\documentclass{amsart}

\setlength{\parindent}{0pt}
\setlength{\parskip}{5pt}
\everymath={\displaystyle}
\pagestyle{empty}

\title{Examples of a Limit Not Existing}

\author{Math 165: Honors Calculus I}

\date{Fall, 1999}

\begin{document}

\maketitle

\thispagestyle{empty}

The definition of $\lim_{x\to p}f(x) = A$ is:

\begin{quote}
For all $\epsilon>0$ there exists a $\delta > 0$ such that if $x$ is any
number satisfying $0 < |x-p| < \delta$ then $|f(x) - A| < \epsilon$.
\end{quote}

The easiest way to show a limit does not exist is to calculate the
one-sided limits, if they exist, and show they are not equal.
For example, if $f(x) = [x]$, then $\lim_{x\to 2^-}f(x) = 1$ and
$\lim_{x\to 2^+}f(x) = 2$, so $\lim_{x\to 2}f(x)$ does not exist.

It is possible that the one-sided limits do not exist either and the
question arises, what exactly do you have to prove to show a limit
does not exist.  The negation of the definition given above is somewhat
complicated and leads to the following:

\begin{quote}
For any $A$, there exists an $\epsilon>0$ such that for all $\delta>0$,
there is an $x$ satisfying $0<|x-p|<\delta$ and $|f(x)-A|\ge\epsilon$.
\end{quote}

The idea is that for any proposed ``limit'' $A$, we can prove that there
are points $x$ close to $p$ ($|x-p|<\delta$), for which $f(x)$ is
far away from $A$ ($|f(x) - A| \ge \epsilon$).

\emph{Example 1:} Let
\[ f(x) = \left\{\begin{array}{ll}
           0 & \mbox{if $x$ is rational}\\
           1 & \mbox{if $x$ is irrational}
           \end{array} \right. \]
For any number $p$,  the limit $\lim_{x\to p}f(x)$ does not exist.

\emph{Proof:} Let $A$ be any number. Case~1): Assume $A \ne 0$. Then
set $\epsilon = |A|/2 > 0$ .  No matter what the value of $\delta$
is, there is a rational number $q$ satisfying $|q-p|<\delta$ (any
interval contains a rational number) and hence $|f(q)-A| = |0-A| = |A| >
|A|/2 = \epsilon$.  Case~2) Assume $A = 0$. Then set $\epsilon = 1/2$.
No matter what the value of $\delta$ is, there is an irrational number
$r$ satisfying $|r-p|<\delta$ (any interval contains a irrational number)
and hence $|f(r)-A| = |1-0| = 1 > 1/2 = \epsilon$.

\emph{Example 2:} $\lim_{x\to0}\sin(1/x)$ does not exist.

\emph{Proof:} Regardless of what $A$ is given, we can take $\epsilon
= 1$.  Case~1): Assume $A\ge0$. For any $\delta > 0$ there is a
positive integer $n$ such that $x = 1/(3\pi/2 + 2n\pi) < \delta$. Then
$\sin(1/x)=\sin(3\pi/2)=-1$ so $|\sin(1/x)-A|\ge1=\epsilon$. Case~2):
Assume $A<0$. For any $\delta>0$, there is a positive integer $n$ such
that $x = 1/(\pi/2 + 2n\pi) < \delta$. Then $\sin(1/x) = \sin(\pi/2) =
1$, so $|\sin(1/x)-A|>1=\epsilon$.

\end{document}
