Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2}. \]

• A. \( -\pi \)
• B. \( \pi \)
• C. \( \frac{\pi}{2} \)
• D. \( 1 \)

Hint:

For small \( x \), use standard approximations \( \cos x \approx 1 - x^2/2 \) and \( \sin x \approx x \) to simplify limits.

Answer 1

The Correct Option is B

Step 1: Expanding \( \cos^2 x \) for small \( x \) Using \( \cos x \approx 1 - \frac{x^2}{2} \), we approximate: \[ \cos^2 x \approx 1 - x^2. \] Step 2: Expanding \( \sin(\pi \cos^2 x) \) \[ \sin(\pi \cos^2 x) = \sin(\pi (1 - x^2)) \approx \sin(\pi - \pi x^2). \] Using \( \sin(\pi - \theta) = \sin\theta \), we get: \[ \sin(\pi - \pi x^2) \approx \sin(\pi x^2) \approx \pi x^2. \] Step 3: Evaluating the limit \[ \lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2} = \lim_{x \to 0} \frac{\pi x^2}{x^2} = \pi. \]