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\begin{document}
%MODULE 1: Squeeze Theorem


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%INPUT THE INSTRUCTIONS PAGE 


%BUILD  TITLE PAGE 



\title{The Squeeze Theorem}


\subtitle{Sometimes, the limit of a function $\ec f$ is difficult to evaluate
using direct methods, but we can compare its values to those of other
functions. Using the other functions as a guide, we can sometimes
determine the limit of $f$. }

\makebeamertitle

\lyxframeend{}\lyxframe{{Sample problem}}

\label{page:sample-problem} \textbf{\color{darkred} \em The problem:}
Evaluate 
\[
\ec\lim_{x\rightarrow0}x\,\sin\left(\frac{1}{x}\right).
\]



\pause{}

\noindent How would you start the problem?

Take a look at the graph of $\ec{\displaystyle x\sin\left(\frac{1}{x}\right)}$.
\begin{figure}[h!]
 

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

\caption{ We see that $\ec f$ \emph{appears} to approach zero as $\ec x$
approaches zero. }
\end{figure}



\lyxframeend{}\lyxframe{{The problem with geometry}}

On the other hand, we can't always rely on geometric intuition. For
instance, $\ec{\displaystyle (x+0.01)\sin\left(\frac{1}{x}\right)}$
also appears to approach zero as $\ec x$ approaches zero: 
\begin{figure}[h!]
 

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

\caption{ We see that $\ec{\displaystyle (x+0.01)\sin\left(\frac{1}{x}\right)}$
\emph{appears} to approach zero as $\ec x$ approaches zero. }
\end{figure}



\lyxframeend{}\lyxframe{\noindent {More of a problem with geometry}}

\noindent If we zoom in on the $\ec x$-axis, we see that there might
\emph{not} be a limit at $\ec x=0$.

\noindent 
\begin{figure}[h!]
 

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

\caption{ Here $\ec{\displaystyle (x+0.01)\sin\left(\frac{1}{x}\right)}$ \emph{appears}
to wiggle wildly as $\ec x$ approaches zero. }
\end{figure}


In this case there is no limit at $\ec x=0$.


\lyxframeend{}\lyxframe{{Squeezing a function}}

\label{pg:squeezing} In many cases, a function is {\hilite trapped}
between two other functions. Another way of saying this is that it
is {\hilite squeezed} between two functions. For example, $\ec{\displaystyle f(x)=x\sin\left(\frac{1}{x}\right)}$
is trapped between the two functions $\ec g(x)=x$ and $\ec h(x)=-x$.

\noindent 
\begin{figure}[h!]
 

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

\caption{ The dashed red lines are $\ec y=\pm x$. }
\end{figure}


In this case, we have $\ec h(x)\le f(x)\le g(x)$ for all $\ec x$.


\lyxframeend{}\lyxframe{{How do we know it is squeezed?}}

\emph{How do we know that the function is squeezed?} All that we saw
there was a graph; zooming in sufficiently might show that something
strange is going on.


\pause{}

We can use simple facts from trigonometry to analyze the function.
Start with 
\[
{\ec-1\leq\sin\left(\frac{1}{x}\right)\leq1\quad\forall x\in\mathbb{R}}.
\]



\pause{}

We want to look at $\ec x\sin(\frac{1}{x})$, not $\sin(\frac{1}{x})$,
so multiply all sides by $\ec x$. If $\ec x{\ecgr\geq}0$, we have
\[
{\ec-x{\ecgr\leq}x\sin\left(\frac{1}{x}\right){\ecgr\leq}1\quad\forall x\geq0}.
\]



\pause{}

We want to look at $\ec x\sin(\frac{1}{x})$, not $\sin(\frac{1}{x})$,
so multiply all sides by $\ec x$. If $\ec x<0$, we change the direction
of the inequalities, and 
\[
{\ec-x{\ecgr\geq}x\sin\left(\frac{1}{x}\right){\ecgr\geq1}\quad\forall x\geq0}.
\]



\lyxframeend{}\lyxframe{{How does this help with the limits?}}

\noindent At this point we describe a well-known result in mathematics.
\begin{theorem}%{}
[(The Squeeze Theorem)] Let $\ec f$, $\ec g$, and $\ec h$ be
functions on $\ec I=(a,b)$ and $\ec c\in I$. If 
\begin{itemize}
\item $\ec f(x)$ lies between $\ec g(x)$ and $\ec h(x)$ for all $\ec x\in I$,
and 
\item $\ec{\displaystyle L=\lim_{x\rightarrow c}g(x)=\lim_{x\rightarrow c}h(x)}$, 
\end{itemize}

then $\ec{\displaystyle \lim_{x\rightarrow c}f(x)=L}$. 

\end{theorem}%{}

\pause{}

Read this carefully. Try to identify which functions from the previous
slide correspond to $\ec f$, $\ec g$, $\ec h$, $\ec c$, and $\ec L$.


\lyxframeend{}\lyxframe{{Applying the theorem to our example}}
\begin{overlayarea}
{\textwidth}{6in}
\begin{theorem}%{}
[(The Squeeze Theorem)] Let $\ec f$, $\ec g$, and $\ec h$ be
functions on $\ec I=(a,b)$ and $\ec c\in I$. If 
\begin{itemize}
\item $\ec f(x)$ lies between $\ec g(x)$ and $\ec h(x)$ for all $\ec x\in I$,
and 
\item $\ec{\displaystyle L=\lim_{x\rightarrow c}g(x)=\lim_{x\rightarrow c}h(x)}$, 
\end{itemize}

then $\ec{\displaystyle \lim_{x\rightarrow c}f(x)=L}$.

\end{theorem}%{}
\begin{onlyenv}%{}
<1>

In our example, we want to show that $\ec\lim_{x\rightarrow0}x\sin(1/x)=0$.
We want to use the fact that $\ec x\sin(1/x)$ always lies between
$\ec x$ and $\ec-x$.

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<2>

These values correspond (${\ec\mapsto}$) in the following way to
the terms of the {\hilite Squeeze Theorem}: 
\begin{eqnarray*}
\ec f(x) & {\ecgr\mapsto} & {\ec x\sin\left(\frac{1}{x}\right)},\\
\ec g(x) & {\ecgr\mapsto} & {\ec x},\\
\ec h(x) & {\ecgr\mapsto} & {\ec-x},\\
\ec c & {\ecgr\mapsto} & {\ec0},\\
\ec L & {\ecgr\mapsto} & {\ec0}.
\end{eqnarray*}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3->

Since $\ec\lim_{x\rightarrow0}x=0$ and $\ec\lim_{x\rightarrow0}(-x)=0$,
what can we conclude about $\ec\lim_{x\rightarrow0}x\sin(1/x)$?

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4>

By the Squeeze Theorem, $\ec\lim_{x\rightarrow0}x\,\sin(1/x)=0$.

\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{{The bad example}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent What about that function with an infinite wiggle? In this
case, 
\[
\ec x+0.01\geq\left(x+0.01\right)\sin\left(\frac{1}{x}\right)\geq-(x+0.01).
\]

\begin{onlyenv}%{}
<1>

\noindent 
\begin{figure}[h!]
\begin{centering}
\includegraphics[scale=0.3]{figures/deceptive_function_not_squeezed} 
\par\end{centering}

\caption{ The dashed red lines correspond to $\ec y=\pm(x+0.01)$. It may look
as if the function crosses those lines, but it does not. }
\end{figure}


\end{onlyenv}%{}

\pause{}


\noindent Doesn't the Squeeze Theorem apply here, too? Not at $\ec x=0$!
Why not?


\pause{}


Here, as $\ec x\rightarrow0$, 
\[
{\ec\lim_{x\rightarrow0}(x+0.01)=0.01},
\]
whereas 
\[
{\ec\lim_{x\rightarrow0}(x-0.01)=-0.01}.
\]
Since the limits are unequal, {\hilite the hypothesis of the Squeeze
Theorem are unsatisfied}. We cannot conclude anything about $\ec{\displaystyle \lim_{x\rightarrow0}(x+0.01)\sin\left(\frac{1}{x}\right)}$.

\end{overlayarea}

\lyxframeend{}\lyxframe{The bad example: putting it together}

\noindent Geometrically, we see this in graph of the function $\ec y=(x+0.01)\sin(1/x)$:
\begin{figure}[h!]
 

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


The limits {}``look'' as if they should be $\ec{\displaystyle \lim_{x\rightarrow0}(x+0.01)=0.01}$,
and $\ec{\displaystyle \lim_{x\rightarrow0}(x-0.01)=-0.01}$, as explained
on the previous slide.


\lyxframeend{}\lyxframe{{Finding functions to squeeze another}}

The {\hilite Squeeze Theorem} allows you to apply your understanding
of a limit in many places. With the sine and cosine functions, you
can usually Start with 
\[
{\ec-1\leq\sin(\mathrm{\mc\/anything})\leq1},\hspace{1em}*{1em}\mbox{ and}\hspace{1em}*{1em}{\ec-1\leq\cos(\mathrm{\mc\/anything})\leq1}.
\]
 Then build this relationship up to the point that we are investigating,
as we did on \RefPage{pg:squeezing}.


\lyxframeend{}\lyxframe{\noindent {The process of estimation}}

\noindent In finding functions $\ec g(x)$ and $\ec h(x)$ such that
some known function $\ec f(x)$ satisfies 
\[
{\ec g(x)\leq f(x)\leq h(x),\qquad\forall x},
\]
 we have managed to {\hilite bound } the behaviour of $\ec f(x)$.
In this case $\ec g(x)$ is a {\hilite lower bound} and $\ec h(x)$
is an {\hilite upper bound} for $\ec f(x)$.

The process of replacing a known function with an estimate, usally
a lower or upper bound, or both, is know as {\hilite estimation}.
A large part of mathematical analysis consists of trying to find estimates
that desribe the behaviour of very complicated functions.


\lyxframeend{}\lyxframe{{Conclusion}}
\begin{itemize}
\item The {\hilite Squeeze Theorem} allows us to evaluate easily the
values of many limits that are otherwise difficult. 
\item To use the Squeeze Theorem, you need to identify upper and lower bounds
for a function. Those upper and lower bounds must have the same limit
at the point in question. 
\end{itemize}

\lyxframeend{}
\end{document}
