Question

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

• 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 \to0$
$\therefore $ Given limit
$ = \displaystyle\lim_{h\to0} \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}} $
$ = \displaystyle\lim_{h\to0} \tan \frac{h }{2} \frac{2\sin^{2} \frac{h}{2}}{8h^{3}} $
$=\displaystyle\lim_{h \to0} \frac{1}{4} . \frac{\tan \frac{h}{2}}{\frac{h}{2} \times2} \left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2} \times\frac{1}{4} $
$ = \displaystyle\lim_{h\to0} \frac{1}{32} . \left(\frac{\tan \frac{h}{2}}{\frac{h}{2} \times2}\right) \left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2} = \frac{1}{32} $