Question

Evaluate $\underset{x\to \pi /2}{lim}\frac{1+\cos 2x}{(\pi -2x{)}^2}$.

Answer 1

$\underset{x\to \pi /2}{lim}\frac{1+\cos 2x}{(\pi -2x{)}^2}$

(Putting $x=\frac{\pi }{2}+h$, when $x\to \frac{\pi }{2},h\to 0)$

$={lim}_{h\to 0}\frac{1+\cos 2(\frac{\pi }{2}+h)}{{\{\pi -2(\frac{\pi }{2}+h)\}}^2}$
$=\underset{h\to 0}{lim}\frac{1+\cos (\pi +2h)}{\left(4h^2\right)}$

$=\underset{h\to 0}{lim}\frac{1-\cos 2h}{4h^2}$

$=\underset{h\to 0}{lim}\frac{2{\sin }^{2}h}{4h^2}$

$=\frac{1}{2}\underset{h\to 0}{lim}{\left(\frac{\sin h}{h}\right)}^2$

$=\frac{1}{2}\times 1=\frac{1}{2}$.