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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


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

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

\end{example}

\subsubsection*{Solution }

This one is more subtle. The problem with this limit, as with the one discussed
at first, is that both the numerator and the denominator go to 0 as \( x \)
goes to \( -1 \). This is what is called an \emph{indeterminate form}, and
in general is a signal that you have more work to do to figure the limit out.
In this case we can see why the limit is \( -3/2 \) by simplifying the expression
for \( f(x) \):\begin{eqnarray*}
f(x) & = & \frac{x^{3}+1}{x^{2}-1}\\
 & = & \frac{(x^{2}-x+1)(x+1)}{(x-1)(x+1)}\\
 & = & \frac{(x^{2}-x+1)}{(x-1)},
\end{eqnarray*}
so\begin{eqnarray*}
\lim _{x\rightarrow -1}f(x) & = & \lim _{x\rightarrow -1}\frac{(x^{2}-x+1)}{(x-1)}\\
 & = & \frac{((-1)^{2}-(-1)+1)}{((-1)-1)}\\
 & = & \frac{3}{-2}\\
 & = & -\frac{3}{2}.
\end{eqnarray*}



\subsubsection*{Back}
\end{document}
