Question

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

• A. 1/8
• B. 0
• C. 1/32
• D. $\infty$

Answer 1

Answer: (C)

Put $x=\frac{\pi }{2}-h$ as $x\to \frac{\pi }{2},h\to 0$
$\therefore$ Given limit
$=\underset{h\to 0}{\lim }\frac{1-\tan \left(\frac{\pi }{4}-\frac{h}{2}\right)}{1+\tan \left(\frac{\pi }{4}-\frac{h}{2}\right)}.\frac{\left(1-\cos h\right)}{{\left(2h\right)}^3}$
$=\underset{h\to 0}{\lim }\tan \frac{h}{2}\frac{2{\sin }^2\frac{h}{2}}{8h^3}$
$=\underset{h\to 0}{\lim }\frac{1}{4}.\frac{\tan \frac{h}{2}}{\frac{h}{2}\times 2}{\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2\times \frac{1}{4}$
$=\underset{h\to 0}{\lim }\frac{1}{32}.\left(\frac{\tan \frac{h}{2}}{\frac{h}{2}\times 2}\right){\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2=\frac{1}{32}$