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

  \begin{exampleblock}{}
    Evaluate the limit\vspace{-3ex}
    \begin{talign}
      \lim_{x\to 0^+} (1 + \sin 4x)^{\cot x}
    \end{talign}
    \pause
    Then
    \quad $\lim_{x\to 0^+} (1 +\sin 4x) = \pause 1$ \pause\quad and \quad $\lim_{x\to 0^+} \cot x = \pause \infty$
    \pause\medskip
    
    We write the limit as:
    \begin{talign}
      \lim_{x\to 0^+} (1 + \sin 4x)^{cot x}
      &\mpause[1]{ = \lim_{x\to 0^+} e^{\ln (1 + \sin 4x)^{cot x}} } \\
      &\mpause[2]{ = e^{\lim_{x\to 0^+} \left( \cot x \cdot \ln (1 + \sin 4x) \right) } } \\
      \mpause[3]{\lim_{x\to 0^+} \left( \cot x \cdot \ln (1 + \sin 4x) \right)}
      &\mpause[4]{= \lim_{x\to 0^+} \frac{\ln (1 + \sin 4x)}{\tan x}}
    \end{talign}
    \pause\pause\pause\pause\pause
    Now
    \quad $\lim_{x\to 0^+} \ln (1 + \sin 4x) = \pause 0$ \pause\quad and \quad $\lim_{x\to 0^+} \tan x = \pause 0$
    \pause\medskip
    
    Hence we can apply l'Hospital's Rule:
    \begin{talign}
      \lim_{x\to 0^+} \frac{\ln (1 + \sin 4x)}{\tan x}
      &\mpause[1]{ = \lim_{x\to 0^+} \frac{\frac{4\cos 4x}{1 + \sin 4x}}{(\sec x)^2} }
      \mpause[2]{ = \frac{\left(\frac{4}{1}\right)}{1}}
      \mpause[3]{ = 4}
    \end{talign}
    \pause\pause\pause\pause
    Thus \quad $\lim_{x\to 0^+} (1 + \sin 4x)^{\cot x} = e^4$
  \end{exampleblock}

\end{frame}