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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{Continuity Examples}

\begin{enumerate}
\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 \( f \) is \textit{not} continuous at \( a=1 \). This is because 
\[
\lim _{x\rightarrow 1}f(x)\]
 does not exist, which you should recall from an earlier lecture.  
\item \noindent Let 
\[
f(x)=\left\{ \begin{array}{ll}
x^{2}, & \mathrm{if}\, x\neq 1\\
5, & \mathrm{if}\, x=1
\end{array}\right. .\]
This function is also not continuous at \( x=1 \), since 
\[
\lim _{x\rightarrow 1}f(x)=1,\]
 even though \( f(1)=5 \). The difference here is that the function \textit{could}
 have been continuous, if it weren't for the fact that it was ``poorly'' defined
at \( x=1 \).  
\item Let 
\[
f(x)=\left\{ \begin{array}{ll}
3x+1, & \mathrm{if}\, x\geq 1\\
x+3, & \mathrm{if}\, x<1
\end{array}\right. .\]
Then \( f \) \textit{}\textbf{is} continuous at \( a=1 \). This is because
\[
\lim _{x\rightarrow 1}f(x)\]
 does exist, and is the same as \( f(1) \).  
\item Since, for any polynomial \( f(x)=a_{n}x^{n}+\dots +a_{1}x+a_{0} \),  and for
any \( x=a \), \( \lim _{x\rightarrow a}f(x)=f(a) \), any polynomial is continuous.
We say that a function \( f \) is \textit{continuous}, instead of continuous
at a point or on an interval, if it is continuous at \textit{every} point of
its domain. 
\item \( f(x)=|x| \) \textit{is} continuous, even though it has ``bad behavior''
at \( x=0 \). Its behavior is just not \textit{that} bad there. The limit of
\( f(x) \), as \( x \) approaches 0, is 0, and since that is the value of
the function, then it is continuous there. At other points, \( f \) is just
like a polynomial in some open interval (either \( x \) or \( -x \)), so it
is continuous. 
\end{enumerate}
\end{document}
