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

  \begin{exampleblock}{}
    Find 
    \begin{talign}
      \lim_{x\to \infty} \frac{e^x}{x^2}
    \end{talign}
    \pause
    We have
    \begin{talign}
      \lim_{x\to \infty} e^x = \infty &&\text{and}&&
      \lim_{x\to \infty} x^2 = \infty
    \end{talign}\pause
    Hence we can apply l'Hospital's Rule:
    \pause
    \begin{talign}
      \lim_{x\to \infty} \frac{e^x}{x^2} = 
      \mpause[1]{ \lim_{x\to \infty} \frac{\frac{d}{dx} e^x}{\frac{d}{dx} x^2} }
      \mpause[2]{ = \lim_{x\to \infty} \frac{e^x}{2x} }
    \end{talign}
    \pause\pause\pause
    Again we have:
    \begin{talign}
      \lim_{x\to \infty} e^x = \infty &&\text{and}&&
      \lim_{x\to \infty} 2x = \infty
    \end{talign}\pause
    So we can again use l'Hospital's Rule:
    \pause
    \begin{talign}
      \lim_{x\to \infty} \frac{e^x}{x^2} = \lim_{x\to \infty} \frac{e^x}{2x} = 
      \mpause[1]{ \lim_{x\to \infty} \frac{\frac{d}{dx} e^x}{\frac{d}{dx} 2x} }
      \mpause[2]{ = \lim_{x\to \infty} \frac{e^x}{2} }
      \mpause[3]{ = \infty}
    \end{talign}\pause
  \end{exampleblock}  

\end{frame}