\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}