\begin{frame} \frametitle{Limits at Infinity} \begin{exampleblock}{} Evaluate \begin{talign} \lim_{x\to \alt<-9>{\infty}{\alert{-\infty}}} \frac{\sqrt{2x^2 + 1}}{3x-5} \end{talign} \pause We have \begin{talign} \lim_{x\to \infty} &\frac{\sqrt{2x^2 + 1}}{3x-5} \mpause[1]{= \lim_{x\to \infty} \left( \frac{\sqrt{2x^2 + 1}}{3x-5} \cdot \frac{(\frac{1}{x})}{(\frac{1}{x})} \right) } \\ &\mpause[2]{= \lim_{x\to \infty} \frac{\frac{\sqrt{2x^2 + 1}}{x}}{3-\frac{5}{x}} } \mpause[3]{= \lim_{x\to \infty} \frac{\frac{\sqrt{2x^2 + 1}}{\alt<-11>{}{\alert{-}}\sqrt{x^2}}}{3-\frac{5}{x}} \quad \alt<-10>{ \text{\textcolor{gray}{since $x>0$, $x = \sqrt{x^2}$}} }{ \text{\textcolor{gray}{since $x<0$, \alert{$x = -\sqrt{x^2}$}}} } }\hspace{1cm} \\ &\mpause[4]{= \lim_{x\to \infty} \alt<-11>{}{\alert{-}} \frac{\sqrt{2 + \frac{1}{x^2}}}{3-\frac{5}{x}} } \mpause[5]{= \alt<-11>{}{\alert{-}} \frac{\lim_{x\to \infty} \sqrt{2 + \frac{1}{x^2}}}{\lim_{x\to \infty} (3-\frac{5}{x})} } \\ &\mpause[6]{= \alt<-11>{}{\alert{-}} \frac{ \sqrt{\lim_{x\to \infty} (2 + \frac{1}{x^2})}}{3} } \mpause[7]{= \alt<-11>{}{\alert{-}} \frac{\sqrt{2}}{3} } \end{talign} \end{exampleblock} \onslide<12>{} \end{frame}