\begin{frame}
  \frametitle{Properties of Limits}

  \begin{exampleblock}{}
    We have
    \begin{talign}
      -x^2 \quad\le\quad x^2 \cdot \sin \frac{1}{x} \quad\le\quad x^2 
    \end{talign}
    \pause
    We take $f(x) = -x^2$ and $h(x) = x^2$. 
    \pause
    \begin{center}
      \scalebox{.7}{
      \begin{tikzpicture}[default,yscale=.6,xscale=2]
        {\def\diaborderx{.3cm}
         \def\diabordery{1.2cm} 
        \diagram{-2}{2}{-3}{3}{1}}
        \diagramannotatez
        \draw[cgreen,ultra thick] plot[smooth,domain=-1.75:1.75,samples=20] function{x**2} node[above] {$x^2$};
        \draw[cblue,ultra thick] plot[smooth,domain=-1.75:1.75,samples=20] function{-x**2} node[below] {$-x^2$};
        \draw[cred,thick] plot[smooth,domain=-1.75:1.68,samples=400] function{x**2 * sin(180/pi/x)} node[right] {$x^2\cdot \sin \frac{1}{x}$};
      \end{tikzpicture}
      }
    \end{center}
    \pause
    We know $\lim_{x\to 0} x^2 = 0$ and $\lim_{x\to 0} -x^2 = 0$.
    \pause\medskip
    
    Hence by the squeeze theorem we get: $\lim_{x \to 0} x^2 \cdot \sin \frac{1}{x} = 0$.
  \end{exampleblock}

\end{frame}