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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{Infinite Limits}

\begin{example}
\[
\lim _{x\rightarrow \infty }\frac{x^{2}+5x-6}{x^{3}+x+5}\]

\end{example}

\subsubsection*{Hints}

For this one, we still can make it clearer what happens to the dominant terms
by dividing out the top and bottom by the largest shared power of \( x \) .


\subsubsection*{More}

In this case, the largest shared power is \( x^{2} \). Diviging out top and
bottom gives:\begin{eqnarray*}
\lim _{x\rightarrow \infty }\frac{x^{2}+5x-6}{x^{3}+x+5} & = & \lim _{x\rightarrow \infty }\frac{\left( x^{2}+5x-6\right) \frac{1}{x^{2}}}{\left( x^{3}+x+5\right) \frac{1}{x^{2}}}\\
 & = & \lim _{x\rightarrow \infty }\frac{1+\frac{5}{x}-\frac{6}{x^{2}}}{x+\frac{1}{x}+\frac{5}{x^{2}}}.
\end{eqnarray*}



\subsubsection*{Finish it}

Now, as \( x \) goes to infinity, all the terms of either the numerator or
denominator with powers of \( x \) in their denominator do go to 0, but there
is that one term of \( x \) in the denominator which goes to infinity, bringing
the whole denominator off to infinity with it, and so the limit goes to 0, \[
\lim _{x\rightarrow \infty }\frac{x^{2}+5x-6}{x^{3}+x+5}=0.\]



\subsubsection*{Back}
\end{document}
