2.6: Limits at Infinity

$\newenvironment {prompt}{}{} \newcommand {\DeclareMathOperator }[2]{\@OldDeclareMathOperator {##1}{##2}\immediate \write \myfile {\unexpanded {\DeclareMathOperator }{\unexpanded {##1}}{\unexpanded {##2}}}} \newcommand {\reserved@a }[2]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{} \newcommand {\AppendGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Append \grfext@Check }{\grfext@Append \grfext@@Add }} \newcommand {\PrependGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Prepend \grfext@Check }{\grfext@Prepend \grfext@@Add }} \newcommand {\RemoveGraphicsExtensions }[1]{\@ifundefined {Gin@extensions}{\def \Gin@extensions {}}{\edef \grfext@tmp {\zap@space ##1 \@empty }\@for \grfext@ext :=\grfext@tmp \do {\def \grfext@next {\let \grfext@tmp \Gin@extensions \@expandtwoargs \@removeelement \grfext@ext \Gin@extensions \Gin@extensions \ifx \grfext@tmp \Gin@extensions \let \grfext@next \relax \fi \grfext@next }\grfext@next }}\grfext@Print \RemoveGraphicsExtensions } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\epstopdfcall }[1]{\ifETE@InsideSetfile \expandafter \@firstoftwo \else \expandafter \@secondoftwo \fi {‘##1}{\Gin@base \Gin@ext }} \newcommand {\epstopdfDeclareGraphicsRule }[4]{\ifx \\##4\\\@PackageError {epstopdf-base}{Conversion command is missing}\@ehc \else \begingroup \@ifundefined {Gin@rule@##1}{}{\@PackageInfo {epstopdf-base}{Redefining graphics rule for ‘##1’}}\endgroup \@namedef {Gin@rule@##1}####1{{##2}{##3}{\epstopdfcall {##4}}}\fi } \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\GetTitleString }[0]{\ifGTS@expand \expandafter \GetTitleStringExpand \else \expandafter \GetTitleStringNonExpand \fi } \newcommand {\GetTitleStringExpand }[1]{\def \GetTitleStringResult {##1}\begingroup \GTS@DisablePredefinedCmds \GTS@DisableHook \edef \x {\endgroup \noexpand \def \noexpand \GetTitleStringResult {\GetTitleStringResult }}\x } \newcommand {\GetTitleStringNonExpand }[1]{\def \GetTitleStringResult {##1}\global \let \GTS@GlobalString \GetTitleStringResult \begingroup \GTS@RemoveLeft \GTS@RemoveRight \endgroup \let \GetTitleStringResult \GTS@GlobalString } \newcommand {\GTS@DisableHook }[0]{} \newcommand {\label@hook }[0]{} \newcommand {\@currentlabelname }[0]{} \newcommand {\reserved@a }[0]{} \newcommand {\@safe@activestrue }[0]{} \newcommand {\@safe@activesfalse }[0]{} \newcommand {\nameref }[0]{\@ifstar \T@nameref \T@nameref } \newcommand {\Sectionformat }[2]{##1} \newcommand {\vnameref }[1]{\unskip ~\nameref {##1}\@vpageref [\unskip ]{##1}} \newcommand {\ref }[0]{\@ifstar \@refstar \T@ref } \newcommand {\pageref }[0]{\@ifstar \@pagerefstar \T@pageref } \newcommand {\nameref }[0]{\@ifstar \@namerefstar \T@nameref } \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{}$

Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below are two videos showing worked examples.

Example 1:

Example 2:

Problems

The limit $\lim _{x \to \infty } f(x)$ is the $y$ -value the function $y = f(x)$ approaches as $x$ gets larger and larger (if it exists).

A similar definition is made for $\lim _{x \to -\infty } f(x)$ .

If $\lim _{x \to \infty } f(x) =L$ or $\lim _{x \to -\infty } f(x) = L$ , then $y=f(x)$ has a horizontal asymptote at $y=L$ .

Consider the graph of a function $f(x)$ shown below.

Based on the graph, we can say $\lim _{x \to \infty } f(x) = \answer {2}$ .

Based on the graph, we can say $\lim _{x \to -\infty } f(x) = \answer {2}$ .

The graph has zero horizontal asymptotesone horizontal asymptotetwo horizontal asymptotes

The equation of the horizontal asymptote is $y = \answer {2}$ .

What is the maximum number of horizontal asymptotes a graph can have? $\answer {2}$ .

If $f(x) = \cos (x)$ , then $\lim _{x \to \infty } = \answer {DNE}$ .

The limit does not exist because:

$\cos (x)$ just oscillates between $-1$ and $1$ and never approaches a single number as $x$ gets big. The limit as $x \to \infty$ doesn’t match the limit as $x \to -\infty$ . Some other reason.

Consider the limit $\lim _{x \to \infty } (x^3-x)$ . The following logic is flawed: $\lim _{x \to \infty } (x^3-x) = \lim _{x \to \infty } x^3 - \lim _{x \to \infty } x = \infty - \infty .$ Why is it that this application of limit laws does not work? Select all that apply.

$\infty$ is not a number The individual limits do not exist Limit laws do not apply when subtracting two functions.

We can factor $x^3-x$ as $x(x+1)(x-1)$ . As $x \to \infty$ , each term gets really big, so the product is really big. Therefore, we can say $\lim _{x \to \infty } (x^3-x) = \answer {\infty }.$

Consider the function $f(x) = x(x+1)^2(x-1)^2$ . The $y$ -intercept of this function is $(0,\answer {0})$ .

The $x$ -intercepts of this function are $x = 0, -1$ , and $\answer {1}$ .

The limit $\lim _{x \to \infty } f(x) = \answer {\infty }$ .

The limit $\lim _{x \to -\infty } f(x) = \answer {-\infty }$ .

Based on all this information, which of the following looks like it could be the graph of $f(x)$ ?

If $\displaystyle f(x) = \frac {x^3+x+1}{x^3+1}$ , evaluate:

$\lim _{x \to \infty } f(x)$ : Notice as $x$ gets large, both the numerator and denominator of $f(x)$ become larger, so we aren’t quite sure what to make of the ratio. To make it easier for us, we will multiply the top and bottom by $1$ over the highest power of $x$ that occurs in the denominator. In this case, that means $1/x^3$ . If we do this, we get $\frac {x^3+x+1}{x^3+1} \cdot \frac {\frac {1}{x^3}}{\frac {1}{x^3}} = \frac {\answer {1+\frac {1}{x^2}+\frac {1}{x^3}}}{1+\frac {1}{x^3}}.$ Now, notice that as $x \to \infty$ , the numerator approaches $\answer {1}$ , as does the denominator. Therefore, $\lim _{x \to \infty } f(x) = \frac {1}{1} = \answer {1}.$
$\lim _{x \to -\infty } f(x)$ : In this case, all the algebra from the previous part remains the same, so in this case, $\lim _{x \to -\infty } f(x) = \lim _{x \to -\infty } \frac {1+\frac {1}{x^2}+\frac {1}{x^3}}{1+\frac {1}{x^3}} = \answer {1}.$

Consider $f(x) = \frac {x^2+\sqrt {x}+1}{2x^2+3}$ . Using similar steps as in the previous problem, we can say $\lim _{x \to \infty } f(x) = \answer {1/2}$ .

We can also compute $\lim _{x \to -\infty } f(x) = \answer {1/2}$ .

How many horizontal asymptotes does the graph of $y = f(x)$ have? $\answer {1}$ .

Consider $f(x) = \frac {-x^3+1}{x^2+x}$ . Then $\lim _{x \to \infty } f(x) = \answer {-\infty }$ .

We can also compute $\lim _{x \to -\infty } f(x) = \answer {\infty }$ .

How many horizontal asymptotes does the graph of $y = f(x)$ have? $\answer {0}$ .

Evaluate $\lim _{x \to \infty } \frac {\sqrt {9x^2+1}}{2x+3}$ .

We will repeat the steps of the previous two problems. Namely, we will multiply by $1$ over the highest power of $x$ that appears in the denominator, meaning $1/x$ in this case. Doing this gives us $\frac {\sqrt {9x^2+1}}{2x+3} = \frac {\sqrt {9x^2+1} \cdot \frac {1}{x}}{(2x+3) \cdot \frac {1}{x}}.$ In the denominator, we can just multiply through by $1/x$ . In the numerator, the trick is to write $x = \sqrt {x^2}$ (which is true because $x > 0$ as $x \to \infty$ ). Therefore, $\frac {1}{x} = \frac {1}{\sqrt {x^2}} = \sqrt {\frac {1}{x^2}}.$ When we write it this way, namely $\frac {\sqrt {9x^2+1} \cdot \sqrt {\frac {1}{x^2}}}{(2x+3) \cdot \frac {1}{x}},$ we can now multiply everything through to get $\frac {\sqrt {\answer {9+\frac {1}{x^2}}}}{2+\frac {3}{x}}.$ Taking a limit as $x \to \infty$ , we get an answer of $\lim _{x \to \infty } f(x) = \frac {\sqrt {9}}{2} = \answer {3/2}.$

Using the same trick, we can get $\lim _{x \to -\infty } \frac {\sqrt {9x^2+1}}{2x+3} = \answer {-3/2}.$ .

Since $x<0$ as $x \to -\infty$ , we have to write $x = -\sqrt {x^2}$ .

The answer should be $-3/2$ .

The $\lim _{x \to \infty } (\sqrt {x^4+5}-x^2) = \answer {0}$ .

Multiply the function by $\frac {\sqrt {x^4+5}+x^2}{\sqrt {x^4+5}-x^2}$ .

The limit $\lim _{x \to -\infty } (\sqrt {x^2+5}-x) = \answer {\infty }$ .

There is no algebra to be done here, since the square root term gets very large and so does $-x$ . When you add these two things together, you still get a very large number.

Evaluate the limit $\lim _{x \to \infty } \frac {\sin ^3(x)}{x^2}$ .

Notice that as $x \to \infty$ , the numerator doesn’t even have a limit. We will get around this by using the squeeze theorem. Again, our problem is the $\sin ^3(x)$ term, so we will first bound that. Since $-1 \leq \sin (x) \leq 1,$ we can say $-1 \leq \sin ^3(x) \leq 1.$ Now we want to make the middle term look like our function, so we will divide everything by $\answer {x^2}$ . Doing this gives $\answer {-\frac {1}{x^2}} \leq \frac {\sin ^3(x)}{x^2} \leq \frac {1}{x^2}.$ Taking a limit as $x \to \infty$ , we see $\lim _{x \to \infty } -\frac {1}{x^2} = \lim _{x \to \infty } \frac {1}{x^2} = \answer {0}.$ Therefore, our limit is $\answer {0}$ .