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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{Examples}


\subsubsection{Examples of basic limits}

\begin{enumerate}
\item Let \( f(x)=3x^{2}+1 \). Then \( \lim _{x\rightarrow 2}f(x)=13 \).  
\item Let 
\[
f(x)=\left\{ \begin{array}{ll}
3x+1 & \mathrm{if}\, x\geq 1\\
x+2 & \mathrm{if}\, x<1
\end{array}\right. .\]
Then \( \lim _{x\rightarrow 1}f(x) \) does not exist. The reason it doesn't
exist is that, as you approach from the left (\( x<1 \)), the values get towards
3, while from the right (\( x>1) \), the values get towards 4.  
\item Let 
\[
f(x)=\left\{ \begin{array}{cc}
x^{2} & \mathrm{if}\, x\neq 1\\
5 & \mathrm{if}\, x=1
\end{array}\right. .\]
Then \( \lim _{x\rightarrow 1}f(x)=1 \), even though \( f(1)=5 \). Here the
function is purposely defined to have the \textit{wrong} value at 1. The limit
\( \lim _{x\rightarrow 1}f(x) \) is what \( f(1) \)  \textit{ought to} be.
 

This example stresses the predictability of values of the function. This function
is bad in that it doesn't have the expected value at 1. The previous example
had no value that \( f(1) \) should be; from one side it should be one thing,
from the other side it should be another. But \( f \) has to have just one
value. 

\item Let \( f(x)=sin(1/x) \). Then \( \lim _{x\rightarrow 0}f(x) \) doesn't exist.
 This one looks much worse than Example \textbf{2.}. Here, as \( x\rightarrow 0 \),
\( f(x) \) is wildly switching between \( -1 \) and 1.  
\item Let \( f(x)=\frac{x^{3}+1}{x^{2}-1} \). Then \( \lim _{x\rightarrow -1}f(x)=\frac{-3}{2} \).
\end{enumerate}

\subsubsection{Limits with infinity}

\begin{enumerate}
\item \( \lim _{x\rightarrow \infty }\frac{\sqrt{x^{2}+1}-1}{x} \)
\item \( \lim _{x\rightarrow \infty }\frac{x^{3}+x^{2}x-6}{x^{3}+5} \)
\item \( \lim _{x\rightarrow 1-}\frac{1}{x-1} \)
\item \( \lim _{x\rightarrow \infty }\frac{x^{2}+5x-6}{x^{3}+x+5} \)
\end{enumerate}

\subsubsection{Examples of limits by the definition}

Examples 

\textbf{1)} Show that \( \lim _{x\rightarrow 1}3x+5=8 \). 

\textbf{2)} Show that \( \lim _{x\rightarrow 2}x^{2}+3x=10 \). 

\textbf{3)} Show that \( \lim _{x\rightarrow 3}\frac{1}{x}=\frac{1}{3} \). 

\textbf{4)} 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 \( |x-1|<\delta  \). 

\begin{example}
Compute \( \lim _{x\rightarrow 3}\frac{x^{2}+2x-1}{x^{3}-3x^{2}+2} \).
\end{example}
%

\begin{example}
Compute \( \lim _{x\rightarrow -1}\sqrt{x^{3}+2x+7} \).
\end{example}
\begin{proof}
To be fair, we really don't have the theory to claim this, but we can argue
simply that, since 
\begin{eqnarray*}
\lim _{x\rightarrow -1}x^{3}+2x+7 & = & \lim _{x\rightarrow -1}\left( \sqrt{x^{3}+2x+7}\right) ^{2}\\
 & = & \lim _{x\rightarrow -1}\sqrt{x^{3}+2x+7}\lim _{x\rightarrow -1}\sqrt{x^{3}+2x+7},
\end{eqnarray*}
and \( \lim _{x\rightarrow -1}x^{3}+2x+7=(-1)^{3}+2(-1)+7=4 \), then \( \lim _{x\rightarrow -1}\sqrt{x^{3}+2x+7}=2 \).
\end{proof}
\begin{example}
Compute \( \lim _{x\rightarrow 9}\frac{x^{2}-81}{\sqrt{x}-3} \).
\end{example}
%

\begin{example}
Define 
\[
f(x):=\left\{ \begin{array}{ll}
\sqrt{x-4}, & \mathrm{if}\, x>4\\
8-2x, & \mathrm{if}\, x<4
\end{array}\right. .\]
Find \( \lim _{x\rightarrow 4}f(x) \), if it exists.

\end{example}
\end{document}
