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 equal to

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

Answer 1

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

Put $x=\frac{\pi }{2}-h$ as $x\to \frac{\pi }{2},h\to 0$

$\therefore$ Given, limit $=\underset{h\to 0}{lim}\frac{1-\tan (\frac{\pi }{4}+\frac{h}{2})(1-\cos h)}{1+\tan (\frac{\pi }{4}+\frac{h}{2})(2h{)}^3}$

$=\underset{h\to 0}{lim}\tan \left(\frac{h}{2}\right)\underset{h\to 0}{lim}\frac{2\cdot {\sin }^2\left(\frac{h}{2}\right)}{8h^3}$

$=\underset{h\to 0}{lim}\frac{1}{4}\left(\frac{\tan \frac{h}{2}}{2\times \frac{h}{2}}\right)\times \underset{h\to 0}{lim}{\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2\times \frac{1}{4}$

$=\frac{1}{32}$.