# 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.

f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz
why do we need to study functions?
to understand how to model one variable as a direct relationship to another variable
Andrew
integrate the root of 1+x²
use the substitution t=1+x. dt=dx √(1+x)dx = √tdt = t^1/2 dt integral is then = t^(1/2 + 1) / (1/2 + 1) + C = (2/3) t^(3/2) + C substitute back t=1+x = (2/3) (1+x)^(3/2) + C
navin
find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
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?