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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}
\pause
Intuitively, when $x$ is close to $3$ but $x\ne 3$ then $f(x)$ is close to $5$.
\pause\bigskip

How close to $3$ does $x$ need to be for $f(x)$ to differ from $5$ less than $0.1$?
}
\begin{itemize}
\pause
\item the distance of $x$ to $3$ is $|x-3|$
\pause
\item the distance of $f(x)$ to $5$ is $|f(x)-5|$
\end{itemize}
\pause
To answer the question we need to find $\delta > 0$ such that
\begin{talign}
|f(x) - 5| < 0.1 \quad\text{ whenever }\quad 0 < |x-3| < \delta
\end{talign}
\pause
For $x \ne 3$ we have
\begin{talign}
|f(x) - 5| \mpause[1]{ = |(2x-1)-5|}
\mpause[2]{ = |2x-6|}
Thus $|f(x) - 5| < 0.1$ whenever $0 < |x-3| < \alert{\alt<-11>{?}{0.05}}$
i.e. $\delta = 0.05$.