Question

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

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $0$
• D. $\frac{1}{32}$

Answer 1

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

Given,

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

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

$=\underset{x\to \frac{\pi }{2}}{lim}\tan (\frac{\pi }{4}-\frac{x}{2})\frac{2{\sin }^2(\frac{\pi }{4}-\frac{x}{2})}{{\left[4(\frac{\pi }{4}-\frac{x}{2})\right]}^3}$

$\underset{x\to 0}{lim}\frac{\sin x}{x}=1,\underset{x\to 0}{lim}\frac{\tan x}{x}=1$

$=\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{32}\frac{\tan (\frac{\pi }{4}-\frac{x}{2})}{\frac{\pi }{4}-\frac{x}{2}}\frac{{\sin }^2(\frac{\pi }{4}-\frac{x}{2})}{{(\frac{\pi }{4}-\frac{x}{2})}^2}$

$=\frac{1}{32}\times 1\times 1^2=\frac{1}{32}$