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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{The idea of a limit}

The idea of a limit is the fundamental distinction between calculus and all
the mathematics that is typically labelled {}``precalculus{}''. Whether we're
dealing with the slope of a curve, the area of a region in the plane, or the
speed of a moving object, the fundamental idea will always be the same; we want
to find what the value of some function \textit{\( f(x) \) ought to} be at
a specific point \( a \), based on its value at all nearby points. That's what
a limit is, what the value of the function should be at the specific point.
We say that that {}``proper{}'' value is the \emph{limit} of the function
\( f(x) \) as \( x \) \emph{approaches} the point \( a \), or \[
\lim _{x\rightarrow a}f(x)=L.\]


As a specific example, let's look at the function \[
f(x)=\frac{x^{2}-1}{x-1}.\]
Here's the graph:

\vspace{0.3cm}
{\par\centering \resizebox*{2in}{2in}{\includegraphics{first-limit.ps}} \par}
\vspace{0.3cm}

The first thing to notice about this function is that it isn't defined at \( x=1 \),
that is, 1 is not in the domain of \( f \). That's what that hole in the graph
is about. It's not defined at \( x=1 \) because both the numerator and the
denominator are 0 there. This is what is called an \emph{indeterminate form},
since a fraction like \( 0/0 \) can't really make sense without some further
work. This is typical with limits, most problems come down to indeterminate
forms, which are a signal that you need to do more work to figure the limit
out.

But that seems silly. All I have to do is divide out the top and bottom by \( (x-1) \)
and it would be defined. Well, there will be other examples where that isn't
so easy, so let's pretend we didn't see that. 

What \textit{should} \( f(1) \) be? Well, the way I want to see that is to
look at various values of \( f(x) \) for \( x \) near 1, and see where things
are heading. here's a small table: \[
\begin{array}{cc}
x & f(x)\\
0.5 & 1.5\\
0.8 & 1.8\\
0.9 & 1.9\\
0.99 & 1.99
\end{array}\]


I think I get the pattern. Of course, \( f(1) \) \emph{ought to} be 2. But,
you can see that just by knowing the values of the function, as \( x \) gets
closer and closer to 2, \textbf{without} knowing how to simplify the expression
so that the division by 0 doesn't appear. Most serious examples of this idea
won't be simple enough so that you can just divide out and have it become obvious. 

We write the result of this ``investigation'' as  \[
\lim _{x\rightarrow 1}\frac{x^{2}-1}{x-1}=2.\]


We need a formal definition of a limit. However, the really formal definition
seems to have a lot of excess baggage that comes with it. It is needed, of  course,
but it is confusing at first. So, we'll start out with a slightly informal definition: 

\begin{defn}
If \( f \) is a function defined on an interval \( (c,a) \), and on \( (a,b) \),
then we say that the limit of \( f(x) \) as \( x \) approaches \( a \), is
\( L \), written\[
\lim _{x\rightarrow a}f(x)=L,\]
if we can be sure that \( f(x) \) is as close as we need to \( L \), by just
taking  \( x \) close enough to \( a \). 

For the formal, rigorous, definition, look at the official definition.{[}Link
to limits-rigor{]}.
\end{defn}
\begin{rem}
Note that the function doesn't have to be defined \textit{at} \( a \), but
it does need to be defined at all points \textit{sufficiently close} to \( a \)
on either side. It can be the case that the function is defined at \( a \),
but that might not be the limit \( L \). 

Here is an example to show the idea of the limit. Even though, technically,
the function represented by this graph is not defined at \( x=2 \), the closer
you take \( x \) to 2 (click on a point of the graph), the closer \( y \)
gets to \( 3 \).
\end{rem}
\vspace{0.3cm}
{\par\centering \includegraphics{limit-toy.ps} \par}
\vspace{0.3cm}

Let's look at some examples before going on.

\begin{example}
Let \( f(x)=3x^{2}+1 \). Then \[
\lim _{x\rightarrow 2}f(x)=\]

\end{example}

\subsubsection*{Answer}


\subsubsection*{Solution }

\begin{example}
Let \[
f(x)=\left\{ \begin{array}{ll}
3x+1 & \mathrm{if}\, x\geq 1\\
x+2 & \mathrm{if}\, x<1
\end{array}\right. .\]


Then \[
\lim _{x\rightarrow 1}f(x)=\]
 
\end{example}

\subsubsection*{Answer}


\subsubsection*{Solution}

\begin{example}
Let \[
f(x)=\left\{ \begin{array}{cc}
x^{2} & \mathrm{if}\, x\neq 1\\
5 & \mathrm{if}\, x=1
\end{array}\right. .\]
Then \[
\lim _{x\rightarrow 1}f(x)=\]
 
\end{example}

\subsubsection*{Answer}


\subsubsection*{Solution}

\begin{example}
Let \( f(x)=sin(1/x) \). Then \[
\lim _{x\rightarrow 0}f(x)\]
 
\end{example}

\subsubsection*{Answer}

\begin{example}
Let \[
f(x)=\frac{x^{3}+1}{x^{2}-1}.\]
 

Then \[
\lim _{x\rightarrow -1}f(x)=\]

\end{example}

\subsubsection*{Answer}


\subsubsection*{Solution}
\end{document}
