70/118
\begin{frame}
  \frametitle{L'Hospital's Rule}

  \begin{exampleblock}{}
    Evaluate the limit
    \begin{talign}
      \lim_{x\to 0^+} x\ln x
    \end{talign}
    \pause
    We have
    \begin{talign}
      \lim_{x\to 0^+} x = 0 &&\text{and}&& \lim_{x\to 0^+} \ln x = \mpause[1]{-\infty}
    \end{talign}
    \pause\pause
    Thus we can choose for rewriting to:
    \begin{talign}
      \lim_{x\to 0^+} \frac{x}{1/\ln x} &&\text{or}&& \lim_{x\to 0^+} \frac{\ln x}{1/x}
    \end{talign}
    \pause
    We choose the 2nd since the derivatives are easier:
    \begin{talign}
      \lim_{x\to 0^+} x\ln x
      = \lim_{x\to 0^+} \frac{\ln x}{1/x}
      = \mpause[1]{ \lim_{x\to 0^+} \frac{1/x}{-1/x^2} }
      \mpause[2]{= \lim_{x\to 0^+} (-x) }
      \mpause[3]{= 0 }
    \end{talign}
  \end{exampleblock}
\end{frame}