\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 \alert{ 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|} \mpause[3]{ = \alert<11->{2|x-3|}} \mpause[4]{ \alert{< 0.1}} \end{talign} \pause\pause\pause\pause\pause Thus $|f(x) - 5| < 0.1$ whenever $0 < |x-3| < \alert{\alt<-11>{?}{0.05}}$ \pause; i.e. $\delta = 0.05$. \end{exampleblock} \vspace{10cm} \end{frame}