Question

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

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

Answer 1

Answer: (D)

$\underset{x\to \frac{\pi }{2}}{\lim }\frac{\tan \left(\frac{\pi }{4}-\frac{x}{2}\right).\left(1-\sin x\right)}{{\left(\pi -2x\right)}^3}$
Let $x=\frac{\pi }{2}+y;y\to 0$
$=\underset{y\to 0}{\lim }\frac{\tan \left(-\frac{y}{2}\right).\left(1-\cos y\right)}{{\left(-2y\right)}^3}=\underset{y\to 0}{\lim }\frac{-\tan \frac{y}{2}2{\sin }^2\frac{y}{2}}{\left(-8\right).\frac{y^3}{8}.8}$
$=\underset{y\to 0}{\lim }\frac{1}{32}\frac{\tan \frac{y}{2}}{\left(\frac{y}{2}\right)}.{\left[\frac{\sin y/2}{y/2}\right]}^2=\frac{1}{32}$