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: Simplify the first trigonometric factor.
We know that \[ \frac{1-\tan A}{1+\tan A}=\tan\left(\frac{\pi}{4}-A\right) \] Here, \[ A=\frac{x}{2} \] Therefore, \[ \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} = \tan\left(\frac{\pi}{4}-\frac{x}{2}\right) \] \[ = \tan\left(\frac{\pi-2x}{4}\right) \]

Step 2: Substitute a new variable.
Let \[ t=\frac{\pi}{2}-x \] Then, \[ x\to \frac{\pi}{2} \implies t\to 0 \] Also, \[ \pi-2x=2\left(\frac{\pi}{2}-x\right)=2t \] and \[ \sin x=\sin\left(\frac{\pi}{2}-t\right)=\cos t \] So, \[ 1-\sin x=1-\cos t \] Also, \[ \tan\left(\frac{\pi-2x}{4}\right) = \tan\left(\frac{2t}{4}\right) = \tan\frac{t}{2} \]

Step 3: Rewrite the limit in terms of \(t\).
The given limit becomes \[ \lim_{t\to 0} \tan\frac{t}{2}\cdot \frac{1-\cos t}{(2t)^3} \] \[ = \lim_{t\to 0} \tan\frac{t}{2}\cdot \frac{1-\cos t}{8t^3} \]

Step 4: Use standard limits.
As \(t\to 0\), \[ \tan\frac{t}{2}\sim \frac{t}{2} \] and \[ 1-\cos t\sim \frac{t^2}{2} \] Therefore, \[ \tan\frac{t}{2}\cdot (1-\cos t) \sim \frac{t}{2}\cdot \frac{t^2}{2} \] \[ = \frac{t^3}{4} \] So, the limit becomes \[ \lim_{t\to 0} \frac{\frac{t^3}{4}}{8t^3} \] \[ = \frac{1}{32} \]

Step 5: Final conclusion.
Hence, \[ \boxed{\frac{1}{32}} \]