Question

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

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

Answer 1

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

$={lim}_{x\to \frac{\pi }{2}}\frac{\tan \frac{\pi }{4}-\tan \frac{x}{2}}{1+\tan \frac{\pi }{4}\tan \frac{x}{2}}\times \frac{1-\sin x}{{(\pi -2x)}^3}$

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

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

$={lim}_{x\to \frac{\pi }{2}}\frac{\tan (\frac{\pi }{4}-\frac{x}{2})}{4.(\frac{\pi }{4}-\frac{x}{2})}.\frac{2{\sin }^2(\frac{\pi }{4}-\frac{x}{2})}{4^2{\left(\frac{\pi -2x}{4}\right)}^2}$

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