\begin{frame}
  \frametitle{L'Hospital's Rule}

  \begin{exampleblock}{}
    Evaluate the limit
    \begin{talign}
      \lim_{x\to (\pi/2)^-} (\sec x - \tan x)
    \end{talign}
    \pause
    Then
    \quad $\lim_{x\to (\pi/2)^-} \sec x = \pause\infty$ \pause\quad and \quad $\lim_{x\to (\pi/2)^-} \tan x = \pause\infty$
    \pause\medskip
    
    We use a common denominator:
    \begin{talign}
      \lim_{x\to (\pi/2)^-} (\sec x - \tan x) 
      &\mpause[1]{ = \lim_{x\to (\pi/2)^-} \frac{1 - \sin x}{\cos x} }
    \end{talign}
    \pause\pause
    Now
    \quad $\lim_{x\to (\pi/2)^-} (1 - \sin x) = \pause 0$ \pause\quad and \quad $\lim_{x\to (\pi/2)^-} \cos x = \pause 0$
    \pause\medskip
    
    Hence we can apply l'Hospital's Rule:
    \begin{talign}
      \lim_{x\to (\pi/2)^-} (\sec x - \tan x) 
      &= \lim_{x\to (\pi/2)^-} \frac{1 - \sin x}{\cos x} \\
      &\mpause[1]{ = \lim_{x\to (\pi/2)^-} \frac{-\cos x}{-\sin x} }
      \mpause[2]{ = 0}
    \end{talign}
  \end{exampleblock}

\end{frame}