\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}