\begin{frame} \frametitle{1st Midterm Exam - Review} \begin{exampleblock}{} Prove $\lim_{x \to 0} g(x) = 0$ where $g(x) = x^{12} \cdot \cos \left( \frac{1+e^{50x}}{13.2 x^2} \right)$. \pause\medskip We know that the range of $\cos$ is $[-1,1]$. Thus \begin{talign} -x^{12} \le g(x) \le x^{12} \end{talign} \pause Moreover $\lim_{x\to 0} -x^{12} = 0 = \lim_{x\to 0} x^{12}$. \pause\medskip Thus we can apply the Squeeze Theorem with \begin{itemize} \item lower bound $-x^{12}$ (i.e. $\le g(x)$), and \item upper bound $x^{12}$ (i.e. $\ge g(x)$) \end{itemize} and it follows that \begin{talign} \lim_{x\to 0} g(x) = 0 \end{talign} \end{exampleblock} \end{frame}