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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{Precise Definition of a Limit}

\begin{example}
Show that \[
\lim _{x\rightarrow 1}3x+5=8.\]
 
\end{example}

\subsubsection*{Solution}

In order to make \( 3x+5 \) be as close as we need to \( 8 \), we just have
to take \( x \) close enough to \( 1 \). That is what we have to show. How
close does \( x \) have to be to \( 1 \)? How close do we need \( 3x+5 \)
to be to \( 8 \)? We presume that some unknown positive distance \( \epsilon  \)
has been chosen as the closeness we need to get \( 3x+5 \) to \( 8 \). That
is, we need to arrange for \[
|(3x+5)-8|<\epsilon .\]
What are we allowed to do to arrange this? We can only make \( x \) closer
to \( 1 \) --- as close as we need to \( 1 \). We can \emph{control} that
distance. The key to figuring these out is to deal with the expression we \emph{need
to control} (the \( |(3x+5)-8| \) in this case), and hope that within that
expression we can see the quantity we \emph{can control} (the \( |x-1| \)).
But, this time,\begin{eqnarray*}
|(3x+5)-8| & = & |3x-3|\\
 & = & 3|x-1|.
\end{eqnarray*}
so we can control the distance from \( 3x+5 \) to \( 8 \) just by making \( x \)
close to 1. No matter what \( \epsilon  \) we have, we can get \( 3x+5 \)
within \( \epsilon  \) of 8 just by taking \( |x-1|<\epsilon /3 \). That is,
we take \( \delta =\epsilon /3 \), and whenever \( )<|x-1|<\delta  \), we
will have \( (3x+5)-8|<\epsilon  \). 

Why divide by 3? We do that to cover for the \( 3 \) in front of the \( |x-1| \). 

So, what is the \emph{answer}? The closest thing to an answer here is the choice
of \( \delta  \) depending on \( \epsilon  \), \( \delta =\epsilon /3 \).
But there is more to be said than just that. You have to justify your {}``answer{}'',
and explain why it works. That is, you need to prove that the limit is as you
say it is.


\subsubsection*{The proof you should write}

\begin{proof}
Let \( \epsilon >0 \) be given. If \( \delta =\epsilon /8 \), then whenever
\[
0<|x-1|<\delta ,\]
\begin{eqnarray*}
|(3x+5)-8| & = & 3|x-1|\\
 & < & 3\left( \frac{\epsilon }{8}\right) \\
 & = & \epsilon .
\end{eqnarray*}


So, \[
\lim _{x\rightarrow 1}3x+5=8.\]

\end{proof}
Notice that the string of equations/inequalities, following from beginning to
end, does say that \( |(3x+5)-8|<\epsilon  \). It is typical to write an equals
sign in an intermediate step if it really is equal to the next step. The total
relation may end up as an inequality, as it did here, though. Note also that
I used the calculations I did earlier in my {}``scratch work{}'' to get from
the |f(x)-L| back to the \( |x-a| \).
\end{document}
