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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{Computing limits}

How do you compute limits? There are a number of tricks to computing
limits, which we'll get to more fully. But, for now, note that we
are using some obvious facts: 

\begin{fact}
If you know the limits of $f(x)$ and $g(x)$ at $x=a$, then:

The limit of a sum is the sum of the limits:\[
\lim _{x\rightarrow a}(f(x)+g(x))=\lim _{x\rightarrow a}f(x)+\lim _{x\rightarrow a}g(x),\]
 

The limit of a product is the product of the limits\[
\lim _{x\rightarrow a}(f(x)g(x))=\lim _{x\rightarrow a}f(x)\lim _{x\rightarrow a}g(x),\]
 

and the limit of the quotient is the quotient of the limits,\[
\lim _{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim _{x\rightarrow a}f(x)}{\lim _{x\rightarrow a}g(x)}.\]
 This last one only makes sense if \[
\lim _{x\rightarrow a}g(x)\neq 0.\]
 If it is zero, we have more work to do to find that limit. 
\end{fact}
At the basis of the computations we did earlier, though, we really
should add a few more facts, which seem kind of obvious, but really
form the basis of the computations with limits: 

\begin{fact}
Obvious limits:\[
\lim _{x\rightarrow a}x=a.\]
 \[
\lim _{x\rightarrow a}1=1.\]
 If $f(x)=g(x)$ for all $x$ near $a$, except perhaps at $a$ itself,
then \[
\lim _{x\rightarrow a}f(x)=\lim _{x\rightarrow a}g(x).\]
 
\end{fact}
The first two in this list are pretty obvious indeed, but the third
one really is the way you first work with limits. let's use it to
do the first limit we did, \[
\lim _{x\rightarrow 1}\frac{x^{2}-1}{x-1}.\]
 If \[
f(x)=\frac{x^{2}-1}{x-1},\]
 then for all $x\neq 1$, $f(x)=x+1$, since \begin{eqnarray*}
\frac{x^{2}-1}{x-1} & = & \frac{(x+1)(x-1)}{x-1}\\
 & = & x+1,\, \textrm{by}\, \textrm{cancelling}\, \textrm{the}\, (x-1)'s,
\end{eqnarray*}
 so the limit  \[
\lim _{x\rightarrow 1}\frac{x^{2}-1}{x-1}=\lim _{x\rightarrow 1}x+1=2.\]


That last part really uses the first two stupid limits and the addition
rule, but that is getting awfully picky.


\subsection{A bit more theory:}

These two results may also seem rather obvious, but they will help
make sense of some rather nasty limits.

\begin{thm}
If $f(x)\leq g(x)$ for all $x$ in an open interval that contains
$a$ (except possibly at $a$ itself, then \[
\lim _{x\rightarrow a}f(x)\leq \lim _{x\rightarrow a}g(x).\]

\end{thm}
%

\begin{thm}
\textbf{\textup{{[}The Squeeze Theorem{]}}} If $f(x)\leq g(x)\leq h(x)$
in an open interval that contains $a$ (except possibly at $a$ itself,
and\[
\lim _{x\rightarrow a}f(x)=L=\lim _{x\rightarrow a}h(x),\]
then\[
\lim _{x\rightarrow a}g(x)=L.\]

\end{thm}
This result is usually used to show results such as 

\begin{example}
\[
\lim _{x\rightarrow 0}x\: \sin \left(\frac{1}{x}\right)=0.\]

\end{example}

\subsubsection*{Solution}

\begin{example}
Compute \[
\lim _{x\rightarrow 3}\frac{x^{2}-x-6}{x^{3}-4x^{2}+3x}.\]

\end{example}

\subsubsection*{Hints}

\begin{example}
Compute \[
\lim _{x\rightarrow -1}\sqrt{x^{3}+2x+7}.\]

\end{example}

\subsubsection*{Solution}

\begin{xca}
Compute \[
\lim _{x\rightarrow 3}\frac{x^{2}-9}{x^{2}-2x-3}.\]

\end{xca}
%

\begin{xca}
Compute\[
\lim _{x\rightarrow 2}\frac{x^{2}-5x+6}{x^{2}-4}=\]

\end{xca}
%

\begin{xca}
Compute \[
\lim _{x\rightarrow 9}\frac{x^{2}-81}{\sqrt{x}-3}=\]

\end{xca}
%

\begin{xca}
Define \[
f(x):=\left\{ \begin{array}{ll}
 \sqrt{x-4}, & \mathrm{if}\, x>4\\
 8-2x, & \mathrm{if}\, x<4\end{array}
\right..\]
Find \[
\lim _{x\rightarrow 4}f(x)=\]
 if it exists.\end{xca}

\end{document}
