(x-1\right)}^{3}\left(x+2\right).


The function f(x) = (x −1)3(x + 2) is graphed.
Hint

ff is a fourth-degree polynomial.

Sketching a Rational Function

Sketch the graph of f(x)=x2(1x2).f\left(x\right)=\frac{{x}^{2}}{\left(1-{x}^{2}\right)}\text{.}

Step 1. The function ff is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers xx except x=±1.x=\text{±}1.

Step 2. Find the intercepts. If x=0,x=0, then f(x)=0,f\left(x\right)=0, so 00 is an intercept. If y=0,y=0, then x2(1x2)=0,\frac{{x}^{2}}{\left(1-{x}^{2}\right)}=0, which implies x=0.x=0. Therefore, (0,0)\left(0,0\right) is the only intercept.

Step 3. Evaluate the limits at infinity. Since ff is a rational function, divide the numerator and denominator by the highest power in the denominator: x2.{x}^{2}. We obtain

limx±x21x2=limx±11x21=1.\underset{x\to \text{±}\infty }{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\underset{x\to \text{±}\infty }{\text{lim}}\frac{1}{\frac{1}{{x}^{2}}-1}=-1.

Therefore, ff has a horizontal asymptote of y=1y=-1 as xx\to \infty and x.x\to \text{−}\infty .

Step 4. To determine whether ff has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when x=±1.x=\text{±}1. To determine whether the lines x=1x=1 or x=1x=-1 are vertical asymptotes of f,f, evaluate limx1f(x)\underset{x\to 1}{\text{lim}}f\left(x\right) and limx1f(x).\underset{x\to \text{−}1}{\text{lim}}f\left(x\right). By looking at each one-sided limit as x1,x\to 1, we see that

limx1+x21x2=andlimx1x21x2=.\underset{x\to {1}^{+}}{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\text{−}\infty \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to {1}^{-}}{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\infty .

In addition, by looking at each one-sided limit as x1,x\to \text{−}1, we find that

limx1+x21x2=andlimx1x21x2=.\underset{x\to \text{−}{1}^{+}}{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\infty \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to \text{−}{1}^{-}}{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\text{−}\infty .

Step 5. Calculate the first derivative:

f(x)=(1x2)(2x)x2(2x)(1x2)2=2x(1x2)2.{f}^{\prime }\left(x\right)=\frac{\left(1-{x}^{2}\right)\left(2x\right)-{x}^{2}\left(-2x\right)}{{\left(1-{x}^{2}\right)}^{2}}=\frac{2x}{{\left(1-{x}^{2}\right)}^{2}}.

Critical points occur at points xx where f(x)=0{f}^{\prime }\left(x\right)=0 or f(x){f}^{\prime }\left(x\right) is undefined. We see that f(x)=0{f}^{\prime }\left(x\right)=0 when x=0.x=0. The derivative f{f}^{\prime } is not undefined at any point in the domain of f.f. However, x=±1x=\text{±}1 are not in the domain of f.f. Therefore, to determine where ff is increasing and where ff is decreasing, divide the interval (,)\left(\text{−}\infty ,\infty \right) into four smaller intervals: (,1),\left(\text{−}\infty ,-1\right), (1,0),\left(-1,0\right), (0,1),\left(0,1\right), and (1,),\left(1,\infty \right), and choose a test point in each interval to determine the sign of f(x){f}^{\prime }\left(x\right) in each of these intervals. The values x=2,x=-2, x=12,x=-\frac{1}{2}, x=12,x=\frac{1}{2}, and x=2x=2 are good choices for test points as shown in the following table.

IntervalTest PointSign of f(x)=2x(1x2)2{f}^{\prime }\left(x\right)=\frac{2x}{{\left(1-{x}^{2}\right)}^{2}}Conclusion
(,1)\left(\text{−}\infty ,-1\right)x=2x=-2−/+=\text{−}\text{/}+=\text{−}ff is decreasing.
(1,0)\left(-1,0\right)x=1/2x=-1\text{/}2−/+=\text{−}\text{/}+=\text{−}ff is decreasing.
(0,1)\left(0,1\right)x=1/2x=1\text{/}2+/+=++\text{/}+=+ff is increasing.
(1,)\left(1,\infty \right)x=2x=2+/+=++\text{/}+=+ff is increasing.

From this analysis, we conclude that ff has a local minimum at x=0x=0 but no local maximum.

Step 6. Calculate the second derivative:

f(x)=(1x2)2(2)2x(2(1x2)(2x))(1x2)4=(1x2)[2(1x2)+8x2](1x2)4=2(1x2)+8x2(1x2)3=6x2+2(1x2)3.\begin{array}{cc}\hfill f\text{″}\left(x\right)& \hfill =\frac{{\left(1-{x}^{2}\right)}^{2}\left(2\right)-2x\left(2\left(1-{x}^{2}\right)\left(-2x\right)\right)}{{\left(1-{x}^{2}\right)}^{4}}\\ & =\frac{\left(1-{x}^{2}\right)\left[2\left(1-{x}^{2}\right)+8{x}^{2}\right]}{{\left(1-{x}^{2}\right)}^{4}}\hfill \\ & =\frac{2\left(1-{x}^{2}\right)+8{x}^{2}}{{\left(1-{x}^{2}\right)}^{3}}\hfill \\ & =\frac{6{x}^{2}+2}{{\left(1-{x}^{2}\right)}^{3}}.\hfill \end{array}

To determine the intervals where ff is concave up and where ff is concave down, we first need to find all points xx where f(x)=0f\text{″}\left(x\right)=0 or f(x)f\text{″}\left(x\right) is undefined. Since the numerator 6x2+206{x}^{2}+2\ne 0 for any x,x, f(x)f\text{″}\left(x\right) is never zero. Furthermore, ff\text{″} is not undefined for any xx in the domain of f.f. However, as discussed earlier, x=±1x=\text{±}1 are not in the domain of f.f. Therefore, to determine the concavity of f,f, we divide the interval (,)\left(\text{−}\infty ,\infty \right) into the three smaller intervals (,1),\left(\text{−}\infty ,-1\right), (1,1),\left(-1,-1\right), and (1,),\left(1,\infty \right), and choose a test point in each of these intervals to evaluate the sign of f(x).f\text{″}\left(x\right). in each of these intervals. The values x=2,x=-2, x=0,x=0, and x=2x=2 are possible test points as shown in the following table.

IntervalTest PointSign of f(x)=6x2+2(1x2)3f\text{″}\left(x\right)=\frac{6{x}^{2}+2}{{\left(1-{x}^{2}\right)}^{3}}Conclusion
(,1)\left(\text{−}\infty ,-1\right)x=2x=-2+/=+\text{/}-=\text{−}ff is concave down.
(1,1)\left(-1,-1\right)x=0x=0+/+=++\text{/}+=+ff is concave up.
(1,)\left(1,\infty \right)x=2x=2+/=+\text{/}-=\text{−}ff is concave down.

Combining all this information, we arrive at the graph of ff shown below. Note that, although ff changes concavity at x=1x=-1 and x=1,x=1, there are no inflection points at either of these places because ff is not continuous at x=1x=-1 or x=1.x=1.

The function f(x) = x2/(1 − x2) is graphed. It has asymptotes y = −1, x = −1, and x = 1.

Sketch a graph of f(x)=(3x+5)(8+4x).f\left(x\right)=\frac{\left(3x+5\right)}{\left(8+4x\right)}.


The function f(x) = (3x + 5)/(8 + 4x) is graphed. It appears to have asymptotes at x = −2 and y = 1.
Hint

A line y=Ly=L is a horizontal asymptote of ff if the limit as xx\to \infty or the limit as xx\to \text{−}\infty of f(x)f\left(x\right) is L.L. A line x=ax=a is a vertical asymptote if at least one of the one-sided limits of ff as xax\to a is \infty or .\text{−}\infty .

Sketching a Rational Function with an Oblique Asymptote

Sketch the graph of f(x)=x2(x1)f\left(x\right)=\frac{{x}^{2}}{\left(x-1\right)}

Step 1. The domain of ff is the set of all real numbers xx except x=1.x=1.

Step 2. Find the intercepts. We can see that when x=0,x=0, f(x)=0,f\left(x\right)=0, so (0,0)\left(0,0\right) is the only intercept.

Step 3. Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, ff must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write

f(x)=x2x1=x+1+1x1.f\left(x\right)=\frac{{x}^{2}}{x-1}=x+1+\frac{1}{x-1}.

Since 1/(x1)01\text{/}\left(x-1\right)\to 0 as x±,x\to \text{±}\infty , f(x)f\left(x\right) approaches the line y=x+1y=x+1 as x±.x\to \text{±}\infty . The line y=x+1y=x+1 is an oblique asymptote for f.f.

Step 4. To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at x=1.x=1. Looking at both one-sided limits as x1,x\to 1, we find

limx1+x2x1=andlimx1x2x1=.\underset{x\to {1}^{+}}{\text{lim}}\frac{{x}^{2}}{x-1}=\infty \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to {1}^{-}}{\text{lim}}\frac{{x}^{2}}{x-1}=\text{−}\infty .

Therefore, x=1x=1 is a vertical asymptote, and we have determined the behavior of ff as xx approaches 11 from the right and the left.

Step 5. Calculate the first derivative:

f(x)=(x1)(2x)x2(1)(x1)2=x22x(x1)2.{f}^{\prime }\left(x\right)=\frac{\left(x-1\right)\left(2x\right)-{x}^{2}\left(1\right)}{{\left(x-1\right)}^{2}}=\frac{{x}^{2}-2x}{{\left(x-1\right)}^{2}}.

We have f(x)=0{f}^{\prime }\left(x\right)=0 when x22x=x(x2)=0.{x}^{2}-2x=x\left(x-2\right)=0. Therefore, x=0x=0 and x=2x=2 are critical points. Since ff is undefined at x=1,x=1, we need to divide the interval (,)\left(\text{−}\infty ,\infty \right) into the smaller intervals (,0),\left(\text{−}\infty ,0\right), (0,1),\left(0,1\right), (1,2),\left(1,2\right), and (2,),\left(2,\infty \right), and choose a test point from each interval to evaluate the sign of f(x){f}^{\prime }\left(x\right) in each of these smaller intervals. For example, let x=1,x=-1, x=12,x=\frac{1}{2}, x=32,x=\frac{3}{2}, and x=3x=3 be the test points as shown in the following table.

IntervalTest PointSign of f(x)=x22x(x1)2=x(x2)(x1)2f\prime \left(x\right)=\frac{{x}^{2}-2x}{{\left(x-1\right)}^{2}}=\frac{x\left(x-2\right)}{{\left(x-1\right)}^{2}}Conclusion
(,0)\left(\text{−}\infty ,0\right)x=1x=-1()()/+=+\left(\text{−}\right)\left(\text{−}\right)\text{/}+=+ff is increasing.
(0,1)\left(0,1\right)x=1/2x=1\text{/}2(+)()/+=\left(\text{+}\right)\left(\text{−}\right)\text{/}+=\text{−}ff is decreasing.
(1,2)\left(1,2\right)x=3/2x=3\text{/}2(+)()/+=\left(\text{+}\right)\left(\text{−}\right)\text{/}+=\text{−}ff is decreasing.
(2,)\left(2,\infty \right)x=3x=3(+)(+)/+=+\left(\text{+}\right)\left(\text{+}\right)\text{/}+=+ff is increasing.

From this table, we see that ff has a local maximum at x=0x=0 and a local minimum at x=2.x=2. The value of ff at the local maximum is f(0)=0f\left(0\right)=0 and the value of ff at the local minimum is f(2)=4.f\left(2\right)=4. Therefore, (0,0)\left(0,0\right) and (2,4)\left(2,4\right) are important points on the graph.

Step 6. Calculate the second derivative:

f(x)=(x1)2(2x2)(x22x)(2(x1))(x1)4=(x1)[(x1)(2x2)2(x22x)](x1)4=(x1)(2x2)2(x22x)(x1)3=2x24x+2(2x24x)(x1)3=2(x1)3.\begin{array}{cc}\hfill f\text{″}\left(x\right)& =\frac{{\left(x-1\right)}^{2}\left(2x-2\right)-\left({x}^{2}-2x\right)\left(2\left(x-1\right)\right)}{{\left(x-1\right)}^{4}}\hfill \\ & =\frac{\left(x-1\right)\left[\left(x-1\right)\left(2x-2\right)-2\left({x}^{2}-2x\right)\right]}{{\left(x-1\right)}^{4}}\hfill \\ & =\frac{\left(x-1\right)\left(2x-2\right)-2\left({x}^{2}-2x\right)}{{\left(x-1\right)}^{3}}\hfill \\ & =\frac{2{x}^{2}-4x+2-\left(2{x}^{2}-4x\right)}{{\left(x-1\right)}^{3}}\hfill \\ & =\frac{2}{{\left(x-1\right)}^{3}}.\hfill \end{array}

We see that f(x)f\text{″}\left(x\right) is never zero or undefined for xx in the domain of f.f. Since ff is undefined at x=1,x=1, to check concavity we just divide the interval (,)\left(\text{−}\infty ,\infty \right) into the two smaller intervals (,1)\left(\text{−}\infty ,1\right) and (1,),\left(1,\infty \right), and choose a test point from each interval to evaluate the sign of f(x)f\text{″}\left(x\right) in each of these intervals. The values x=0x=0 and x=2x=2 are possible test points as shown in the following table.

IntervalTest PointSign of f(x)=2(x1)3f\text{″}\left(x\right)=\frac{2}{{\left(x-1\right)}^{3}}Conclusion
(,1)\left(\text{−}\infty ,1\right)x=0x=0+/=+\text{/}-=\text{−}ff is concave down.
(1,)\left(1,\infty \right)x=2x=2+/+=++\text{/}+=+ff is concave up.

From the information gathered, we arrive at the following graph for f.f.

The function f(x) = x2/(x − 1) is graphed. It has asymptotes y = x + 1 and x = 1.

Find the oblique asymptote for f(x)=(3x32x+1)(2x24).f\left(x\right)=\frac{\left(3{x}^{3}-2x+1\right)}{\left(2{x}^{2}-4\right)}.

y=32xy=\frac{3}{2}x

Hint

Use long division of polynomials.

Sketching the Graph of a Function with a Cusp

Sketch a graph of f(x)=(x1)2/3.f\left(x\right)={\left(x-1\right)}^{2\text{/}3}.

Step 1. Since the cube-root function is defined for all real numbers xx and (x1)2/3=(x13)2,{\left(x-1\right)}^{2\text{/}3}={\left(\sqrt[3]{x-1}\right)}^{2}, the domain of ff is all real numbers.

Step 2: To find the yy-intercept, evaluate f(0).f\left(0\right). Since f(0)=1,f\left(0\right)=1, the yy-intercept is (0,1).\left(0,1\right). To find the xx-intercept, solve (x1)2/3=0.{\left(x-1\right)}^{2\text{/}3}=0. The solution of this equation is x=1,x=1, so the xx-intercept is (1,0).\left(1,0\right).

Step 3: Since limx±(x1)2/3=,\underset{x\to \text{±}\infty }{\text{lim}}{\left(x-1\right)}^{2\text{/}3}=\infty , the function continues to grow without bound as xx\to \infty and x.x\to \text{−}\infty .

Step 4: The function has no vertical asymptotes.

Step 5: To determine where ff is increasing or decreasing, calculate f.{f}^{\prime }. We find

f(x)=23(x1)1/3=23(x1)1/3.{f}^{\prime }\left(x\right)=\frac{2}{3}{\left(x-1\right)}^{-1\text{/}3}=\frac{2}{3{\left(x-1\right)}^{1\text{/}3}}.

This function is not zero anywhere, but it is undefined when x=1.x=1. Therefore, the only critical point is x=1.x=1. Divide the interval (,)\left(\text{−}\infty ,\infty \right) into the smaller intervals (,1)\left(\text{−}\infty ,1\right) and (1,),\left(1,\infty \right), and choose test points in each of these intervals to determine the sign of f(x){f}^{\prime }\left(x\right) in each of these smaller intervals. Let x=0x=0 and x=2x=2 be the test points as shown in the following table.

IntervalTest PointSign of f(x)=23(x1)1/3{f}^{\prime }\left(x\right)=\frac{2}{3{\left(x-1\right)}^{1\text{/}3}}Conclusion
(,1)\left(\text{−}\infty ,1\right)x=0x=0+/=+\text{/}-=\text{−}ff is decreasing.
(1,)\left(1,\infty \right)x=2x=2+/+=++\text{/}+=+ff is increasing.

We conclude that ff has a local minimum at x=1.x=1. Evaluating ff at x=1,x=1, we find that the value of ff at the local minimum is zero. Note that f(1){f}^{\prime }\left(1\right) is undefined, so to determine the behavior of the function at this critical point, we need to examine limx1f(x).\underset{x\to 1}{\text{lim}}{f}^{\prime }\left(x\right). Looking at the one-sided limits, we have

limx1+23(x1)1/3=andlimx123(x1)1/3=.\underset{x\to {1}^{+}}{\text{lim}}\frac{2}{3{\left(x-1\right)}^{1\text{/}3}}=\infty \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to {1}^{-}}{\text{lim}}\frac{2}{3{\left(x-1\right)}^{1\text{/}3}}=\text{−}\infty .

Therefore, ff has a cusp at x=1.x=1.

Step 6: To determine concavity, we calculate the second derivative of f:f\text{:}

f(x)=29(x1)4/3=29(x1)4/3.f\text{″}\left(x\right)=-\frac{2}{9}{\left(x-1\right)}^{-4\text{/}3}=\frac{-2}{9{\left(x-1\right)}^{4\text{/}3}}.

We find that f(x)f\text{″}\left(x\right) is defined for all x,x, but is undefined when x=1.x=1. Therefore, divide the interval (,)\left(\text{−}\infty ,\infty \right) into the smaller intervals (,1)\left(\text{−}\infty ,1\right) and (1,),\left(1,\infty \right), and choose test points to evaluate the sign of f(x)f\text{″}\left(x\right) in each of these intervals. As we did earlier, let x=0x=0 and x=2x=2 be test points as shown in the following table.

IntervalTest PointSign of f(x)=29(x1)4/3f\text{″}\left(x\right)=\frac{-2}{9{\left(x-1\right)}^{4\text{/}3}}Conclusion
(,1)\left(\text{−}\infty ,1\right)x=0x=0−/+=\text{−}\text{/}+=\text{−}ff is concave down.
(1,)\left(1,\infty \right)x=2x=2−/+=\text{−}\text{/}+=\text{−}ff is concave down.

From this table, we conclude that ff is concave down everywhere. Combining all of this information, we arrive at the following graph for f.f.

The function f(x) = (x − 1)2/3 is graphed. It touches the x axis at x = 1, where it comes to something of a sharp point and then flairs out on either side.

Consider the function f(x)=5x2/3.f\left(x\right)=5-{x}^{2\text{/}3}. Determine the point on the graph where a cusp is located. Determine the end behavior of f.f.

The function ff has a cusp at (0,5)\left(0,5\right) limx0f(x)=,\underset{x\to {0}^{-}}{\text{lim}}{f}^{\prime }\left(x\right)=\infty , limx0+f(x)=.\underset{x\to {0}^{+}}{\text{lim}}{f}^{\prime }\left(x\right)=\text{−}\infty . For end behavior, limx±f(x)=.\underset{x\to \text{±}\infty }{\text{lim}}f\left(x\right)=\text{−}\infty .

Hint

A function ff has a cusp at a point aa if f(a)f\left(a\right) exists, f(a)f\prime \left(a\right) is undefined, one of the one-sided limits as xax\to a of f(x)f\prime \left(x\right) is +,+\infty , and the other one-sided limit is .\text{−}\infty .

Key Concepts

  • The limit of f(x)f\left(x\right) is LL as xx\to \infty (or as x\to \text{−}\infty \right) if the values f(x)f\left(x\right) become arbitrarily close to LL as xx becomes sufficiently large.
  • The limit of f(x)f\left(x\right) is \infty as xx\to \infty if f(x)f\left(x\right) becomes arbitrarily large as xx becomes sufficiently large. The limit of f(x)f\left(x\right) is \text{−}\infty as xx\to \infty if f(x)<0f\left(x\right)<0 and f(x)|f\left(x\right)| becomes arbitrarily large as xx becomes sufficiently large. We can define the limit of f(x)f\left(x\right) as xx approaches \text{−}\infty similarly.
  • For a polynomial function p(x)=anxn+an1xn1++a1x+a0,p\left(x\right)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\text{…}+{a}_{1}x+{a}_{0}, where an0,{a}_{n}\ne 0, the end behavior is determined by the leading term anxn.{a}_{n}{x}^{n}. If n0,n\ne 0, p(x)p\left(x\right) approaches \infty or \text{−}\infty at each end.
  • For a rational function f(x)=p(x)q(x),f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}, the end behavior is determined by the relationship between the degree of pp and the degree of q.q. If the degree of pp is less than the degree of q,q, the line y=0y=0 is a horizontal asymptote for f.f. If the degree of pp is equal to the degree of q,q, then the line y=anbny=\frac{{a}_{n}}{{b}_{n}} is a horizontal asymptote, where an{a}_{n} and bn{b}_{n} are the leading coefficients of pp and q,q, respectively. If the degree of pp is greater than the degree of q,q, then ff approaches \infty or \text{−}\infty at each end.

For the following exercises, examine the graphs. Identify where the vertical asymptotes are located.

The function graphed decreases very rapidly as it approaches x = 1 from the left, and on the other side of x = 1, it seems to start near infinity and then decrease rapidly.

x=1x=1

The function graphed increases very rapidly as it approaches x = −3 from the left, and on the other side of x = −3, it seems to start near negative infinity and then increase rapidly to form a sort of U shape that is pointing down, with the other side of the U being at x = 2. On the other side of x = 2, the graph seems to start near infinity and then decrease rapidly.
The function graphed decreases very rapidly as it approaches x = −1 from the left, and on the other side of x = −1, it seems to start near negative infinity and then increase rapidly to form a sort of U shape that is pointing down, with the other side of the U being at x = 2. On the other side of x = 2, the graph seems to start near infinity and then decrease rapidly.

x=1,x=2x=-1,x=2

The function graphed decreases very rapidly as it approaches x = 0 from the left, and on the other side of x = 0, it seems to start near infinity and then decrease rapidly to form a sort of U shape that is pointing up, with the other side of the U being at x = 1. On the other side of x = 1, there is another U shape pointing down, with its other side being at x = 2. On the other side of x = 2, the graph seems to start near negative infinity and then increase rapidly.
The function graphed decreases very rapidly as it approaches x = 0 from the left, and on the other side of x = 0, it seems to start near infinity and then decrease rapidly to form a sort of U shape that is pointing up, with the other side being a normal function that appears as if it will take the entirety of the values of the x-axis.

x=0x=0

For the following functions f(x),f\left(x\right), determine whether there is an asymptote at x=a.x=a. Justify your answer without graphing on a calculator.

f(x)=x+1x2+5x+4,a=1f\left(x\right)=\frac{x+1}{{x}^{2}+5x+4},a=-1

f(x)=xx2,a=2f\left(x\right)=\frac{x}{x-2},a=2

Yes, there is a vertical asymptote

f(x)=(x+2)3/2,a=2f\left(x\right)={\left(x+2\right)}^{3\text{/}2},a=-2

f(x)=(x1)1/3,a=1f\left(x\right)={\left(x-1\right)}^{-1\text{/}3},a=1

Yes, there is vertical asymptote

f(x)=1+x2/5,a=1f\left(x\right)=1+{x}^{-2\text{/}5},a=1

For the following exercises, evaluate the limit.

limx13x+6\underset{x\to \infty }{\text{lim}}\frac{1}{3x+6}

00

limx2x54x\underset{x\to \infty }{\text{lim}}\frac{2x-5}{4x}

limxx22x+5x+2\underset{x\to \infty }{\text{lim}}\frac{{x}^{2}-2x+5}{x+2}

\infty

limx3x32xx2+2x+8\underset{x\to \text{−}\infty }{\text{lim}}\frac{3{x}^{3}-2x}{{x}^{2}+2x+8}

limxx44x3+122x27x4\underset{x\to \text{−}\infty }{\text{lim}}\frac{{x}^{4}-4{x}^{3}+1}{2-2{x}^{2}-7{x}^{4}}

17-\frac{1}{7}

limx3xx2+1\underset{x\to \infty }{\text{lim}}\frac{3x}{\sqrt{{x}^{2}+1}}

limx4x21x+2\underset{x\to \text{−}\infty }{\text{lim}}\frac{\sqrt{4{x}^{2}-1}}{x+2}

2-2

limx4xx21\underset{x\to \infty }{\text{lim}}\frac{4x}{\sqrt{{x}^{2}-1}}

limx4xx21\underset{x\to \text{−}\infty }{\text{lim}}\frac{4x}{\sqrt{{x}^{2}-1}}

4-4

limx2xxx+1\underset{x\to \infty }{\text{lim}}\frac{2\sqrt{x}}{x-\sqrt{x}+1}

For the following exercises, find the horizontal and vertical asymptotes.

f(x)=x9xf\left(x\right)=x-\frac{9}{x}

Horizontal: none, vertical: x=0x=0

f(x)=11x2f\left(x\right)=\frac{1}{1-{x}^{2}}

f(x)=x34x2f\left(x\right)=\frac{{x}^{3}}{4-{x}^{2}}

Horizontal: none, vertical: x=±2x=\text{±}2

f(x)=x2+3x2+1f\left(x\right)=\frac{{x}^{2}+3}{{x}^{2}+1}

f(x)=sin(x)sin(2x)f\left(x\right)=\text{sin}\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{sin}\left(2x\right)

Horizontal: none, vertical: none

f(x)=cosx+cos(3x)+cos(5x)f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x+\text{cos}\left(3x\right)+\text{cos}\left(5x\right)

f(x)=xsin(x)x21f\left(x\right)=\frac{x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(x\right)}{{x}^{2}-1}

Horizontal: y=0,y=0, vertical: x=±1x=\text{±}1

f(x)=xsin(x)f\left(x\right)=\frac{x}{\text{sin}\left(x\right)}

f(x)=1x3+x2f\left(x\right)=\frac{1}{{x}^{3}+{x}^{2}}

Horizontal: y=0,y=0, vertical: x=0x=0 and x=1x=-1

f(x)=1x12xf\left(x\right)=\frac{1}{x-1}-2x

f(x)=x3+1x31f\left(x\right)=\frac{{x}^{3}+1}{{x}^{3}-1}

Horizontal: y=1,y=1, vertical: x=1x=1

f(x)=sinx+cosxsinxcosxf\left(x\right)=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x}{\text{sin}\phantom{\rule{0.1em}{0ex}}x-\text{cos}\phantom{\rule{0.1em}{0ex}}x}

f(x)=xsinxf\left(x\right)=x-\text{sin}\phantom{\rule{0.1em}{0ex}}x

Horizontal: none, vertical: none

f(x)=1xxf\left(x\right)=\frac{1}{x}-\sqrt{x}

For the following exercises, construct a function f(x)f\left(x\right) that has the given asymptotes.

x=1x=1 and y=2y=2

Answers will vary, for example: y=2xx1y=\frac{2x}{x-1}

x=1x=1 and y=0y=0

y=4,y=4,x=1x=-1

Answers will vary, for example: y=4xx+1y=\frac{4x}{x+1}

x=0x=0

For the following exercises, graph the function on a graphing calculator on the window x=[5,5]x=\left[-5,5\right] and estimate the horizontal asymptote or limit. Then, calculate the actual horizontal asymptote or limit.

[T]f(x)=1x+10f\left(x\right)=\frac{1}{x+10}

y=0y=0

[T]f(x)=x+1x2+7x+6f\left(x\right)=\frac{x+1}{{x}^{2}+7x+6}

[T]limxx2+10x+25\underset{x\to \text{−}\infty }{\text{lim}}{x}^{2}+10x+25

\infty

[T]limxx+2x2+7x+6\underset{x\to \text{−}\infty }{\text{lim}}\frac{x+2}{{x}^{2}+7x+6}

[T]limx3x+2x+5\underset{x\to \infty }{\text{lim}}\frac{3x+2}{x+5}

y=3y=3

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.

y=3x2+2x+4y=3{x}^{2}+2x+4

y=x33x2+4y={x}^{3}-3{x}^{2}+4


The function starts in the third quadrant, increases to pass through (−1, 0), increases to a maximum and y intercept at 4, decreases to touch (2, 0), and then increases to (4, 20).

y=2x+1x2+6x+5y=\frac{2x+1}{{x}^{2}+6x+5}

y=x3+4x2+3x3x+9y=\frac{{x}^{3}+4{x}^{2}+3x}{3x+9}


An upward-facing parabola with minimum between x = 0 and x = −1 with y intercept between 0 and 1.

y=x2+x2x23x4y=\frac{{x}^{2}+x-2}{{x}^{2}-3x-4}

y=x25x+4y=\sqrt{{x}^{2}-5x+4}


This graph starts at (−2, 4) and decreases in a convex way to (1, 0). Then the graph starts again at (4, 0) and increases in a convex way to (6, 3).

y=2x16x2y=2x\sqrt{16-{x}^{2}}

y=cosxx,y=\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x}, on x=[2π,2π]x=\left[-2\pi ,2\pi \right]


This graph has vertical asymptote at x = 0. The first part of the function occurs in the second and third quadrants and starts in the third quadrant just below (−2π, 0), increases and passes through the x axis at −3π/2, reaches a maximum and then decreases through the x axis at −π/2 before approaching the asymptote. On the other side of the asymptote, the function starts in the first quadrant, decreases quickly to pass through π/2, decreases to a local minimum and then increases through (3π/2, 0) before staying just above (2π, 0).

y=exx3y={e}^{x}-{x}^{3}

y=xtanx,x=[π,π]y=x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x,x=\left[\text{−}\pi ,\pi \right]


This graph has vertical asymptotes at x = ±π/2. The graph is symmetric about the y axis, so describing the left hand side will be sufficient. The function starts at (−π, 0) and decreases quickly to the asymptote. Then it starts on the other side of the asymptote in the second quadrant and decreases to the the origin.

y=xln(x),x>0y=x\phantom{\rule{0.1em}{0ex}}\text{ln}\left(x\right),x>0

y=x2sin(x),x=[2π,2π]y={x}^{2}\text{sin}\left(x\right),x=\left[-2\pi ,2\pi \right]


This function starts at (−2π, 0), increases to near (−3π/2, 25), decreases through (−π, 0), achieves a local minimum and then increases through the origin. On the other side of the origin, the graph is the same but flipped, that is, it is congruent to the other half by a rotation of 180 degrees.

For f(x)=P(x)Q(x)f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)} to have an asymptote at y=2y=2 then the polynomials P(x)P\left(x\right) and Q(x)Q\left(x\right) must have what relation?

For f(x)=P(x)Q(x)f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)} to have an asymptote at x=0,x=0, then the polynomials P(x)P\left(x\right) and Q(x).Q\left(x\right). must have what relation?

Q(x).Q\left(x\right). must have have xk+1{x}^{k+1} as a factor, where P(x)P\left(x\right) has xk{x}^{k} as a factor.

If f(x){f}^{\prime }\left(x\right) has asymptotes at y=3y=3 and x=1,x=1, then f(x)f\left(x\right) has what asymptotes?

Both f(x)=1(x1)f\left(x\right)=\frac{1}{\left(x-1\right)} and g(x)=1(x1)2g\left(x\right)=\frac{1}{{\left(x-1\right)}^{2}} have asymptotes at x=1x=1 and y=0.y=0. What is the most obvious difference between these two functions?

limx1f(x)andlimx1g(x)\underset{x\to {1}^{-}}{\text{lim}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to {1}^{-}}{\text{lim}}g\left(x\right)

True or false: Every ratio of polynomials has vertical asymptotes.

Glossary

end behavior
the behavior of a function as xx\to \infty and xx\to \text{−}\infty
horizontal asymptote
if limxf(x)=L\underset{x\to \infty }{\text{lim}}f\left(x\right)=L or limxf(x)=L,\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=L, then y=Ly=L is a horizontal asymptote of ff
infinite limit at infinity
a function that becomes arbitrarily large as x becomes large
limit at infinity
the limiting value, if it exists, of a function as xx\to \infty or xx\to \text{−}\infty
oblique asymptote
the line y=mx+by=mx+b if f(x)f\left(x\right) approaches it as xx\to \infty or xx\to \text{−}\infty