\begin{frame} \frametitle{Precise Definition of Limits - Example} \begin{exampleblock}{} Find $\delta > 0$ such that \begin{talign} \text{if} \quad 0<|x-1| < \delta \quad\text{then}\quad |(x^2 - 5x+6) - 2| < 0.2 \end{talign} \pause Note that $\delta$ is a bound on the distance of $x$ from $1$.\\ \pause Lets say \alert{$x = 1 + \delta$}. \pause Then \begin{talign} (x^2 - 5x + 6) - 2 &\mpause[1]{= (1+\delta)^2 - 5(1+\delta) + 4}\\ &\mpause[2]{= (1+2\delta+\delta^2) - (5 + 5\delta) +4}\\ &\mpause[3]{= \delta^2 - 3\delta} \end{talign}\vspace{-4ex} \pause\pause\pause\pause Thus \begin{talign} |(x^2 - 5x + 6) - 2| < 0.2 \quad &\iff\quad |\delta^2 - 3\delta| < 0.2 \end{talign}\pause Assume that $|\delta| < 1$ \pause(we can make it as small as we want)\pause, then: \begin{talign} |\delta^2 - 3\delta| \;\;\le\;\; \mpause[1]{|\delta^2| + |3\delta| \;\;\le\;\;} \mpause[2]{|\delta| + |3\delta| \;\;\le\;\;} \mpause[3]{4|\delta|} \end{talign} \pause\pause\pause\pause Thus: $\quad\text{if} \quad 4|\delta| < 0.2 \quad\text{then}\quad |(x^2 - 5x + 6) - 2| < 0.2$\;. \pause\smallskip Hence \alert{$\delta = 0.04$ is a possible choice}. \end{exampleblock} \end{frame}