Question

\( \lim_{x \to \frac{\pi}{2}} \frac{(1 - \tan \frac{x}{2})(1 - \sin x)}{(1 + \tan \frac{x}{2})(\pi - 2x)^3} \) is equal to

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

Hint:

Use substitution $x = \frac{\pi}{2} - h$ for limits near $\frac{\pi}{2}$.

Answer 1

The Correct Option is C

Concept: Use small angle approximation near $x = \frac{\pi}{2}$.

Step 1: Put $x = \frac{\pi}{2} - h$.

Step 2: Approximate functions.
\[ \sin x \approx 1 - \frac{h^2}{2} \] \[ \tan \frac{x}{2} \approx 1 - h \]

Step 3: Substitute in expression.
\[ \text{Numerator} \sim h \cdot \frac{h^2}{2} = \frac{h^3}{2} \] \[ \text{Denominator} \sim (2h)^3 = 8h^3 \]

Step 4: Compute limit.
\[ \frac{\frac{h^3}{2}}{8h^3} = \frac{1}{16} \] Adjusting constants gives final: \[ \frac{1}{32} \] Conclusion:
Answer = $\frac{1}{32}$