\begin{frame}
  \frametitle{Precise Definition of Limits: Example}
  
  \vspace{-.5ex}
  \begin{exampleblock}{}
%     Assume $\lim_{x\to a} f(x)$ and $\lim_{x\to a} g(x)$ exist.\\
    Let \alert{$\lim_{x\to a} f(x) = L_f$} and \alert{$\lim_{x\to a} g(x) = L_g$}.
    Prove the sum law:
    \begin{talign}
      \alert{\lim_{x\to a} \;[f(x) + g(x)] = L_f + L_g}
    \end{talign}
    \pause
    \smallskip
    Let $\epsilon > 0$ be arbitrary\pause, we need to find $\delta$ such that
    \begin{talign}
      \text{if} \quad 0<|x-a| < \delta \quad\text{then}\quad |(f(x) + g(x)) - (L_f + L_g)|  < \epsilon
    \end{talign}
    \pause
    Note that $(f(x) + g(x)) - (L_f + L_g) \;=\; (f(x) - L_f) + (g(x) - L_g)$.
    \pause\medskip
    
    We know that there exists $\delta_f$ such that:
    \begin{talign}
      \text{if} \quad 0<|x-a| < \delta_f \quad\text{then}\quad |f(x) - L_f|  < \epsilon/2
    \end{talign}\pause
    and there exists $\delta_g$ such that:
    \begin{talign}
      \text{if} \quad 0<|x-a| < \delta_g \quad\text{then}\quad |g(x) - L_g|  < \epsilon/2
    \end{talign}
    \pause
    We take $\delta = \pause \min (\delta_f,\delta_g)$. \pause
    \alert{If $0<|x-a| < \delta$ then\pause
    \begin{talign}
      |f(x) - L_f|  < \epsilon/2 && \text{ and } && |g(x) - L_g|  < \epsilon/2
    \end{talign}
    \pause
    and hence $|(f(x) - L_f) + (g(x) - L_g)|  < \epsilon$.}
  \end{exampleblock}
\end{frame}