\begin{frame} \frametitle{L'Hospital's Rule} \begin{exampleblock}{} Find \begin{talign} \lim_{x\to \infty} \frac{e^x}{x^2} \end{talign} \pause We have \begin{talign} \lim_{x\to \infty} e^x = \infty &&\text{and}&& \lim_{x\to \infty} x^2 = \infty \end{talign}\pause Hence we can apply l'Hospital's Rule: \pause \begin{talign} \lim_{x\to \infty} \frac{e^x}{x^2} = \mpause[1]{ \lim_{x\to \infty} \frac{\frac{d}{dx} e^x}{\frac{d}{dx} x^2} } \mpause[2]{ = \lim_{x\to \infty} \frac{e^x}{2x} } \end{talign} \pause\pause\pause Again we have: \begin{talign} \lim_{x\to \infty} e^x = \infty &&\text{and}&& \lim_{x\to \infty} 2x = \infty \end{talign}\pause So we can again use l'Hospital's Rule: \pause \begin{talign} \lim_{x\to \infty} \frac{e^x}{x^2} = \lim_{x\to \infty} \frac{e^x}{2x} = \mpause[1]{ \lim_{x\to \infty} \frac{\frac{d}{dx} e^x}{\frac{d}{dx} 2x} } \mpause[2]{ = \lim_{x\to \infty} \frac{e^x}{2} } \mpause[3]{ = \infty} \end{talign}\pause \end{exampleblock} \end{frame}