\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}