\begin{frame} \frametitle{L'Hospital's Rule} \begin{exampleblock}{} Evaluate the limit \begin{talign} \lim_{x\to 0^+} x\ln x \end{talign} \pause We have \begin{talign} \lim_{x\to 0^+} x = 0 &&\text{and}&& \lim_{x\to 0^+} \ln x = \mpause[1]{-\infty} \end{talign} \pause\pause Thus we can choose for rewriting to: \begin{talign} \lim_{x\to 0^+} \frac{x}{1/\ln x} &&\text{or}&& \lim_{x\to 0^+} \frac{\ln x}{1/x} \end{talign} \pause We choose the 2nd since the derivatives are easier: \begin{talign} \lim_{x\to 0^+} x\ln x = \lim_{x\to 0^+} \frac{\ln x}{1/x} = \mpause[1]{ \lim_{x\to 0^+} \frac{1/x}{-1/x^2} } \mpause[2]{= \lim_{x\to 0^+} (-x) } \mpause[3]{= 0 } \end{talign} \end{exampleblock} \end{frame}