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

  \begin{block}{L'Hospital's Rule}
    Suppose $f$ and $g$ are differentiable and $g'(x) \ne 0$ near $a$, and
    \begin{talign}
      \lim_{x\to a} \frac{f(x)}{g(x)}
    \end{talign}
    is an indeterminate form of type \alert<3->{$\frac{0}{0}$} or \alert<3->{$\frac{\infty}{\infty}$}.
    Then
    \begin{talign}
      \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}
    \end{talign}
    if the limit on the right side exists or is $-\infty$ or $\infty$.
  \end{block}
  \pause
  (near $a$ = on an open interval containing $a$ except possibly $a$ itself)\hspace*{-20ex}
  \pause\bigskip
  
  \alert{Before applying L'Hospital's Rule it is important to verify that:}\vspace{-.25ex}
  \begin{exampleblock}{}\vspace{-1.1ex}
  \begin{malign}
    \lim_{x \to a} f(x) = 0 \text{\quad and \quad} \lim_{x \to a} g(x) = 0
  \end{malign}\vspace{-.5ex}
  \end{exampleblock}\vspace{-.8ex}
  
  or\vspace{-.7ex}
  \begin{exampleblock}{}\vspace{-1.1ex}
  \begin{malign}
    \lim_{x \to a} f(x) = \pm\infty \text{\quad and \quad} \lim_{x \to a} g(x) = \pm\infty
  \end{malign}\vspace{-.5ex}
  \end{exampleblock}
\end{frame}