\begin{frame} \frametitle{L'Hospital's Rule} \begin{block}{L'Hospital's Rule} Suppose $f$ and $g$ are differentiable and $g'(x) \ne 0$ near $a$, and \begin{talign} \lim_{x\to a} \frac{f(x)}{g(x)} \end{talign} is an indeterminate form of type \alert<3->{$\frac{0}{0}$} or \alert<3->{$\frac{\infty}{\infty}$}. Then \begin{talign} \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)} \end{talign} if the limit on the right side exists or is $-\infty$ or $\infty$. \end{block} \pause (near $a$ = on an open interval containing $a$ except possibly $a$ itself)\hspace*{-20ex} \pause\bigskip \alert{Before applying L'Hospital's Rule it is important to verify that:}\vspace{-.25ex} \begin{exampleblock}{}\vspace{-1.1ex} \begin{malign} \lim_{x \to a} f(x) = 0 \text{\quad and \quad} \lim_{x \to a} g(x) = 0 \end{malign}\vspace{-.5ex} \end{exampleblock}\vspace{-.8ex} or\vspace{-.7ex} \begin{exampleblock}{}\vspace{-1.1ex} \begin{malign} \lim_{x \to a} f(x) = \pm\infty \text{\quad and \quad} \lim_{x \to a} g(x) = \pm\infty \end{malign}\vspace{-.5ex} \end{exampleblock} \end{frame}