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\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}