# 4.6 Limits at infinity and asymptotes  (Page 9/14)

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

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

## Key concepts

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

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

$x=1$

$x=-1,x=2$

$x=0$

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

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

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

Yes, there is a vertical asymptote

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

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

Yes, there is vertical asymptote

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

For the following exercises, evaluate the limit.

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

$0$

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

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

$\infty$

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

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

$-\frac{1}{7}$

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

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

$-2$

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

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

$-4$

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

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

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

Horizontal: none, vertical: $x=0$

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

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

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

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

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

Horizontal: none, vertical: none

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

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

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

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

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

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

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

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

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

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

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

Horizontal: none, vertical: none

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

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

$x=1$ and $y=2$

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

$x=1$ and $y=0$

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

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

$x=0$

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

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

$y=0$

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

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

$\infty$

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

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

$y=3$

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.

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

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

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

$y=\frac{{x}^{3}+4{x}^{2}+3x}{3x+9}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

I don't understand the formula
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funny
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solve number one step by step
x-xcosx/sinsq.3x
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x-xcosx/sin^23x
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1-sin x/cos x= cos x/-1+sin x
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how to prove tan^2 x=csc^2 x tan^2 x-1?
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sin x - sin x cos^2 x sin x (1-cos^2 x) note the identity:sin^2 x + cos^2 x = 1 thus, sin^2 x = 1 - cos^2 x now substitute this into the above: sin x (sin^2 x), now multiply, yielding: sin^3 x Q.E.D.
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an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
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a term/algebraic expression raised to a non-negative integer power and a multiple of co-efficient,,,,,, T^n where n is a non-negative,,,,, 4x^2
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find volume of solid about y axis and y=x^3, x=0,y=1
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5x-5
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x
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The set of inputs of a function. x goes in the function, y comes out.
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where u from verna
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If you differentiate then answer is not x
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domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
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what is functions
give different types of functions.
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pls solve it i Want to see the answer
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ok
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differentiate each term
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