\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}