Question

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

• A. \( \frac{1}{32} \)
• B. 0
• C. \( \frac{1}{16} \)
• D. \( \frac{1}{8} \)

Hint:

Use identity \( \frac{1-\tan A}{1+\tan A} = \tan(\frac{\pi}{4}-A) \).
For limits approaching a value \(a\), substitute \(x=a-h\) or \(x=a+h\) where \(h \to 0\).
Use small angle approximations: \(\tan\theta \approx \theta\), \(1-\cos\theta \approx \theta^2/2\).

Answer 1

The Correct Option is A

Let the limit be L. First term: \( \frac{1-\tan(x/2)}{1+\tan(x/2)} = \tan(\pi/4 - x/2) \). Let \(x = \pi/2 - h\). As \(x \to \pi/2\), \(h \to 0\). Then \(x/2 = \pi/4 - h/2\). So, \(\pi/4 - x/2 = \pi/4 - (\pi/4 - h/2) = h/2\). First term becomes \(\tan(h/2)\). Second term: \(1-\sin x = 1-\sin(\pi/2-h) = 1-\cos h\). Denominator: \(\pi-2x = \pi-2(\pi/2-h) = \pi - (\pi-2h) = 2h\). So \((\pi-2x)^3 = (2h)^3 = 8h^3\). The expression for the limit is: \( L = \lim_{h \to 0} \tan(h/2) \cdot \frac{1-\cos h}{8h^3} \) Using approximations for small \(h\): \(\tan(h/2) \approx h/2\) and \(1-\cos h \approx h^2/2\). \( L = \lim_{h \to 0} \frac{(h/2) \cdot (h^2/2)}{8h^3} = \lim_{h \to 0} \frac{h^3/4}{8h^3} \) \( L = \frac{1/4}{8} = \frac{1}{32} \). This matches option (a). \[ \boxed{\frac{1}{32}} \]