\begin{frame}
  \frametitle{Properties of Limits}
  
  \begin{exampleblock}{}
    Show that $\lim_{x \to 0} g(x) = 0$ where $g(x) = x^2 \cdot \sin \frac{1}{x}$.
    \pause\medskip
    
    The application of limit laws 
    \begin{talign}
      \lim_{x \to 0} ( x^2 \cdot \sin \frac{1}{x} ) = (\lim_{x \to 0} x^2) \cdot (\alert<3->{\lim_{x \to 0} \sin \frac{1}{x}})
    \end{talign}\pause
    does not work since $\lim_{x \to 0} \sin \frac{1}{x}$ does not exist.
    \pause\bigskip

    To apply the squeeze theorem we need:
    \begin{itemize}
    \pause
      \item a function $f$ smaller ($\le$) than $g$, and
    \pause
      \item a function $h$ bigger ($\ge$) than $g$
    \end{itemize}
    \pause
    for which $\lim_{x\to 0} f(x) = 0$ and $\lim_{x\to 0} h(x) = 0$.
    \pause\bigskip
    
    We know that $-1 \le \sin \frac{1}{x} \le 1$ \pause
    and hence% (we multiply everything with the non-negative $x^2$)
    \begin{talign}
      -x^2 \quad\le\quad x^2 \cdot \sin \frac{1}{x} \quad\le\quad x^2
    \end{talign}
    
  \end{exampleblock}
  \vspace{10cm}
\end{frame}