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

To solve the given limit, we need to evaluate:

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

This expression is in an indeterminate form (0/0) as \(x \to \frac{\pi}{2}\). Therefore, we will use L'Hôpital's Rule, which is appropriate for handling such indeterminate forms.

Steps:

1. Substitute \(x = \frac{\pi}{2}-h\) and let \(h \to 0\) as \(x \to \frac{\pi}{2}\). Therefore:

• \((1 - \tan \frac{x}{2}) = 1 - \tan(\frac{\pi}{4} - \frac{h}{2}) \approx 1 - \left(1 - \frac{h}{2}\right) = \frac{h}{2}\)
• \((1 - \sin x) = 1 - \sin(\frac{\pi}{2} - h) = 1 - \cos(h) \approx \frac{h^2}{2}\)
• \((1 + \tan \frac{x}{2}) = 1 + \tan(\frac{\pi}{4} - \frac{h}{2}) \approx 1 + \left(1 - \frac{h}{2}\right) = 2 - \frac{h}{2}\approx 2\)
• \((\pi - 2x)^3 = (2h)^3 = 8h^3\)

2. Substitute these approximations into the original limit:

\[\lim_{h \to 0} \frac{\frac{h}{2} \cdot \frac{h^2}{2}}{2 \cdot 8h^3} = \lim_{h \to 0} \frac{\frac{h^3}{4}}{16h^3} = \lim_{h \to 0} \frac{1}{64} = \frac{1}{32}\]

3. Thus, the limit evaluates to:

\[\frac{1}{32}\]

The correct answer is $\frac{1}{32}$.