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\begin{document}
%MODULE: Continuity


%USER DEFINED COMMANDS
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%INPUT THE INSTRUCTIONS PAGE 


%BUILD  TITLE PAGE 



\title{Properties of the Limit}


\subtitle{We use limits a great deal in Calculus. As a result, we need to analyze
the limit carefully and identify any properties that might prove useful.
We will in fact apply some of the properties discussed here in future
modules. }

\makebeamertitle

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

\noindent We would like to evaluate the following limits \emph{without}
numerical or graphical approximation, and \emph{without} verifying
$\ec\epsilon$ and $\ec\delta$. 
\begin{itemize}
\item $\ec\lim_{x\rightarrow0}c$, where $\ec c$ is any real number. 
\item $\ec\lim_{h\rightarrow0}\left(e^{x}f(h)\right)$, where $\ec x$ is
any real number and $\ec f$ is any function in terms of $\ec h$
\emph{and not $\ec x$}. 
\item $\ec\lim_{h\rightarrow0}\frac{f(x)}{g(x)}$ when we know $\ec\lim_{h\rightarrow0}f(x)$
and $\ec\lim_{h\rightarrow0}g(x)$. 
\item Other algebraic combinations of $\ec f$ and $\ec g$. 
\end{itemize}

\lyxframeend{}\lyxframe{{SAGE worksheets}}

\noindent You should open the SAGElet {}``Limits: $\epsilon$-$\delta$''.


\lyxframeend{}\lyxframe{{A word on continuity}}

\noindent If you have not studied the module on {\hilite continuity}
yet, you can ignore this slide. Otherwise, read on.

You instructor may have chosen to discuss with you the topic of {\hilite
continuity} before covering this material. If so, you may notice
that some of the properties we will give seem redundant with what
was discussed in that slide. This is not really true. We gave in that
module a theorem that certain functions were continuous. We did not
prove the theorem, however, and with good reason: the proof of that
theorem relies on properties of the limit described below.


\lyxframeend{}\lyxframe{{A word on proofs}}

\noindent If you have not studied the module on {\hilite limits
using $\epsilon$-$\delta$}, you can ignore any slide that talks
about $\epsilon$-$\delta$. Many instructors choose not to dwell
on this approach.

It is possible to prove all of the properties that we describe in
this module using an $\epsilon$-$\delta$ argument. We will prove
only a few of them this way; the others we will state without any
argument at all.


\lyxframeend{}\lyxframe{{First properties}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent The first properties that we describe deal with a {\hilite
constant} or a {\hilite constant multiple}.
\begin{theorem}%{}
Let $\ec c$ be a constant with respect to $\ec x$. Then for any
$\ec a\in\mathbb{R}$ and for any function $\ec f(x)$, 
\begin{itemize}
\item $\ec\lim_{x\rightarrow a}c=c$; and 
\item $\ec\lim_{x\rightarrow a}f(x)=c\lim_{x\rightarrow a}f(x)$, if the
latter limit exists. 
\end{itemize}
\end{theorem}%{}

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

For an example of the first property, 
\[
{\ec\lim_{x\rightarrow3}4=4}.
\]


\end{onlyenv}%{}

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

For an example of the second property,
\[
{\ec\lim_{x\rightarrow1}4x^{2}=4\lim_{x\rightarrow1}x^{2}=4\cdot1=4}.
\]


\end{onlyenv}%{}

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

Another example of the second property that we will use in a future
module is
\[
{\ec\lim_{h\rightarrow0}\left(e^{x}f(h)\right)=e^{x}\lim_{h\rightarrow0}f(h)}
\]
 where $\ec f$ is a function in terms of $\ec h$ \emph{and not $\ec x$}.
Notice that $\ec e^{x}$ depends only on $\ec h$, so it is constant
\emph{with respect to $\ec h$}.

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

\lyxframeend{}\lyxframe{{Proof of the limit of a constant: concept}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent It is relatively easy to prove these two properties using
an $\epsilon$-$\delta$ argument.
\begin{onlyenv}%{}
<1>

We remind you that the $\epsilon$-$\delta$ argument comes from the
precise definition of a limit:
\begin{definition}%{}
[(Precise definition of the limit)]$\ec\lim_{x\rightarrow a}f(x)=L$
if 
\begin{itemize}
\item for all $\ec\epsilon>0$, 
\item there exists $\ec\delta>0$ 
\item such that for all $\ec x$ satisfying $\ec|x-a|<\delta$ (except maybe
$\ec x=a$), 
\item $\ec|f(x)-L|<\epsilon$. 
\end{itemize}
\end{definition}%{}
\end{onlyenv}%{}

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

\noindent In the SAGE worksheet {}``Limits: $\epsilon$-$\delta$'',
set $\ec f$ to be the constant $\ec1$. (To do this, you must type
\texttt{\ttcolor lambda x: 1} in the \texttt{\ttcolor f} box. This
strange notation is due to a bug which ought to be resolved soon.)
You should see the following plot. 
\begin{figure}[h!]
 

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


\end{onlyenv}%{}

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

Look carefully at the plot. Notice that the graph of $\ec f$ does
not pass through the pink regions. This means that you have found
an appropriate value of $\ec\delta$ to guarantee that the distance
between $\ec L$ and $\ec f(x)$ is smaller than $\ec\epsilon$, satisfying
the definition of the limit.

\end{onlyenv}%{}

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

Now look more carefully. Experiment with different values of $\ec\delta$.
Can you find any value of $\ec\delta$ such that $\ec f(x)$ passes
through the pink regions?

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<5>

You should not be able to find such a value. In fact, $\ec|f(x)-L|=|1-1|=0<\epsilon$
\emph{regardless of the value of $\ec\epsilon$}.

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

\lyxframeend{}\lyxframe{{Proof of the limit of a constant}}

\noindent We can generalize this argument to show that for any constant
function $\ec f(x)=c$ and for any real number $\ec a$, $\ec\lim_{x\rightarrow a}f(x)=c$.
Let $\ec\epsilon$ be any positive real number, and then
\[
{\ec|f(x)-L|=|c-c|=0<\epsilon}.
\]
 You can choose \emph{any} value of $\ec\delta$ and satisfy the precise
definition of the limit.


\lyxframeend{}\lyxframe{Proof of the limit of a constant multiple: concept}
\begin{overlayarea}
{\textwidth}{6in}

\noindent To determine how to prove the limit of a constant multiple,
we'll take a similar approach, and try an example function and an
example constant. Let $\ec f(x)=x^{2}$, $\ec a=1$, and $\ec c=-2$.
We want to see if we can show that 
\[
{\ec\lim_{x\rightarrow1}(-2x^{2})=-2\lim_{x\rightarrow1}x^{2}}.
\]



\pause{}
\begin{onlyenv}%{}
<2-3>

A glance at the graph suggests that $\ec\lim_{x\rightarrow1}x^{2}=1$.
Look again at the SAGElet, and start with $\ec f(x)=x^{2}$. Using
$\ec\epsilon=0.1$, find $\ec\delta$ satisfying the definition of
the limit.


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

\vspace{-2ex}
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.3]{figures/x2_good} 
\end{figure}



\vspace{-3ex}
You should find that $\ec\delta\approx0.05$ works. (In fact the correct
answer is $\ec\delta=-1+\sqrt{1.1}\approx0.0488$.)

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

\pause{}
\begin{onlyenv}%{}
<4-5>

Now try the SAGElet with $\ec cf(x)=-2x^{2}$. Keep $\ec\delta=0.05$.
What value of $\ec\epsilon$ works with this $\ec\delta$ to guarantees
that $\ec|cf(x)-cL|<\epsilon$? (Note $\ec cL=-2\cdot1=-2$.)


\pause{}
\begin{onlyenv}%{}
<5>

\vspace{-2ex}
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.3]{figures/neg2x2_good} 
\end{figure}



\vspace{-3ex}
You should find that $\ec\epsilon=0.2$. This is twice the value of
the previous epsilon.

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

\pause{}


{\dims You should find that $\epsilon=0.2$. This is twice the value
of the previous epsilon.}


Perhaps the $\ec\delta$ that makes $\ec\epsilon$ work for $\ec cf(x)$,
is the same as the $\ec\delta$ that makes $\ec\epsilon/|c|$ work
for $\ec f(x)$.

\end{overlayarea}

\lyxframeend{}\lyxframe{{Proof of the limit of a constant multiple}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent We try this in a proof. We want to show that $\ec\lim_{x\rightarrow a}cf(x)=c\lim_{x\rightarrow a}f(x)$,
if the latter limit exists. Assume that the latter limit \emph{does}
exist; that is, $\ec\lim_{x\rightarrow a}f(x)=L$.
\begin{onlyenv}%{}
<1>

We need to show that $\ec\lim_{x\rightarrow a}cf(x)=cL$.

\end{onlyenv}%{}

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

If $\ec c=0$ then $\ec cf(x)=0$, and we are talking about the limit
of a constant. We have already discussed this case, so we can assume
that $\ec c\neq0$.

\end{onlyenv}%{}

\pause{}


By the definition of a limit, we must show that for every positive
$\ec\epsilon$ we can find a positive $\ec\delta$ satisfying\vspace{-1ex}
\[
{\ec|x-a|<\delta}\qquad{\ec\Longrightarrow}\qquad{\ec|cf(x)-cL|<\epsilon}.
\]
\vspace{-1ex}
Let $\ec\epsilon'=\epsilon/|c|$; this is a positive number. Since
$\ec L$ is the limit of $\ec f(x)$ as $\ec x\rightarrow a$, the
precise definition tells us that there exists $\ec\delta$ such that\vspace{-1ex}
\[
{\ec|x-a|<\delta}\qquad{\ec\Longrightarrow}\qquad{\ec|f(x)-L|<\epsilon'}.
\]



\pause{}


\vspace{-1ex}
Then\vspace{-1ex}
\[
{\ec|cf(x)-cL|=|c(f(x)-L)|=|c|\cdot|f(x)-L|<|c|\cdot\epsilon'=\epsilon}.
\]



\pause{}


\vspace{-1ex}
We have found $\ec\delta$ such that if $\ec|x-a|<\delta$, then $\ec|cf(x)-cL|<\epsilon$.
We have proved the second part of the theorem:\vspace{-1ex}
\[
{\ec\lim_{x\rightarrow a}cf(x)=cL=c\lim_{x\rightarrow a}f(x)}.
\]


\end{overlayarea}

\lyxframeend{}\lyxframe{{The limit of $x$}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent Our next property is very convenient.
\begin{theorem}%{}
\noindent For all $\ec a\in\mathbb{R}$,
\[
{\ec\lim_{x\rightarrow a}x=a}.
\]
 
\end{theorem}%{}

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

The graph of $\ec f(x)=x$ gives credence to the theorem. 
\begin{figure}[h!]
 

\begin{centering}
\includegraphics[scale=0.3]{figures/x_around_0} 
\par\end{centering}

\caption{ If you approach $\ec x=0$ from both sides, it looks as if the limit
is $\ec0$. }
\end{figure}


\end{onlyenv}%{}

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

Let's experiment a little with the SAGElet to see how $\ec\delta$
corresponds to $\ec\epsilon$. Change \texttt{\ttcolor f} to \texttt{\ttcolor
x}, and pick any $x$-value you like.
\begin{onlyenv}%{}
<3-4>

What is the largest value of $\ec\delta$ that satisfies the precise
definition of the limit for $\ec\epsilon=\hilite0.1$?


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

\begin{figure}[h!]
\centering{}\includegraphics[scale=0.2]{figures/x_epsilon1} 
\end{figure}



It looks here as if $\ec\delta=0.1$ works. What if we change $\ec\epsilon$
to {\ec 0.01}?

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

\pause{}
\begin{onlyenv}%{}
<5-6>

What is the value of $\delta$ that satisfies the precise definition
of the limit for $\epsilon={\hilite0.01}$? 
\begin{figure}[h!]
\centering{}\includegraphics[scale=0.2]{figures/x_epsilon2} 
\end{figure}



\pause{}
\begin{onlyenv}%{}
<6>

It looks here as if $\ec\delta=0.01$ works. What if we change $\ec\epsilon$
to {\ec 0.5}?

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

\pause{}
\begin{onlyenv}%{}
<7-8>
\end{onlyenv}%{}

What is the value of $\delta$ that satisfies the precise definition
of the limit for $\epsilon={\hilite0.5}$? 
\begin{figure}[h!]
\centering{}\includegraphics[scale=0.2]{figures/x_epsilon3} 
\end{figure}

\begin{onlyenv}%{}
<8>

It looks here as if $\ec\delta=0.5$ works.

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

\lyxframeend{}\lyxframe{The limit of $x$}

\noindent Our next property is very convenient.
\begin{theorem}%{}
\noindent For all $a\in\mathbb{R}$,
\[
{\lim_{x\rightarrow a}x=a}.
\]
 
\end{theorem}%{}
In each case, $\ec\delta$ had the same value as $\ec\epsilon$. Let's
try that in a proof.


\pause{}

Let $\ec\epsilon$ be any positive real number, and set $\ec\delta=\epsilon$.
Assume that $\ec|x-a|<\epsilon$. Then
\[
{\ec|f(x)-L|=|x-a|}.
\]
 By hypothesis, $\ec|x-a|<\delta=\epsilon$, so
\[
{\ec|f(x)-L|<\epsilon}.
\]
 We have thus proved the theorem.


\lyxframeend{}\lyxframe{{More properties}}

\noindent Here are most remaining properties. We do not prove them.
\begin{theorem}%{}
\noindent Let $\ec a\in\mathbb{R}$, and $\ec f$ and $\ec g$ functions
such that $\ec\lim_{x\rightarrow a}f(x)=L$ and $\ec\lim_{x\rightarrow a}g(x)=M$. 
\begin{itemize}
\item The limit of a sum is the sum of the limits:


\begin{center}
\vspace{-1ex}
$\ec\lim_{x\rightarrow a}\left(f(x)\pm g(x)\right)=L+M$.
\par\end{center}

\item \vspace{-1ex}
The limit of a product is the product of the limits:


\begin{center}
\vspace{-1ex}
$\ec\lim_{x\rightarrow a}\left(f(x)g(x)\right)=LM$.
\par\end{center}

\item \vspace{-1ex}
The limit of a quotient is the quotient of the limits:


\begin{center}
\vspace{-1ex}
If $\ec M\neq0$, then $\ec\lim_{x\rightarrow a}\left(f(x)/g(x)\right)=L/M$.
\par\end{center}

\item \vspace{-1ex}
If $\ec n\in\mathbb{N}$ then $\ec\lim_{x\rightarrow a}x^{n}=a^{n}$
and $\ec\lim_{x\rightarrow a}\sqrt[n]{x}=\sqrt[n]{L}$. 
\item \vspace{-1ex}
If $\ec n\in\mathbb{N}$ then $\ec\lim_{x\rightarrow a}f(x)^{n}=L^{n}$
and $\ec\lim_{x\rightarrow a}\sqrt[n]{f(x)}=\sqrt[n]{L}$,\\
 \emph{provided that} $\ec\sqrt[n]{L}$ exists. 
\item \vspace{-1ex}
If $\ec f$ is a \emph{polynomial} or \emph{rational expression} and
$\ec a$ is in the domain of $\ec f$,\\
 then you can substitute the value $\ec x=a$ into $\ec f$ to find
the limit:


\begin{center}
\vspace{-1ex}
$\ec\lim_{x\rightarrow a}f(x)=f(a)$.
\par\end{center}

\end{itemize}
\end{theorem}%{}

\lyxframeend{}\lyxframe{{Examples}}
\begin{overlayarea}
{\textwidth}{6in}
\begin{onlyenv}%{}
<1>

\noindent Suppose we have two probes whose value at $\ec x$ is modeled
by the functions $\ec f(x)$ and $\ec g(x)$, and we know that $\ec\lim_{x\rightarrow3}f(x)=-2.5$
and $\ec\lim_{x\rightarrow3}g(x)=0$. Then\vspace{-1ex}
\begin{align*}
\ec\lim_{x\rightarrow3}\left(f(x)+g(x)\right) & \ec=-2.5\\
\ec\lim_{x\rightarrow3}\frac{g(x)}{f(x)} & \ec=0\\
\ec\lim_{x\rightarrow3}\sqrt[4]{-f(x)} & {\ec=\sqrt[4]{-\lim_{x\rightarrow3}f(x)}\approx1.2574}.
\end{align*}


\end{onlyenv}%{}


\begin{onlyenv}%{}
<1->

\noindent We have to be careful with division by zero. In particular,
we \emph{cannot} say that\vspace{-1ex}
\[
\ec\lim_{x\rightarrow3}\frac{f(x)}{g(x)}
\]
\vspace{-1ex}
does not exist.

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3-6>

Limits \emph{can} exist even when the denominator of a quotient approaches
zero. This is not the same as division by zero, because we are considering
$\ec x\rightarrow a$ and \emph{not} $\ec x=a$.
\begin{onlyenv}%{}
<3>

You saw an example of this in a previous module, when we showed that\vspace{-1ex}
\[
\ec\lim_{x\rightarrow0}\frac{x^{2}-x}{x}=-1
\]
 \emph{even though} $\ec\lim_{x\rightarrow0}x=0$!

\end{onlyenv}%{}

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

Algebraically, we can show this as follows:\vspace{-1ex}
\begin{align*}
\ec\lim_{x\rightarrow0}\frac{x^{2}-x}{x} & \ec=\lim_{x\rightarrow0}\frac{\cancel{x}(x-1)}{\cancel{x}}
\end{align*}
\begin{explanation}Factored common $x$ in numerator.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<5>

Algebraically, we can show this as follows:\vspace{-1ex}
\begin{align*}
\dims\lim_{x\rightarrow0}\frac{x^{2}-x}{x} & \dims=\lim_{x\rightarrow0}\frac{\cancel{x}(x-1)}{\cancel{x}}\\
 & \ec=\lim_{x\rightarrow0}(x-1)\\
 & \ec=\lim_{x\rightarrow0}x-\lim_{x\rightarrow0}(-1)=-1.
\end{align*}
 \begin{explanation}Notice that we used the properties of\\
 (1) the sum of a limit,\\
 (2) the limit of a constant, and\\
 (3) the limit of $x$.\end{explanation}

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

\pause{}
\begin{onlyenv}%{}
<6>

On the other hand, limits do not \emph{have} to exist, either. If
$\ec f(x)=\frac{|x|}{x}$, then
\[
{\ec\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0^{-}}\frac{-x}{x}=-1}\ldots
\]
 


\begin{center}
(If $\ec x\rightarrow0^{-}$, the value of $\ec x$ is negative, so
$\ec|x|=-x$.\\
 This puzzles people who read it the first time,\\
 so think about it carefully---it is \emph{not} a typo!) 
\par\end{center}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<7>

On the other hand, limits do not \emph{have} to exist, either. If
$\ec f(x)=\frac{|x|}{x}$, then
\[
{\dims\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0^{-}}\frac{-x}{x}=-1}
\]
but
\[
{\ec\lim_{x\rightarrow0^{+}}f(x)=\lim_{x\rightarrow0^{+}}\frac{x}{x}=1}.
\]


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<8>

On the other hand, limits do not \emph{have} to exist, either. If
$\ec f(x)=\frac{|x|}{x}$, then
\[
{\dims\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0^{-}}\frac{-x}{x}=-1}
\]
but
\[
{\ec\lim_{x\rightarrow0^{+}}f(x)=\lim_{x\rightarrow0^{+}}\frac{x}{x}=1}.
\]
 So
\[
{\ec\lim_{x\rightarrow0^{-}}f(x)\neq\lim_{x\rightarrow0^{+}}f(x)}
\]
and there is no two-sided limit.


\begin{center}
\vspace{-2ex}
When division by zero appears in a limit, \emph{further investigation
is needed}.
\par\end{center}

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

\lyxframeend{}\lyxframe{\noindent {Conclusion}}
\begin{itemize}
\item We identified (and used!) a number of properties of the limit. These
properties will be useful in future modules. 
\item We proved some properties using the precise definition of the limit
($\epsilon$-$\delta$). The others can also be proved, of course. 
\item If we are using the quotient property of limits and encounter division
by zero, then further investigation is needed. A useful tool for such
investigation is factoring. Another useful tool is analyzing the one-sided
limits. 
\end{itemize}

\lyxframeend{}
\end{document}
