\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}