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\problemgrouptitle{Squeeze Theorem}

\item If $2\sin x\le f(x)\le 2$ evaluate $\displaystyle\lim_{x\to\pi/2}f(x)$\,.
  \answer{Note $\displaystyle \lim_{x\to\pi/2}2\sin x=2\sin(\pi/2)=2(1)=2$, $\displaystyle\lim_{x\to\pi/2}2=2$ and apply the Squeeze Theorem to conclude $\displaystyle\lim_{x\to\pi/2}f(x)=2$.}

\item Evaluate $\displaystyle \lim_{x\to0}x^2\cos\frac{1}{x^5}$\,.
  \answer{Since $-1\le\cos\theta\le1$ it follows that $\displaystyle-1\le\cos\frac{1}{x^5}\le1$ and hence (multiplying the inequalities by~$x^2$) that $\displaystyle-x^2\le x^2\cos\frac{1}{x^5}\le x^2$.  Observing that $\displaystyle \lim_{x\to0}-x^2=0$ and $\displaystyle \lim_{x\to0}x^2=0$ and applying the Squeeze Theorem implies $\displaystyle \lim_{x\to0}x^2\cos\frac{1}{x^5}=0$.}

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