\begin{frame}
  \frametitle{Continuity on Intervals}

  \begin{block}{}
    A function $f$ is \emph{continuous} on an interval if
    it is continuous on every number in the interval.
  \end{block}
  \pause\medskip
  If the interval is left- and/or right-closed, then
  \begin{itemize}
    \item At the left-end we are only interested in right-continuity.
    \item At the right-end we are only interested in left-continuity.
  \end{itemize}
  (the values outside of the interval do not matter)
  \pause
  
  \begin{exampleblock}{}
    Show that $f(x) = 1 - \sqrt{1-x^2}$ is continuous on $[-1,1]$.
    \pause\medskip
    
    For $-1 < a < 1$ we have by the limit laws:
    \begin{talign}
      \lim_{x\to a} f(x)
%       = 1 - \lim_{x\to a} \sqrt{1-x^2}
      \mpause[1]{ = 1 - \sqrt{\lim_{x\to a}(1-x^2)} }
      \mpause[2]{ = 1 - \sqrt{1-a^2} = f(a) }
    \end{talign}
    \pause\pause\pause
    Similar calculations show
    \begin{itemize}
    \pause
      \item 
        $\lim_{x\to -1^+} f(x) = 1 = f(-1)$
    \pause
      \item 
        $\lim_{x\to 1^-} f(x) = 1 = f(1)$
    \end{itemize}
    \pause
    Therefore $f$ is continuous on $[-1,1]$.
  \end{exampleblock}
\end{frame}