\begin{frame} \frametitle{L'Hospital's Rule} \begin{exampleblock}{} Evaluate the limit \begin{talign} \lim_{x\to (\pi/2)^-} (\sec x - \tan x) \end{talign} \pause Then \quad $\lim_{x\to (\pi/2)^-} \sec x = \pause\infty$ \pause\quad and \quad $\lim_{x\to (\pi/2)^-} \tan x = \pause\infty$ \pause\medskip We use a common denominator: \begin{talign} \lim_{x\to (\pi/2)^-} (\sec x - \tan x) &\mpause[1]{ = \lim_{x\to (\pi/2)^-} \frac{1 - \sin x}{\cos x} } \end{talign} \pause\pause Now \quad $\lim_{x\to (\pi/2)^-} (1 - \sin x) = \pause 0$ \pause\quad and \quad $\lim_{x\to (\pi/2)^-} \cos x = \pause 0$ \pause\medskip Hence we can apply l'Hospital's Rule: \begin{talign} \lim_{x\to (\pi/2)^-} (\sec x - \tan x) &= \lim_{x\to (\pi/2)^-} \frac{1 - \sin x}{\cos x} \\ &\mpause[1]{ = \lim_{x\to (\pi/2)^-} \frac{-\cos x}{-\sin x} } \mpause[2]{ = 0} \end{talign} \end{exampleblock} \end{frame}