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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{Precise Definition of a Limit}

\begin{example}
Show that \[
\lim _{x\rightarrow \infty }\frac{2x+3}{x-1}=2.\]

\end{example}

\subsubsection*{Hints}

Assume that \( \epsilon >0 \) is given. We need to be able to get \[
\frac{2x+3}{x-1}\]
 within \( \epsilon  \) of 2, by taking \( x \) large enough. However, that
distance can be re-written in such a way that it involves the size of \( x \)
(this, or analogous tricks, are standard for these calculations). 


\subsubsection*{More }

\[
\left| \frac{2x+3}{x-1}-2\right| =\left| \frac{(2x+3)-2(x-1)}{x-1}\right| =\left| \frac{5}{x-1}\right| .\]


Then,  \[
\left| \frac{2x+3}{x-1}-2\right| <\epsilon \, \Leftrightarrow \, \left| \frac{5}{x-1}\right| <\epsilon ,\]


which means that  \[
\frac{1}{|x-1|}<\frac{\epsilon }{5},\]


or, solving for \( x \), and taking \( x>0 \) (since \( x\rightarrow \infty  \),
this is not a  severe restriction): \[
|x-1|>\frac{5}{\epsilon }\, \Leftrightarrow \, x>\frac{5}{\epsilon }+1\]
 So, take \[
D=\frac{5}{\epsilon }+1.\]
 Then, if \( x>D \),  \( |f(x)-2|<\epsilon  \), as required. 
\end{document}
