\begin{frame}
  \frametitle{Precise Definition of Limits - Example}
  
  \begin{exampleblock}{}
    Proof that 
    \begin{talign}
      \lim_{x\to 3} (4x-5) = 7
    \end{talign}
    \pause
    Let $\epsilon > 0$ be arbitrary (the error tolerance).
    \pause\bigskip
    
    We need to find $\delta$ such that
    \begin{talign}
      \text{if} \quad 0<|x-3| < \delta \quad\text{then}\quad |(4x-5) - 7|  < \epsilon
    \end{talign}
    \pause
    We have
    \begin{talign}
      |(4x-5) - 7| < \epsilon \quad
      &\mpause[1]{\iff\quad |4x - 12| < \epsilon} \\
      &\mpause[2]{\iff\quad -\epsilon < 4x - 12 < \epsilon} \\
      &\mpause[3]{\iff\quad -\frac{\epsilon}{4} < x - 3 < \frac{\epsilon}{4}} \\
      &\mpause[4]{\iff\quad |x - 3| < \frac{\epsilon}{4}}
    \end{talign}
    \pause\pause\pause\pause\pause
    Thus $\delta = \frac{\epsilon}{4}$.\pause \quad
    If \quad $0<|x-3| <\frac{\epsilon}{4}$  \quad then \quad $|(4x-5) - 7|  < \epsilon$.
  \end{exampleblock}
\end{frame}