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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}

A function \( f(x) \) is \textit{continuous} at a point \( a \) if it behaves
as it  \textit{should} there, that is: 

\begin{defn}
Assume that \( f(x) \) is defined in some open interval  that contains \( a \).
Then, \( f \) is \textit{continuous at} \( a \) if: \[
\lim _{x\rightarrow a}f(x)=f(a).\]

\end{defn}
\( f \) is \textit{continuous on the interval} \( (b,c) \) if it is continuous
at all  points \( a\in (b,c) \). 

Note that this says two things, that the limit exists, and that the function's
value at \( a \) is the same as the limit. 

The usual idea of continuity is that a graph is \textit{continuous} if you can
draw the picture without lifting your pencil from the paper. That's a little
limited,  but gives the idea well enough. 

\vspace{0.3cm}
{\par\centering \includegraphics{continuous-discontinuous.ps} \par}
\vspace{0.3cm}

\begin{rem*}
There is also a definition of one-sided continuity (continuous with respect
to limits from one side or the other). This concept is primarily useful only
to discuss whether or not the function is continuous at the endpoints of its
domain. If \( f \) is defined on \( [a,b] \) it will be \textit{continuous
from the  right at} \( a \) if if: \[
\lim _{x\downarrow a}f(x)=f(a).\]

\end{rem*}
A similar definition can be made for continuity from the left.


\subsection*{Examples}

\begin{example}
Let \[
f(x)=\left\{ \begin{array}{ll}
3x+1, & \mathrm{if}\, x\geq 1\\
x+2, & \mathrm{if}\, x<1
\end{array}\right. .\]

\end{example}
\vspace{0.3cm}
{\par\centering \includegraphics{cont-ex1.ps} \par}
\vspace{0.3cm}

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 section.  

%

\begin{example}
\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 \( x=1 \).  
\end{example}
%

\begin{example}
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
\[
\lim _{x\rightarrow 1}f(x)=4\]
 does exist, and is the same as \( f(1) \).  
\end{example}
\vspace{0.3cm}
{\par\centering \includegraphics{cont-ex3.ps} \par}
\vspace{0.3cm}

%

\begin{example}
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. 
\end{example}
%

\begin{example}
\( f(x)=|x| \) \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{example}

\subsection*{Exercises}

\begin{xca}
Find the values of \( a \) for which the function \[
f(x):=\left\{ \begin{array}{ll}
x^{2}-2x, & \textrm{if}\, \textrm{x}\leq 2\\
ax+4, & \textrm{if}\, \textrm{x}>2,
\end{array}\right. \]
will be continuous at \( x=2 \). \textbf{Answer:} \( a= \).
\end{xca}
%

\begin{xca}
Is there a way to define the value of \( f(0) \) so that the function \[
f(x)=\frac{x}{|x|}\]
 will be continuous at \( x=0 \)? \textbf{Answer:} \( f(0)= \).\end{xca}

\end{document}
