# 4.6 Limits at infinity and asymptotes  (Page 4/14)

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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 $f\left(x\right)={x}^{n}$ where $n$ is a positive integer. From [link] and [link] , we see that

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

and

$\underset{x\to \text{−}\infty }{\text{lim}}{x}^{n}=\left\{\begin{array}{c}\infty ;n=2,4,6\text{,…}\hfill \\ \text{−}\infty ;n=1,3,5\text{,…}\hfill \end{array}.$

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

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

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

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

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

## Limits at infinity for power functions

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

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

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

$\text{−}\infty$

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

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

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

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

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

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

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

 $x$ $10$ $100$ $1000$ $f\left(x\right)=5{x}^{3}-3{x}^{2}+4$ $4704$ $4,970,004$ $4,997,000,004$ $g\left(x\right)=5{x}^{3}$ $5000$ $5,000,000$ $5,000,000,000$ $x$ $-10$ $-100$ $-1000$ $f\left(x\right)=5{x}^{3}-3{x}^{2}+4$ $-5296$ $-5,029,996$ $-5,002,999,996$ $g\left(x\right)=5{x}^{3}$ $-5000$ $-5,000,000$ $-5,000,000,000$

## End behavior for algebraic functions

The end behavior for rational functions and functions involving radicals is a little more complicated than for polynomials. In [link] , we show that the limits at infinity of a rational function $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ 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 $x$ appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of $x.$

## Determining end behavior for rational functions

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

1. $f\left(x\right)=\frac{3x-1}{2x+5}$ (Note: The degree of the numerator and the denominator are the same.)
2. $f\left(x\right)=\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}$ (Note: The degree of numerator is less than the degree of the denominator.)
3. $f\left(x\right)=\frac{3{x}^{2}+4x}{x+2}$ (Note: The degree of numerator is greater than the degree of the denominator.)
1. The highest power of $x$ in the denominator is $x.$ Therefore, dividing the numerator and denominator by $x$ and applying the algebraic limit laws, we see that
$\begin{array}{cc}\hfill \underset{x\to \text{±}\infty }{\text{lim}}\frac{3x-1}{2x+5}& =\underset{x\to \text{±}\infty }{\text{lim}}\frac{3-1\text{/}x}{2+5\text{/}x}\hfill \\ & =\frac{\underset{x\to \text{±}\infty }{\text{lim}}\left(3-1\text{/}x\right)}{\underset{x\to \text{±}\infty }{\text{lim}}\left(2+5\text{/}x\right)}\hfill \\ & =\frac{\underset{x\to \text{±}\infty }{\text{lim}}3-\underset{x\to \text{±}\infty }{\text{lim}}1\text{/}x}{\underset{x\to \text{±}\infty }{\text{lim}}2+\underset{x\to \text{±}\infty }{\text{lim}}5\text{/}x}\hfill \\ & =\frac{3-0}{2+0}=\frac{3}{2}.\hfill \end{array}$

Since $\underset{x\to \text{±}\infty }{\text{lim}}f\left(x\right)=\frac{3}{2},$ we know that $y=\frac{3}{2}$ is a horizontal asymptote for this function as shown in the following graph.
2. Since the largest power of $x$ appearing in the denominator is ${x}^{3},$ divide the numerator and denominator by ${x}^{3}.$ After doing so and applying algebraic limit laws, we obtain
$\underset{x\to \text{±}\infty }{\text{lim}}\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}=\underset{x\to \text{±}\infty }{\text{lim}}\frac{3\text{/}x+2\text{/}{x}^{2}}{4-5\text{/}{x}^{2}+7\text{/}{x}^{3}}=\frac{3.0+2.0}{4-5.0+7.0}=0.$

Therefore $f$ has a horizontal asymptote of $y=0$ as shown in the following graph.
3. Dividing the numerator and denominator by $x,$ we have
$\underset{x\to \text{±}\infty }{\text{lim}}\frac{3{x}^{2}+4x}{x+2}=\underset{x\to \text{±}\infty }{\text{lim}}\frac{3x+4}{1+2\text{/}x}.$

As $x\to \text{±}\infty ,$ the denominator approaches $1.$ As $x\to \infty ,$ the numerator approaches $+\infty .$ As $x\to \text{−}\infty ,$ the numerator approaches $\text{−}\infty .$ Therefore $\underset{x\to \infty }{\text{lim}}f\left(x\right)=\infty ,$ whereas $\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=\text{−}\infty$ as shown in the following figure.

can you give me a problem for function. a trigonometric one
state and prove L hospital rule
I want to know about hospital rule
Faysal
If you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)
Faysal
I don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning
Amor
if you want some help on l hospital rule ask me
it's spelled hopital
Connor
hi
BERNANDINO
you are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'Hôpital.
Leo
I had no clue this was an online app
Connor
Total online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were$48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach \$60 billion?
what is the derivative of x= Arc sin (x)^1/2
y^2 = arcsin(x)
Pitior
x = sin (y^2)
Pitior
differentiate implicitly
Pitior
then solve for dy/dx
Pitior
thank you it was very helpful
morfling
questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio