Question

$\underset{x\to \frac{\pi }{2}}{lim}\frac{(1-\tan \frac{x}{2})(1-\sin x)}{(1+\tan \frac{x}{2}){(\pi -2x)}^3}$ is

• A. $0$
• B. $\frac{1}{32}$
• C. $\infty$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $\frac{1}{32}$

$\because \tan (\frac{\pi }{4}-\frac{x}{2})=\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}$
$\therefore \underset{x\to \frac{\pi }{2}}{lim}\frac{\tan (\frac{\pi }{4}-\frac{x}{2})(1-\sin x)}{4\left(\frac{\pi -2x}{4}\right){(\pi -2x)}^2}$

$\therefore \underset{x\to \frac{\pi }{2}}{lim}\frac{\tan (\frac{\pi }{2}-\frac{x}{2})(1-\cos (\frac{\pi }{2}-x))}{4.(\frac{\pi }{4}-\frac{x}{2}){(\pi -2x)}^2}$

$\therefore \underset{x\to \frac{\pi }{2}}{lim}\frac{\tan (\frac{\pi }{2}-\frac{x}{2})\cdot 2\cdot {\sin }^2(\frac{\pi }{4}-\frac{x}{2})}{4\cdot (\frac{\pi }{4}-\frac{x}{2})\cdot 4^2\cdot {\left(\frac{\pi -2x}{4}\right)}^2}$

$=\frac{1}{4}\times \frac{2}{16}=\frac{1}{32}$