Question

How do you find the limit $\frac{\cos x}{\frac{\pi }{2}-x}$ as $x\to \frac{\pi }{2}$

Answer 1

Using L-Hospital's rue we compute the derivative of the numerator and the denominator
$\frac{d\left(\cos x\right)}{dx}=-\sin x$
$\frac{d(\frac{\pi }{2}-x)}{dx}=-1$
Assemble the new expression and evaluate at the limit:
${lim}_{x\to \frac{\pi }{2}}\frac{-\sin x}{-1}=1$
According to the rule the original limit goes to the same value
${lim}_{x\to \frac{\pi }{2}}\frac{\cos x}{\frac{\pi }{2}-x}=1$