Q126 / 225 · JEE Main 2003 · Mathematics
$$
\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left[1-\tan \left(\frac{x}{2}\right)\right][1-\sin x]}{\left[1+\tan \left(\frac{x}{2}\right)\right]\left[\pi-2 x^3\right]}
$$
is
- A$\infty$
- B$\frac{1}{8}$
- C0
- Correct answerD$\frac{1}{32}$
Solution
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot(1-\sin x)}{(\pi-2 x)^3}$
Let $\mathrm{x}=\frac{\pi}{2}+\mathrm{y} ; \mathrm{y} \rightarrow 0$
$=\lim _{y \rightarrow 0} \frac{-\tan \left(-\frac{y}{2}\right) \cdot(1-\cos y)}{(-2 y)^3}=\lim _{y \rightarrow 0} \frac{-\tan \frac{y}{2} 2 \sin ^2 \frac{y}{2}}{(-8) \cdot \frac{y^3}{8} \cdot 8}$
$=\lim _{y \rightarrow 0} \frac{1}{32} \frac{\tan \frac{y}{2}}{\left(\frac{y}{2}\right)} \cdot\left[\frac{\sin y / 2}{y / 2}\right]^2=\frac{1}{32}$
Hence, the correct answer is option: 'D'.