\begin{frame}
  \frametitle{Computing Limits: Function Replacement}
  
  \begin{exampleblock}{}
    Find \vspace{-2ex}
    \begin{talign}
      \lim_{t\to 0} \frac{\sqrt{t^2 + 9} - 3}{t^2}
    \end{talign}
    \pause
    We have:
    \begin{talign}
      \frac{\sqrt{t^2 + 9} - 3}{t^2}
      &\mpause[1]{ = \frac{\sqrt{t^2 + 9} - 3}{t^2} \cdot \frac{\sqrt{t^2 + 9} + 3}{\sqrt{t^2 + 9} + 3}}
      \mpause[2]{ = \frac{t^2 + 9 - 9}{t^2 \cdot (\sqrt{t^2 + 9} + 3)}}\\
      &\mpause[3]{ = \frac{t^2}{t^2 \cdot (\sqrt{t^2 + 9} + 3)} }
      \mpause[4]{ \stackrel{\text{\alert{for $t \ne 0$}}}{=} \frac{1}{\sqrt{t^2 + 9} + 3} }
    \end{talign}
    \pause\pause\pause\pause\pause
    As a consequence:
    \begin{talign}
      \lim_{t\to 0} \frac{\sqrt{t^2 + 9} - 3}{t^2}  &= \lim_{t\to 0} \frac{1}{\sqrt{t^2 + 9} + 3} \\
      &\mpause[1]{ = \frac{1}{\sqrt{\lim_{t\to 0} (t^2 + 9)} + 3} \quad \text{by laws 5, 1, 9, 7}  }\\
      &\mpause[2]{ = \frac{1}{\sqrt{9} + 3} }
      \mpause[3]{ = \frac{1}{6} }
    \end{talign}
  \end{exampleblock}
\end{frame}