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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


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

\maketitle

\subsubsection{Examples of limits by the definition}

\begin{example}
Show that 
\[
\lim _{x\rightarrow 1}3x+5=8.\]
 
\end{example}
%

\begin{example}
Show that 
\[
\lim _{x\rightarrow 2}x^{2}+3x=10.\]
 
\end{example}
%

\begin{example}
Show that 
\[
\lim _{x\rightarrow 3}\frac{1}{x}=\frac{1}{3}.\]
 
\end{example}
%

\begin{example}
Find a number \( \delta  \) sufficiently small so that the distance from \( f(x)=2x^{2}+3x-1 \)
to \( 4 \) is less than \( 1/100 \) if \( |x-1|<\delta  \). 
\end{example}

\subsubsection{Examples of rigorous infinite limits}

\begin{example}
Show that 
\[
\lim _{x\rightarrow \infty }\sqrt{x}=\infty .\]

\end{example}

\subsubsection*{Solution }

Let \( E \) be given. We can assume that  \( E\geq 0 \), because if it weren't,
you could replace it by the larger number 0. Now, let's find the number \( D \).
But how, you might ask, can we find \( D \) if we don't know what \( E \)
is, really? The idea is to find a \( D \) expressed  \textbf{in terms of \( E \)}
that works, so that, no matter what \( E \) is, \( D \) is adjusted automatically.
You figure that out by a sort of reverse-engineering: In order to make \( \sqrt{x}>E \),
try to see if you can get an inequality involving \( x \) from the one involving
\( f(x) \). Here the way to do it is to square both sided of the inequality.
Since everything in the inequality is positive, that won't change the inequality
(you have to worry about that sort of thing, sometimes). Then: \( \sqrt{x}>E\iff x>E^{2}, \)

so, if \( x>E^{2} \), we'll guarantee that \( \sqrt{x}>E \). So, \( E^{2} \)
will do as our \( D \), \( D=E^{2} \). 

Note that I said that that choice of \( D \) \textit{will do}. There isn't
any  \textbf{one} answer here, you just need to find a way to get a large enough
\( x \);  \textit{larger} will also do nicely. 

\begin{example}
Show that 
\[
\lim _{x\rightarrow \infty }(x/2+\sin x)=\infty .\]
 
\end{example}
%

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

\end{example}

\subsubsection*{Solution }

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


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


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}
