\begin{frame}
  \frametitle{Precise Definition of One-Sided Limits - Example}
  
  \preciseright
  \pause
  
  \begin{exampleblock}{}
    Proof that $\lim_{x\to 0^+} \sqrt{x} = 0$.
    \pause\bigskip
    
    Let $\epsilon > 0$. \pause We look for $\delta > 0$ such that
    \begin{talign}
      \text{if } \quad 0 < x < 0+\delta \quad\text{ then }\quad |\sqrt{x} - 0| < \epsilon 
    \end{talign}%
    \pause% 
    We have \mpause[2]{(since $0 < x$)}
    \begin{talign}
      |\sqrt{x} - 0| \mpause[1]{= |\sqrt{x}|} \mpause[2]{= \sqrt{x}}
      \mpause[3]{< \epsilon}
      \mpause[4]{\quad\implies\quad x < \epsilon^2}
    \end{talign}
    \pause\pause\pause\pause\pause
%     We know
%     $\quad\sqrt{x} = \epsilon \mpause[1]{\quad\implies\quad x = \epsilon^2}$.
%     \pause\medskip
    
    Thus $\delta = \epsilon^2$.
    \quad\pause $\text{If } \quad 0 < x < 0+\epsilon^2 \quad\text{ then }\quad |\sqrt{x}-0| < \epsilon$. 
  \end{exampleblock}
\end{frame}