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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{Curve Sketching}

\textbf{Asymptotes:} An \textit{asymptote} is a line, vertical, horizontal,
or slanted, which the graph of the function approximates {}``near infinity{}''.
This was already discussed somewhat back in <a href="../Unit1/limits-infinite.html">Chapter 1</a>,
when we were dealing with infinite limits. 

\emph{More} \textbf{\emph{about Asymptotes}}

\begin{enumerate}
\item A \textit{vertical asymptote} is a vertical line \( x=a \) , where \( a \)
is a point at which the function has an infinite limit. If \[
\lim _{x\rightarrow a}f(x)=\infty ,\]
 then \( x=a \) is a vertical asymptote for the graph. This holds for one-sided
limits as well. Usually, but not always, vertical asymptotes are places where
the denominator of the expression for \( f(x) \) has value \( 0 \). \( f(x)=\ln x \)
also has a vertical asymptote at \( x=0 \).


\vspace{0.3cm}
{\par\centering \includegraphics{vert-asymptotes.ps} \par}
\vspace{0.3cm}

\end{enumerate}
\begin{enumerate}
\item A \textit{horizontal asymptote} is a horizontal line \( y=c \), where \[
\lim _{x\rightarrow \pm \infty }f(x)=c.\]
 The limit could approach \( c \) at either \( +\infty \, \, \, -\infty  \),
or both.


\vspace{0.3cm}
{\par\centering \includegraphics{horiz-asymptote.ps} \par}
\vspace{0.3cm}

\item A \textit{slant asymptote} is a situation where the graph \( y=f(x) \) approaches
some other line as \( x\rightarrow \infty  \) (or \( -\infty  \)). You can
see when this occurs by finding when there is some line \( y=mx+b \) for which
\[
\lim _{x\rightarrow \infty }\left( f(x)-(mx+b)\right) =0.\]



\vspace{0.3cm}
{\par\centering \includegraphics{slant-asymptote.ps} \par}
\vspace{0.3cm}

\end{enumerate}
Roughly, you should look for situations where \( f(x) \) has an expression
which is linear except for terms of lower order, like\[
f(x)=\frac{x^{3}+2x+3}{x^{2}-x+1}.\]
 This function has a slant asymptote of \( y=x+1 \), since \begin{eqnarray*}
\lim _{x\rightarrow \infty }\left( f(x)-(x+1)\right)  & = & \lim _{x\rightarrow \infty }\frac{x^{3}+2x+3-(x+1)\left( x^{2}-x+1\right) }{x^{2}-x+1}\\
 & = & \lim _{x\rightarrow \infty }\frac{x^{3}+2x+3-\left( x^{3}+x^{2}-x^{2}-x+x+1\right) }{x^{2}-x+1}\\
 & = & \lim _{x\rightarrow \infty }\frac{2x+2}{x^{2}-x+1}\\
 & = & 0.
\end{eqnarray*}

\end{document}
