\begin{frame}
  \frametitle{Computing Limits: Function Replacement}
  
  \begin{block}{Function Replacement}
    If $f(x) = g(x)$ for all $x \ne a$, then $\lim_{x \to a} f(x) = \lim_{x \to a} g(x)$
    (provided that the limit exists).
  \end{block}
  \pause
  Actually it suffices $f(x) = g(x)$ when $x$ is close to $a$.
  \pause
  
  \begin{exampleblock}{}
    Find $\lim_{x \to 1} \frac{x^2 - 1}{x-1}$. 
    \pause
    \begin{itemize}
      \item 
     Direct substitution is not applicable because $x = 1$ is not in the domain.
    \end{itemize}
    \pause We replace the function:
    \begin{talign}
      \frac{x^2 - 1}{x-1} 
      \mpause[1]{ = \frac{(x+1)(x-1)}{x-1}}
      \mpause[2]{ \stackrel{\text{\alert{for $x\ne 1$}}}{=} x+1}
    \end{talign}
    \pause\pause\pause
    As a consequence
    \begin{talign}
      \lim_{x\to 1} \frac{x^2 - 1}{x-1} 
      = \lim_{x\to 1} x+1 \mpause[1]{ = 1 + 1 = 2 }
    \end{talign}
  \end{exampleblock}
\end{frame}