\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}