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Prove It Are you allowed to use L'Hospital's Rule?
If so, it's very easy...
$\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1}{\cos{x}} = \lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}(1)}{\frac{d}{dx}(\cos{x})}$
$\displaystyle = \lim_{x \to \frac{\pi}{2}} \frac{0}{-\sin{x}}$
$\displaystyle = \frac{0}{-\sin{\frac{\pi}{2}}}$
$\displaystyle = \frac{0}{-1}$
$\displaystyle = 0$.
Without L'Hospital, not so easy...
All I can think of is the Sandwich Theorem, but I can't think of what to sandwich it inside.
I know that $\displaystyle \frac{\sin{x}}{x}\leq \frac{1}{\cos{x}}$.
We just need to find something that $\displaystyle \frac{1}{\cos{x}}$ is less than or equal to...