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

  \begin{exampleblock}{}
    Find 
    \begin{talign}
      \lim_{x\to \pi^-} \frac{\sin x}{1 -\cos x}
    \end{talign}
    \pause
    If we were to apply l'Hospital's Rule:
    \pause
    \begin{talign}
      \lim_{x\to \pi^-} \frac{\sin x}{1 -\cos x} =
      \mpause[1]{ \lim_{x\to \pi^-} \frac{\cos x}{\sin x} }
      \mpause[2]{ = -\infty }
    \end{talign}
    \pause\pause\pause
    \alert{However, this is wrong!}
    \pause\medskip
    
    We have \quad \alert{$\lim_{x\to \pi^-} (1-\cos x) = 1 - (-1) = 2$}.\pause
    \begin{talign}
      \lim_{x\to \pi^-} \frac{\sin x}{1 -\cos x} = 
      \mpause[1]{ \frac{0}{1-(-1)} = 0 }
    \end{talign}
  \end{exampleblock}  
  \pause\pause
  
  \begin{alertblock}{}
    Before applying l'Hospital's Rule, always check that the limit is an 
    indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
  \end{alertblock}
\end{frame}