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\begin{center} \begin{large} The product theorem for limits\end{large}\\
Matchett, \today
\end{center}

The proof of the theorem in this short paper is more complicated than the 
proof of the  ``sum theorem'' for limits, which is the second little project
that I have assigned to my calculus students this semester.  So those who can 
understand the proof here will have no great difficulty producing a proof of 
the sum theorem.  That proof is due Monday, Feb. 11.  As with all assignments
to turn in, the paper should be typed using standard math symbol processing 
software like LaTeX.  However, those who used LaTeX successfully on the
first assignment, with 
a grade of 7 or higher, are not required to use LaTeX.  Anyone who can make
some other software product work nicely for typing up mathematical papers, is
welcome to do so.

There is a second proof of the product theorem in Appendix F of
our calculus textbook.  That proof is similar to mine. Both 
proofs show the level of detail that I would like to see in the 
proof of the sum theorem.  However, since the
sum theorem is a little easier, that proof will be shorter than the one here.

With that lengthy preamble, I shall now state and prove the afore mentioned 
theorem.
\begin{flushleft}\textbf{Theorem} If $\lim_{x\rightarrow a} f(x) = L$ and
$\lim_{x\rightarrow a} g(x) = M$, then 
\begin{displaymath}
                \lim_{x\rightarrow a} f(x)g(x) = LM\ .
\end{displaymath}
\end{flushleft}
\noindent \textbf{Proof}.  For all $x$ in the domains of $f$ and $g$,
\begin{displaymath}  
    |f(x)g(x) - LM| = |f(x)g(x) - f(x)M + f(x)M - LM|\ .
\end{displaymath}
By the triangle inequality,
\begin{displaymath}
 |f(x)g(x) - f(x)M + f(x)M - LM|  \le  |f(x)g(x) - f(x)M| + |f(x)M - LM|\ .
\end{displaymath}
The right hand side of the last inequality may be rewritten 
\begin{displaymath} 
|f(x)|\,|g(x) - M| + |M|\,|f(x) - L|\ ,
\end{displaymath}
and we conclude that 
\begin{equation}|f(x)g(x) - LM| \le |f(x)|\,|g(x) - M| + 
|M|\,|f(x) - L|\: .\label{key}
\end{equation}

Now let $\epsilon > 0$, and let $\delta_{1} > 0$ be such that if $0 < |x-a| < 
\delta_{1}$, then
\begin{equation} |f(x) - L| < \mbox{Min}\left\{\frac{\epsilon}{3(|M| + 1)}
\ , \ 1\right\}\ .  \label{special}
\end{equation}
Such a $\delta_{1}$ exists because $\lim_{x\rightarrow a} f(x) = L$.
Similarly, let $\delta_{2}$ be a real number greater than 0 such that
if $0 < |x-a| < \delta_{2}$, then
\begin{displaymath} |g(x) - M| < \frac{\epsilon}{3(|L|+1)}\ .
\end{displaymath}
Such a number $\delta_{2}$ exists, because $\lim_{x\rightarrow a}g(x) = M$.
Let $\delta = \mbox{Min}\{\delta_{1}, \delta_{2}\}$.\quad  
We will now finish the proof by showing that 
if $0 < |x-a| < \delta$, then $|f(x)g(x) - LM| < \epsilon$.

So assume $0 < |x-a| < \delta$.    Then
\begin{eqnarray*} 
|M|\,|f(x) - L| & \le & (|M|+1)\,|f(x) - L|\\
      & < & (|M|+1)\, \left(\frac{\epsilon}{3(|M|+1)}\right)  \\
      & = & \epsilon/3 \ .
\end{eqnarray*}

Since $\delta = Min\{\, \delta_1\: , \: \delta_2\,\}$, we know that
$|x-a|$ is less than both $\delta_1$ and $\delta_2$.  In particular,
$|x-a| < \delta_1$.  Therefore from
 line~(\ref{special}) above it follows that
 $|f(x) - L | < 1\, .$\ \ Next, notice 
that by the 
triangle inequality, $||f(x)| - |L|| \le |f(x) - L\,|$.  Therefore
$||f(x)| - |L|| \le 1$.  From this we can conclude that 
$|f(x)| \le |L| + 1$.

Now  
\begin{eqnarray*}
|f(x)|\,|g(x) - M| & \le & (|L| + 1)|g(x) - M|\\
        & < & (|L| + 1) \frac{\epsilon}{3(|L| + 1)} \\
      & = & \epsilon/3
\end{eqnarray*}
We now have good inequalities for $|f(x)|\,|g(x) - M|$ and $|M|\,|f(x) - L|$.
Putting these together with the inequality in line~(\ref{key}), we see that
\begin{displaymath}  |f(x)g(x) - LM|\quad \le \quad \epsilon/3 + 
\epsilon/3\ .
\end{displaymath}
Since $\epsilon/3 + \epsilon/3 \ < \ \epsilon$, it follows that 
$|f(x)g(x) - LM| \ < \ \epsilon$, and the proof is complete.\\ 
QED

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