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%BUILD  TITLE PAGE 



\title{Asymptotes}


\subtitle{\noindent In this module we give a new definition to an idea you
have seen before by defining {\hilite asymptotes} in terms of limits.}

\makebeamertitle

\lyxframeend{}\lyxframe{Defining the problem}

\noindent Do the following functions have {\hilite asymptotes}? 
\begin{itemize}
\item $\ec y=1/x$ 
\item $\ec y=\ln x$ 
\item $\ec y=(3x^{2}+x+1)/(2x^{2}-2)$ 
\end{itemize}
If so, what kind of asymptote, and where? If not, why not?


\lyxframeend{}\lyxframe{Horizontal asymptotes}

In the module on limits at infinity, we noticed that some functions
have a finite value for 
\[
{\ec\lim_{x\rightarrow\pm\infty}f(x)}.
\]
 One example was 
\[
{\ec\lim_{x\rightarrow\pm\infty}\frac{1}{x}=0}.
\]
 
\begin{figure}[h!]
 

\centering{}{ \includegraphics[scale=0.3]{figures/plot_1_over_x_horiz_as}
}
\end{figure}



\lyxframeend{}\lyxframe{\noindent Horizontal asymptotes}

\noindent \begin{definition}[(Horizontal asymptote)]

\noindent If $\ec\lim_{x\rightarrow\infty}f(x)=L$ where $\ec L$
is a real number, or if $\ec\lim_{x\rightarrow-\infty}f(x)=L$ where
$\ec L$ is a real number, then the line $\ec y=L$ is a \textbf{\hilite
horizontal asymptote} of $\ec f(x)$. \end{definition}

\noindent In the previous example, $\ec y=0$ is a horizontal asymptote
of $\ec1/x$.


\lyxframeend{}\lyxframe{\noindent Horizontal asymptotes: $\ln x$}

\noindent On the other hand, what is $\ec\lim_{x\rightarrow\infty}\ln x$?
\begin{figure}[h!]
 

\begin{centering}
{ \includegraphics[scale=0.3]{figures/plot_lnx_plain} }
\par\end{centering}

\caption{Here $\ec L=1$ and $\ec\varepsilon=1$; $\ec x>e^{2}$ implies that
$\ec\ln x>L+\varepsilon$. }
\end{figure}


\textcolor{white}{Let $\varepsilon=1$. No matter what value you choose
for $L$, eventually $|\ln x-L|>\varepsilon$ for }\textcolor{white}{\emph{every}}\textcolor{white}{{}
value of $x$.}


\lyxframeend{}\lyxframe{Horizontal asymptotes: $\ln x$}

\noindent On the other hand, what is $\ec\lim_{x\rightarrow\infty}\ln x$?
\begin{figure}[h!]
 

\centering{ \includegraphics[scale=0.3]{figures/lnx_and_epsilon}
}

\caption{Here $\ec L=1$ and $\ec\varepsilon=1$; $\ec x>e^{2}$ implies that
$\ec\ln x>L+\varepsilon$. }
\end{figure}


Let $\ec\varepsilon=1$. No matter what value you choose for $\ec L$,
eventually $\ec|\ln x-L|>\varepsilon$ for \emph{every} value of $\ec x$.


\lyxframeend{}\lyxframe{Horizontal asymptotes: $\ln x$}

\noindent On the other hand, what is $\ec\lim_{x\rightarrow\infty}\ln x$?
Let $\ec\varepsilon=1$. No matter what value you choose for $\ec L$,
eventually $\ec|\ln x-L|>\varepsilon$ for \emph{every} value of $\ec x$.
\begin{figure}[h!]
 

\begin{centering}
{ \includegraphics[scale=0.3]{figures/lnx_and_epsilon} }
\par\end{centering}

\caption{ Here $\ec L=1$ and $\ec\varepsilon=1$; $\ec x>e^{2}$ implies
that $\ec\ln x>L+\varepsilon$. }
\end{figure}


Since $\ec\ln x$ is a one-to-one function, if $\ec x>\exp(L+\varepsilon)$
then $\ec|\ln x-L|>\varepsilon$.


\lyxframeend{}\lyxframe{Horizontal asymptotes: third example}

\noindent The third example is
\[
\ec{f(x)=\frac{3x^{2}+x+1}{2x^{2}-1}}.
\]



\pause{}

\noindent To find the limit, we rewrite this expression, multiplying
numerator and denominator by $1/x^{2}$ to obtain
\[
\ec f(x)=\frac{3x^{2}+x+1}{2x^{2}-1}\cdot\frac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}}=\frac{3+\frac{1}{x}+\frac{1}{x^{2}}}{2-\frac{1}{x^{2}}}.
\]



\pause{}

\noindent This allows us to say that
\[
\ec\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow0}\frac{3+\cancelto{0}{\frac{1}{x}}+\cancelto{0}{\frac{1}{x^{2}}}}{2-\cancelto{0}{\frac{1}{x^{2}}}}=\frac{3}{2}.
\]



\lyxframeend{}\lyxframe{Vertical asymptotes}

\noindent Now we move to {\hilite vertical asymptotes}. The graph
of $\ec1/x$ has a vertical asymptote at $\ec x=0$. 
\begin{figure}[h!]
 

\centering{}{ \includegraphics[scale=0.3]{figures/plot_1_over_x_vert_as}
}
\end{figure}


\noindent How can we describe this using limits?


\lyxframeend{}\lyxframe{Vertical asymptotes}

\noindent {\dims Now we move to vertical asymptotes. The graph of
$1/x$ has a vertical asymptote at $x=0$. 
\begin{figure}[h!]
 

\centering{}{ \includegraphics[scale=0.3]{figures/plot_1_over_x_vert_as}
}
\end{figure}


\noindent How can we describe this using limits?} Notice that the
$y$-values increase dramatically when the $x$-values approach 0.


\lyxframeend{}\lyxframe{Vertical asymptotes: definition}

That suggests the following definition. \begin{definition}[(Vertical asymptotes)]

A \textbf{\hilite vertical asymptote} occurs at $\ec x=a$ when
\[
{\ec\lim_{x\rightarrow a^{+}}f(x)=\pm\infty}\quad\mathrm{or}\quad{\ec\lim_{x\rightarrow a^{-}}f(x)=\pm\infty}.
\]
 \end{definition} 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.3]{figures/plot_1_over_x_vert_as} 
\end{figure}



\lyxframeend{}\lyxframe{\noindent {Limit \emph{of} infinity}}

\noindent Here we encounter a new idea:
\[
{\ec\lim_{x\rightarrow a}f(x)=\pm\infty}.
\]
 Intuitively it makes sense, but we'd best give a precise definition
for it.


\lyxframeend{}\lyxframe{\noindent A limit \emph{of} infinity?}
\begin{overlayarea}
{\textwidth}{6in}

\noindent {\dims Here we encounter a new idea:
\[
{\lim_{x\rightarrow a}f(x)=\pm\infty}.
\]
 Intuitively it makes sense, but we'd best give a precise definition
for it.}


\noindent We want to capture the notion that 
\begin{itemize}
\item when the $x$ values draw arbitrarily close to $\ec a$, 
\item the $y$ values increase (or decrease) without bound. \end{itemize}
\begin{onlyenv}%{}
<1>

\begin{figure}[h!]
 

\centering{}{ \includegraphics[scale=0.3]{figures/plot_1_over_x_vert_as}
}
\end{figure}


\end{onlyenv}%{}

\pause{}


\noindent Go back and review the precise definitions of (a) a finite
limit, and (b) a limit \emph{at} infinity. See if you can combine
those two ideas into the definition of a limit \emph{of} infinity.


\pause{\noindent }


\noindent \begin{definition}[(Limit of )] $\lim_{x\rightarrow a}=\infty$
if 
\begin{itemize}
\item for all $\ec N\in\mathbb{R}$, 
\item there exists $\ec\delta>0$ such that 

\begin{itemize}
\item for all $\ec x$ satisfying $\ec|x-a|<\delta$ (except maybe $\ec x=a$),
then 
\item $\ec f(x)>N$ for all $\ec f(x)$ (except maybe $\ec f(a)$). 
\end{itemize}
\end{itemize}

\end{definition}


Study the definition to make sure you understand how it captures the
notion.

\end{overlayarea}

\lyxframeend{}\lyxframe{{A limit of $-\infty$?}}

We can build a similar definition for a limit of $\ec-\infty$. \begin{definition}[(Limit of
$-\infty$)] $\lim_{x\rightarrow a}=-\infty$ if 
\begin{itemize}
\item for all $\ec N\in\mathbb{R}$, 
\item there exists $\ec\delta>0$ such that 

\begin{itemize}
\item for all $\ec x$ satisfying $\ec|x-a|<\delta$ (except maybe $\ec x=a$),
then 
\item $\ec f(x)<N$ for all $\ec f(x)$ (except maybe $\ec f(a)$). 
\end{itemize}
\end{itemize}
\noindent \end{definition}


\lyxframeend{}\lyxframe{Detecting vertical asymptotes: division by zero?}

\begin{definition}[(Vertical asymptotes)] A \textbf{\hilite vertical
asymptote} occurs at $\ec x=a$ when
\[
{\ec\lim_{x\rightarrow a^{+}}f(x)=\pm\infty}\quad\mathrm{or}\quad{\ec\lim_{x\rightarrow a^{-}}f(x)=\pm\infty}.
\]
 \end{definition} In the case of $\ec1/x$, we can notice this from
division by zero. If a function has the form $\ec f(x)/g(x)$ such
that $\ec x\rightarrow a$ implies that 
\begin{itemize}
\item $\ec g(x)$ approaches zero, and 
\item $\ec f(x)$ approaches a nonzero constant, 
\end{itemize}
\noindent then the function has a vertical asymptote at $\ec x=a$.
We won't prove this observation, but you should think about why it's
true.

\begin{theorem} If $\ec f(x)\rightarrow1/0$ when $\ec x\rightarrow a$
from the left or from the right,\\
 then $\ec\lim_{x\rightarrow a}f(x)=\infty$ or $\ec\lim_{x\rightarrow a}f(x)=-\infty$.
\end{theorem}


\lyxframeend{}\lyxframe{{Detecting vertical asymptotes: division by zero?}}

\begin{theorem} If $\ec f(x)\rightarrow1/0$ when $\ec x\rightarrow a$
from the left or from the right,\\
 then $\ec\lim_{x\rightarrow a}f(x)=\infty$ or $\ec\lim_{x\rightarrow a}f(x)=-\infty$.
\end{theorem} It's important that the numerator approaches a \emph{nonzero}
constant. If both the numerator and the denominator approach zero,
then there may be a hole instead of an asymptote, as with the example
\[
\ec\frac{3x+1}{x}
\]
in the module on limits at infinity.


\lyxframeend{}\lyxframe{\noindent Detecting vertical asymptotes}

\noindent Not all functions with vertical asymptotes fit this mold.
In other cases, you must perform a careful analysis before you can
decide whether there is a vertical asymptote. This is the case with
$\ec\ln x$: graphical and numerical checks suggest that it has a
vertical asymptote at $\ec x=0$. Since $\ec\ln x$ is undefined for
$\ec x<0$, we need check only that $\ec\lim_{x\rightarrow0^{+}}\ln x=-\infty$. 

\begin{center}
{%
\begin{tabular}{c|ccccc}
$x$  & 1  & 0.1  & 0.01  & 0.001  & 0.0001 \tabularnewline
\hline 
$\ln x$  & 0  & -2.3026  & -4.6052  & -6.9078  & -9.2103 \tabularnewline
\end{tabular}}
\par\end{center}

\begin{figure}[h!]
 

\centering{}{ %
\parbox[t]{8cm}{%
\begin{center}
\includegraphics[scale=0.2]{figures/plot_lnx}
\par\end{center}%
}}
\end{figure}



\lyxframeend{}\lyxframe{Detecting vertical asymptotes}

\noindent If we want to use the precise definition and show that $\ec\ln x<N$
for any real number $\ec N$ once we are close enough to zero, the
choice $\ec x=e^{-N}$ suffices. 
\begin{figure}[h!]
 

\begin{centering}
{ %
\parbox[t]{13cm}{%
\begin{center}
\includegraphics[scale=0.2]{figures/plot_lnx_smallerthan-3}
\par\end{center}%
}}
\par\end{centering}

\caption{ If $\ec x<e^{-1}$, then $\ec\ln x<-1$. }
\end{figure}



\lyxframeend{}\lyxframe{{Other asymptotes}}

\noindent Some functions have asymptotes that are neither horizontal
nor vertical. These can be {\hilite oblique} asymptotes, or {\hilite
nonlinear} asymptotes. In each case, the function has the form $\ec a(x)+z(x)$
where as $\ec x\rightarrow\infty$, $\ec z(x)\rightarrow0$.


\lyxframeend{}\lyxframe{Other asymptotes}

\noindent {\dims Some functions have asymptotes that are neither
horizontal nor vertical. These can be { oblique} asymptotes, or
{ nonlinear} asymptotes. In each case, the function has the form
$a(x)+z(x)$ where as $x\rightarrow\infty$, $z(x)\rightarrow0$.}

For example, if 
\[
{\ec f(x)=\frac{x^{2}+1}{x+1}},
\]
 polynomial division shows that
\[
{\ec f(x)=x-1+\frac{2}{x+1}}.
\]
 Here $\ec a(x)=x-1$ and $\ec z(x)=\frac{2}{x+1}$. As $\ec x\rightarrow\pm\infty$,
$\ec\frac{2}{x+1}\rightarrow0$ so $\ec x-1$ is an {\hilite oblique
asymptote}.


\lyxframeend{}\lyxframe{{Other asymptotes}}

\noindent {\dims Some functions have asymptotes that are neither
horizontal nor vertical. These can be { oblique} asymptotes, or
{ nonlinear} asymptotes. In each case, the function has the form
$a(x)+z(x)$ where as $x\rightarrow\infty$, $z(x)\rightarrow0$.

For example, 
\[
{f(x)=\frac{x^{2}+1}{x+1}=x-1+\frac{2}{x+1}}.
\]
 Here $a(x)=x-1$ and $z(x)=\frac{2}{x+1}$. As $x\rightarrow\pm\infty$,
$\frac{2}{x+1}\rightarrow0$ so $x-1$ is an {oblique asymptote}.}

\begin{figure}[h!]
 

\begin{centering}
{ %
\parbox[t]{8cm}{%
\begin{center}
\includegraphics[scale=0.2]{figures/plot_oblique_asymptote}
\par\end{center}%
}}
\par\end{centering}

\caption{ Oblique asymptote $\ec x-1$. }
\end{figure}



\lyxframeend{}\lyxframe{{Other asymptotes}}

\noindent {\dims Some functions have asymptotes that are neither
horizontal nor vertical. These can be { oblique} asymptotes, or
{ nonlinear} asymptotes. In each case, the function has the form
$a(x)+z(x)$ where as $x\rightarrow\infty$, $z(x)\rightarrow0$.}

For a nonlinear example, take 
\[
{\ec f(x)=\ln x+\frac{1}{x}}.
\]
 Here $\ec a(x)=\ln x$ and $\ec z(x)=\frac{1}{x}$. As $\ec x\rightarrow\pm\infty$,
$\ec\frac{1}{x}\rightarrow0$ so $\ec\ln x$ is a {\hilite nonlinear
asymptote}.

\begin{figure}[h!]
 

\begin{centering}
{ %
\parbox[t]{8cm}{%
\begin{center}
 \includegraphics[scale=0.2]{figures/plot_nonlinear_asymptote} 
\par\end{center}%
}}%
\ovalbox{\begin{minipage}[t]{1\columnwidth}%
%
\end{minipage}}
\par\end{centering}

\caption{ The nonlinear asymptote $\ec\ln x$. }
\end{figure}



\lyxframeend{}\lyxframe{{A complete example}}

Find the asymptotes of
\[
{\ec\frac{3x^{2}+x+1}{2x^{2}-2}}.
\]



\lyxframeend{}\lyxframe{\noindent {A complete example}}

{\dims Find the asymptotes of
\[
{\frac{3x^{2}+x+1}{2x^{2}-2}}.
\]
}

We start by looking for {vertical asymptotes}. The only place this
can occur is when the denominator is zero (all the other $x$-values
are defined); that is,
\begin{align*}
\ec2x^{2}-2 & \ec=0\\
\ec x^{2} & \ec=1\\
\ec x & {\ec=\pm1}.
\end{align*}



\lyxframeend{}\lyxframe{{A complete example}}

{\dims Find the asymptotes of
\[
{\frac{3x^{2}+x+1}{2x^{2}-2}}.
\]


It has vertical asymptotes at $x=\pm1$.}

\noindent Now we consider {horizontal}, {oblique}, or {nonlinear
asymptotes}. 
\begin{align*}
\ec\lim_{x\rightarrow\pm\infty}\frac{3x^{2}+x+1}{2x^{2}-2} & \ec=\lim_{x\rightarrow\pm\infty}\frac{\left(3x^{2}+x+1\right)\cdot\frac{1}{x^{2}}}{\left(2x^{2}-2\right)\cdot\frac{1}{x^{2}}}\\
\end{align*}
 \begin{explanation}Multiply top and bottom by the hightest power
of $x$.\end{explanation}


\lyxframeend{}\lyxframe{{A complete example}}

{\dims Find the asymptotes of
\[
{\frac{3x^{2}+x+1}{2x^{2}-2}}.
\]


It has vertical asymptotes at $x=\pm1$.}

\noindent Now we consider horizontal, oblique, or nonlinear asymptotes.
\begin{align*}
\dims\lim_{x\rightarrow\pm\infty}\frac{3x^{2}+x+1}{2x^{2}-2} & \dims=\lim_{x\rightarrow\pm\infty}\frac{\left(3x^{2}+x+1\right)\cdot\frac{1}{x^{2}}}{\left(2x^{2}-2\right)\cdot\frac{1}{x^{2}}}\\
 & \ec=\lim_{x\rightarrow\pm\infty}\frac{3+\frac{1}{x}+\frac{1}{x^{2}}}{2-\frac{2}{x^{2}}}\\
\end{align*}
 \begin{explanation}Distributed $1/x^{2}$.\end{explanation}


\lyxframeend{}\lyxframe{{A complete example}}

{\dims Find the asymptotes of
\[
{\frac{3x^{2}+x+1}{2x^{2}-2}}.
\]


It has vertical asymptotes at $x=\pm1$.}

\noindent Now we consider horizontal, oblique, or nonlinear asymptotes.
\begin{align*}
\dims\lim_{x\rightarrow\pm\infty}\frac{3x^{2}+x+1}{2x^{2}-2} & \dims=\lim_{x\rightarrow\pm\infty}\frac{3+\frac{1}{x}+\frac{1}{x^{2}}}{2-\frac{2}{x^{2}}}\\
 & \ec=\lim_{x\rightarrow\infty}\frac{3+\cancelto{0}{\frac{1}{x}}+\cancelto{0}{\frac{1}{x^{2}}}}{2-\cancelto{0}{\frac{2}{x^{2}}}}
\end{align*}
 \begin{explanation}Using $\lim_{x\rightarrow\pm\infty}1/x=0$ and
limit properties\ldots{}\end{explanation}


\lyxframeend{}\lyxframe{{A complete example}}

{\dims Find the asymptotes of
\[
{\frac{3x^{2}+x+1}{2x^{2}-2}}.
\]


It has vertical asymptotes at $x=\pm1$.}

\noindent Now we consider horizontal, oblique, or nonlinear asymptotes.
\begin{align*}
\dims\lim_{x\rightarrow\pm\infty}\frac{3x^{2}+x+1}{2x^{2}-2} & \dims=\lim_{x\rightarrow\infty}\frac{3+\cancelto{0}{\frac{1}{x}}+\cancelto{0}{\frac{1}{x^{2}}}}{2-\cancelto{0}{\frac{2}{x^{2}}}}\\
 & {\ec=\frac{3}{2}}.
\end{align*}



\lyxframeend{}\lyxframe{{A complete example}}

{\dims Find the asymptotes of
\[
{\frac{3x^{2}+x+1}{2x^{2}-2}}.
\]


It has vertical asymptotes at $x=\pm1$. It has a horizontal asymptote
at $y=3/2.$}

\noindent There are no oblique or non-linear asymptotes; long division
of polynomials gives us
\[
\ec{\frac{3x^{2}+x+1}{2x^{2}-2}=\frac{3}{2}+\frac{x+4}{2x^{2}-2}},
\]
 and as $\ec x\rightarrow\pm\infty$ this approaches $\ec3/2$, which
corresponds to the horizontal asymptote.


\lyxframeend{}\lyxframe{{A complete example}}

{\dims Find the asymptotes of
\[
{\frac{3x^{2}+x+1}{2x^{2}-2}}.
\]


It has vertical asymptotes at $x=\pm1$. It has a horizontal asymptote
at $y=3/2.$}

\noindent Combining this information and plotting a few points, we
end up with this graph: 
\begin{figure}[h!]
 

\centering{}{ \includegraphics[scale=0.2]{figures/complete_example}}
\end{figure}


(The graph crosses $\ec y=-3/2$ on the left, but still approaches
it eventually.)


\lyxframeend{}\lyxframe{{Conclusion}}

\noindent We considered different ways to find the {\hilite asymptotes}
of a function $\ec g(x)$. All of these come from limits: 
\begin{itemize}
\item A {horizontal asymptote} exists at $\ec x=a$ if $\ec\lim_{x\rightarrow a}g(x)=\pm\infty$.
This is true even if the limit is only from one side. 
\item A {vertical asymptote} exists at $\ec y=L$ if $\ec\lim_{x\rightarrow\infty}g(x)=L$.
This is true for $\ec x\rightarrow-\infty$ as well. 
\item An {oblique} or {nonlinear asymptote} $\ec f(x)$ exists if $\ec g(x)=a(x)+z(x)$
where $\ec z\rightarrow0$ when $\ec x\rightarrow0$. 
\end{itemize}

\lyxframeend{}
\end{document}
