\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}