Question

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

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

Hint:

For limits near \(0\), use the standard approximations: \[ \tan t\sim t \] and \[ 1-\cos t\sim \frac{t^2}{2}. \] Also, \[ \frac{1-\tan A}{1+\tan A} = \tan\left(\frac{\pi}{4}-A\right). \]

Answer 1

The Correct Option is A

Step 1: Substitute \(t = \pi/2 - x\) so \(t \to 0\).
Then \(x = \pi/2 - t\), \(\sin x = \cos t\), \(\pi - 2x = 2t\), and \(\tan(x/2) = \tan(\pi/4 - t/2)\).
First factor: \(\dfrac{1-\tan(x/2)}{1+\tan(x/2)} = \tan\!\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right) = \tan\!\left(\dfrac{t}{2}\right) \approx \dfrac{t}{2}\).

Step 2: Simplify the second factor.
\(1-\sin x = 1-\cos t \approx \dfrac{t^2}{2}\). \((\pi-2x)^3 = (2t)^3 = 8t^3\). Second factor \(\approx \dfrac{t^2/2}{8t^3} = \dfrac{1}{16t}\).

Step 3: Multiply and take the limit.
\(\dfrac{t}{2} \cdot \dfrac{1}{16t} = \dfrac{1}{32}\).
\[ \boxed{\dfrac{1}{32}} \]