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\begin{frame}
\frametitle{Precise Definition of Limits: Example}

\begin{exampleblock}{}
\vspace{-1ex}
\begin{talign}
f(x) = \begin{cases}
2x - 1 &\text{for $x \ne 3$}\\
6 &\text{for $x = 3$}
\end{cases}
\end{talign}
Similarly, we find
\begin{talign}
\end{talign}
\pause
The distances \structure{$0.1$},\; \structure{$0.01$}, \structure{\ldots} are called \emph{error tolerance}.
\pause\medskip

We have: \alert<-5>{$\delta(0.1) = 0.05$}\pause,\; \alert<-5>{$\delta(0.01) = 0.005$}\pause,\; \alert<-5>{$\delta(0.001) = 0.0005$}
\pause\medskip

Thus \alert{\emph{$\boldsymbol{\delta(\epsilon)}$ is a function of the error tolerance $\boldsymbol{\epsilon}$}}!
\pause\bigskip

We need to define $\delta(\epsilon)$ for arbitrary error tolerance $\epsilon > 0$:
\begin{talign}
|f(x) - 5| < \epsilon \quad&\text{ whenever }\quad 0 < |x-3| < \delta(\epsilon)
\end{talign}\pause
We want $|f(x) - 5| = 2|x-3| < \epsilon$. \pause We define \alert{$\delta(\epsilon) = \pause\epsilon/2$}.
\end{exampleblock}
\vspace{10cm}
\end{frame}