Question

$\lim_{x \rightarrow \frac{\pi}{2}}\frac{\left\lbrack 1 - \tan\left( \frac{x}{2} \right) \right\rbrack\lbrack 1 - \sin x\rbrack}{\left\lbrack 1 + \tan\left( \frac{x}{2} \right) \right\rbrack\lbrack\pi - 2x\rbrack^{3}}$ is

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

Answer 1

Answer: (C)

$\frac{1}{32}$

Answer 2

$\lim_{x \rightarrow \frac{\pi}{2}}\frac{\tan\left( \frac{\pi}{4} - \frac{x}{2} \right)(1 - \sin x)}{(\pi - 2x)^{3}}$

Let $x = \frac{\pi}{2} + y,\text{then }y \rightarrow 0$ ⇒ $\lim_{y \rightarrow 0}\frac{\tan\left( \frac{- y}{2} \right)(1 - \cos y)}{( - 2y)^{3}}$

= $\lim_{y \rightarrow 0}\frac{- \tan\frac{y}{2}.2\sin^{2}\frac{y}{2}}{( - 8)y^{3}} = \lim_{y \rightarrow 0}\frac{1}{32}\frac{\tan\frac{y}{2}}{\left( \frac{y}{2} \right)}.\left\lbrack \frac{\sin\frac{y}{2}}{\frac{y}{2}} \right\rbrack^{2} = \frac{1}{32}$.