Question

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1-\sin x}{(\pi -2x{)}^2}$=

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{6}$
• D. $\frac{1}{8}$

Answer 1

The correct option is D $\frac{1}{8}$

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1-\sin x}{(\pi -2x{)}^2}$
$=\underset{h\to 0}{lim}\frac{1-\sin (\frac{\pi }{2}-h)}{(\pi -2(\frac{\pi }{2}-h){)}^2}$......(x-->h-pi/2)
$=\underset{h\to 0}{lim}\frac{1-\cos h}{4h^2}$

$=\frac{1}{8}$