Question

The limit: \[ \lim_{x \to 0} \frac{\sin \left( \pi \sin^2 x \right)}{x^2} \] is equal to:

• A. \( 2\pi \)
• B. \( \pi^2 \)
• C. \( 2\pi^2 \)
• D. \( \frac{\pi}{2} \)
• E. \( \pi \)

Hint:

For small values of \( x \), use approximations like \( \sin x \approx x \) to simplify trigonometric expressions. These approximations help in evaluating limits.

Answer 1

Answer: (E)

We are given the following limit: \[ \lim_{x \to 0} \frac{\sin \left( \pi \sin^2 x \right)}{x^2}. \] We need to evaluate this limit as \( x \to 0 \).
Step 1: Use the approximation for small \( x \), which is \( \sin x \approx x \) when \( x \) is close to 0. Therefore, for small \( x \), we have: \[ \sin^2 x \approx x^2. \] Thus, \( \pi \sin^2 x \approx \pi x^2 \). 
Step 2: Now, substitute \( \pi x^2 \) into the sine function: \[ \sin \left( \pi \sin^2 x \right) \approx \sin \left( \pi x^2 \right). \] For small \( x \), we can use the approximation \( \sin y \approx y \) when \( y \) is small. Therefore: \[ \sin \left( \pi x^2 \right) \approx \pi x^2. \] Step 3: Substituting this approximation into the original limit expression, we get: \[ \lim_{x \to 0} \frac{\sin \left( \pi x^2 \right)}{x^2} \approx \lim_{x \to 0} \frac{\pi x^2}{x^2} = \pi. \] Thus, the correct answer is option (E).