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\begin{document}
%MODULE 1: Optimization


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%INPUT THE INSTRUCTIONS PAGE 


%BUILD  TITLE PAGE 



\title{Continuity of Functions}


\subtitle{We look at some harder cases of deciding when a function is continuous.}

\makebeamertitle

\lyxframeend{}\lyxframe{Introduction}

{ Determining whether a function is continuous on a given domain,
requires showing that at each point in the domain that the functions
is continuous. }
\begin{itemize}
\item We know that all polynomials are continuous on $\ec\reals$, and thus
they are continuous on any interval $\ec[a,b]\subset\reals$.
\item Similarly, the elementary functions such as $\ec\sin(x)$, $\ec\cos(x)$
and $\ec\exp(x)$ are all continuous on $\ec\reals$.
\item To determine whether a function $\ec f$ is continuous at a point
$\ec x_{0}$ in the domain, we have to show that the limit of the
function as it approaches $\ec x_{0}$ from the left equals the limit
of the functions as it approaches $\ec x_{0}$ from the right, and
these must equal the value of the function at $\ec x_{0}$. Formally
this is given by, 
\[
\ec\lim_{x\rightarrow x_{0}^{-}}f(x)=\lim_{x\rightarrow x_{0}^{+}}f(x)=f(x_{0}).
\]
 Note that this means the function must be defined at $\ec x_{0}$.
For example the function $\ec\ln(x)$ is not continuous at $\ec x=0$
because the logarithm is not defined there.


\bigskip{}


\item If you need to review this, go to \CalcText\ and search for {\hilite
continuity}.
\end{itemize}
%1



\lyxframeend{}\lyxframe{{Defining the problem}}

\label{page:modf-problem} \textbf{ \em The Problem:} Determine on
which domain 
\[
\ec g(x)=\left\{ \begin{array}{ll}
(x^{2}-4)/(x+2) & \mbox{ if }x<0\\
5 & \mbox{ if }x=0\\
(x^{3}+2x^{2}-2x-4)/(x^{2}+4x+4) & \mbox{ if }x>0
\end{array}\right.
\]
 is continuous. Can the function $\ec g(x)$ be redefined to have
a greater domain of continuity?

\noindent %2



\pause{}

{ What does it mean to have a greater, i.e., a larger domain of continuity?}

%3



\pause{}

\bigskip{}

\begin{itemize}
\item If $\ec g(x)$ is continuous on the {\hilite set} $\ec S\subset\reals$,
to have a greater domain of discontinuity means that the {\hilite
set} of points on which $\ec g(x)$ is continuous can be extended
to a larger set $\ec\tilde{S}$ with $\ec S\subset\tilde{S}$, such
that there is a function, we'll call it $\ec\tilde{g}(x)$, that is
continuous on on $\ec\tilde{S}$, \textit{and} $\ec\tilde{g}(x)=g(x)$
for all $\ec x\in S$. 
\item First, if $\ec g(x)$ is continuous on $\ec\reals$, then $\ec S=\reals$,
and it is impossible to increase the size of this set.
\end{itemize}
{Are there any other points of discontinuity?}

%1



\lyxframeend{}\lyxframe{{Examining the function} }
\begin{itemize}
\item The possible points of discontinuity are when $\ec x=0$, (the break
point in the definition of $\ec g(x)$), and at the points at which
the denominators are zero, i.e., where $\ec(x+2)=0$, and when $\ec(x^{2}+4x+4)=0$.
\item The first of equations can be solved immediately to give $\ec x=-2$,
while the second requires only a bit more work in {\hilite factoring}
the quadratic polynomial into the terms $\ec(x+2)(x+2)$. Thus $\ec x=-2$
and $\ec x=2$ are possible points where $\ec g(x)$ may fail to be
defined. Thus, 
\begin{align*}
 & \textrm{Possible point}\\
 & \textrm{ of discontinuity}\\
\ec g(x)=\left\{ \begin{array}{ll}
(x^{2}-4)/(x+2), & \mbox{ if }x<0\\
5, & \mbox{ if }x=0\\
\frac{x^{3}+2x^{2}-2x-4}{x^{2}+4x+4}, & \mbox{ if }x>0
\end{array}\right. & \begin{array}{l}
\implies\mbox{{\rm at }}x=-2\\
\implies\mbox{{\rm at }}x=0\\
\implies\mbox{{\rm at }}x=-2
\end{array}
\end{align*}
 
\end{itemize}
%2



\pause{}

{We should graph each {\hilite branch} of the function $\ec g(x)$
so that we have a sense of what his going on.}


\lyxframeend{}\lyxframe{{Plotting of $g(x)$}}

\label{page:cont-graph} 
\begin{itemize}
\item We graph the the branches of the function, $\ec f(x)=g(x)$,


$\ec g_{1}(x)=(x^{2}-4)/(x+2)$, \\
 $\ec g_{2}(x)=5$, and\\
 $\ec g_{3}(x)=(x^{3}+2x^{2}-2x-4)/(x^{2}+4x+4)$


using \Plot\ to input the value {\ttcolor (x{*}{*}3 + 2{*}x{*}{*}2
- 2{*}x -4)/(x{*}{*}2 + 4{*}x + 4), 5, (x{*}x-4)/(x+2)} where the
{\hilite independent} variable is {\ttcolor x}. 

\end{itemize}
{ You will need to examine the figure you obtain closely. Try to
understand what each line represents, and how they relate to the three
branches of the function $\ec g(x)$. Try to identify the possible
singularities, and points of discontinuity of the function $\ec g(x)$.}


\lyxframeend{}\lyxframe{{The graph of $g(x)$}}

\setlength{\unitlength}{1cm} 
\begin{figure}[h!]
 

\begin{centering}
\centering{ \includegraphics[bb=0bp 0bp 300bp 300bp,scale=0.5]{figures/cont1}} 
\par\end{centering}

\caption{ Note that function $\ec g(x)=g_{1}(x)$ for all $\ec x<0$ appears
to be the equation of a line, i.e., a function of the form $y=mx+b$,
and that the function $\ec g_{3}(x)$ has a singularity at $\ec x=-2$.
The value of $\ec g(x)$ at $\ec x=0$, i.e., $\ec g(0)$, is a single
point that is not connected to the values of $\ec g(x)$ to the left
or to the right of this point. \label{fig:cont1} }
\end{figure}



\lyxframeend{}\lyxframe{{Examining $g(x)$ on $(-\infty,0)$} }
\begin{itemize}
\item If $\ec x<0$ then: 
\[
\ec g(x)=\frac{(x^{2}-4)}{(x+2)}=\frac{(x-2)(x+2)}{(x+2)}=g_{1}(x).
\]
 
\item This function is not continuous since the function is not defined
at $\ec x=-2$. However, since 
\begin{align*}
\ec\lim_{x\rightarrow-2^{-}}g(x) & =\frac{(x-2)(x+2)}{(x+2)}=\lim_{x\rightarrow-2^{-}}(x-2)=-4\\
\ec\lim_{x\rightarrow-2^{+}}g(x) & =\frac{(x-2)(x+2)}{(x+2)}=\lim_{x\rightarrow-2^{+}}(x-2)=-4
\end{align*}
 the { limit from the left} equals the { limit from the right}.
This is typical of a {\hilite removable discontinuity}.
\item Thus, if we define $\ec\tilde{g}(x)=(x-2)$, then $\ec\tilde{g}(x)=g(x)$
on $\ec(-\infty,0)$ except at $\ec x=-2$, and $\ec\tilde{g}$ is
continuous on $\ec(-\infty,0)$.


{Note that $\ec g_{1}(x)=x-2$ is the equation of a line, as we observed
in the graph on \RefPage{page:cont-graph}


}

\end{itemize}

\lyxframeend{}\lyxframe{{The details of taking the limit}}
\begin{itemize}
\item The evaluation the limit from below ${\displaystyle \ec\lim_{x\rightarrow-2^{-}}\frac{(x-2)(x+2)}{(x+2)}=-4}$,
in \RefEq{eq:blim} and similarly the limit in \RefEq{eq:alim}
seems obvious. We simply cancel the terms ${\displaystyle \ec\frac{(x+2)}{(x+2)}=1}$.


\bigskip{}


\item There is a technical issue here, i.e., at the limit point, $\ec x=-2$,
we have that ${\displaystyle \ec\frac{(x+2)}{(x+2)}=\frac{0}{0}}$,
and we know that we cannot evaluate this indeterminate form.
\item We can argue that for each $\ec x_{0}$ in any {\hilite neighborhood}
of $\ec x=-2$ we have that $\ec(x_{0}+2)/(x_{0}+2)=1$ and thus the
limit must be $\ec1$, and thus in any neighborhood of $\ec x=-2$
the value of $\ec{\displaystyle \frac{(x-2)(x+2)}{(x+2)}=1\cdot(x-2)-4}$.


{As long as we are not at $\ec x=-2$, we can evaluate the expression
$\ec(x+2)/(x+2)=1$, and this is all that matters in taking the limit.}

\end{itemize}

\lyxframeend{}\lyxframe{{Examining $g(x)$ on $(0,\infty)$} }
\begin{itemize}
\item If $\ec x>0$ then 
\[
\ec g(x)=(x^{3}+2x^{2}-2x-4)/(x^{2}+4x+4)=\frac{(x^{2}-2)(x+2)}{(x+2)^{2}}=\frac{x^{2}-2}{x+2}
\]
 and $\ec g(x)$ has no {\hilite singularities} on the domain $\ec(0,\infty)$
since the point $\ec x=-2$, i.e., where the denominator vanishes,
is outside the domain of definition of this branch of the function.
\item Note that 
\[
\ec(x^{3}+2x^{2}-2x-4)/(x^{2}+4x+4)=\frac{x^{2}-2}{x+2}
\]
 represent the same functions on $\ec(0,\infty)$, i.e., at each point
$\ec x\in(0,\infty)$ both of these are algebraically equivalent yielding
the same value.
\item Thus $\ec g(x)$ is continuous on $\ec(0,\infty)$ since $\ec\frac{(x^{2}-2)(x+2)}{(x+2)^{2}}$
is a {\hilite rational function} with no singularities in it its
domain.
\end{itemize}
%1



\lyxframeend{}\lyxframe{{Examining $g(x)$ in a neighborhood of $x=0$} }
\begin{itemize}
\item If $\ec x=0$, then consider 
\[
\ec\lim_{x\rightarrow0^{-}}g(x)=\lim_{x\rightarrow0^{-}}x-2=-2,
\]
 and 
\[
\ec\lim_{x\rightarrow0^{+}}g(x)=\lim_{x\rightarrow0^{+}}\frac{x^{2}-2}{x+2}=-1.
\]

\item Thus $\ec\lim_{x\rightarrow0^{-}}g(x)\ne\lim_{x\rightarrow0^{+}}g(x)$
and $\ec g(x)$ is not continuous at $\ec x=0$. This is not a {\hilite
removable discontinuity} since we cannot redefine $\ec g(0)$ in
a manner which makes the limit from the left equal the limit from
the right, and which also makes this limit equal to some value of
$\ec g$ that we can assign at $\ec x=0$.
\end{itemize}
%2



\pause{}
\begin{itemize}
\item The point $\ec x=0$ is an {\hilite essential} singularity of $\ec g(x)$.
\end{itemize}
{From everything we have done so far, the conclusion is that $\ec g(x)$
as defined can be made continuous on the domain $\ec(-\infty,0)\bigcup(0,\infty)$.

\vfil

\hfil Let's pull this all together.}

%2



\lyxframeend{}\lyxframe{{Redefining $g(x)$}}
\begin{overlayarea}
{\textwidth}{6in}
\begin{itemize}
\item We have ascertained that $\ec g(x)$ is continuous on $\ec\reals$,
except at $\ec x=-2$, and $\ec x=0$.
\item We showed that if $\ec g(x)$ is redefined to be $\ec\tilde{g}(x)$
on $\ec(-\infty,0)$, then $\ec g(x)$ is continuous on $\ec(-\infty,0)$,
but that at $\ec x=0$, there was no way to redefine $\ec g$ so as
to make it continuous at that point.
\end{itemize}

\pause{}
\begin{onlyenv}%{}
<2>
\begin{itemize}
\item If we define 
\[
\ec\tilde{g}(x)=\left\{ \begin{array}{ll}
\ec x-2 & \ec\mbox{ if }x<0\\
\ec-2 & \ec\mbox{ if }x=0\\
\ec(x^{2}-2)/(x+2) & \ec\mbox{ if }x>0
\end{array}\right.
\]
 then $\ec\tilde{g}(x)=g(x)$ except at $\ec x=-2$ and $\ec x=0$
and $\ec\tilde{g}$ is continuous on $\ec(-\infty,0]\bigcup(0,\infty)$.
\end{itemize}
\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3>
\begin{itemize}
\item If we define 
\[
\ec\overline{g}(x)=\left\{ \begin{array}{ll}
\ec x-2 & \ec\mbox{ if }x<0\\
\ec-1 & \ec\mbox{ if }x=0\\
\ec(x^{2}-2)/(x+2) & \ec\mbox{ if }x>0
\end{array}\right.
\]
 then $\ec\overline{g}(x)=g(x)$ except at $\ec x=-2$ and $\ec x=0$
and $\ec\overline{g}$ is continuous on $\ec(-\infty,0)\bigcup[0,\infty)$.
\end{itemize}
\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4>
\begin{itemize}
\item There is no way to redefine $\ec g$ to be continuous on $\ec(-\infty,\infty)$,
and thus $\ec\tilde{g}$ or $\ec\overline{g}$ are the only {\hilite
continuous extensions} of the function $\ec g$.
\end{itemize}
\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{{Conclusion}}
\begin{itemize}
\item Some functions can be \emph{extended} in such a way that some, or
even all, of their discontinuities are {}``removed''. 
\item To determine which discontinuities are \emph{removable} and which
are \emph{essential}, we must examine the limit from each side: from
the left or from the right. 
\end{itemize}

\lyxframeend{}
\end{document}
