Question

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

• A. 1
• B. 0
• C. \infty
• D. Does not exist}

Hint:

When dealing with small-angle approximations, replace trigonometric functions like \( \sin x \) with their approximations to simplify the limit.

Answer 1

The Correct Option is B

We start by recognizing that as \( x \to 0 \), \( \sin x \approx x \). Therefore: \[ \sin^2 x \approx x^2 \] Thus, \( \pi \sin^2 x \approx \pi x^2 \), and the limit becomes: \[ \lim_{x \to 0} \frac{\sin(\pi x^2)}{x^2} \] Since \( \sin(\pi x^2) \approx \pi x^2 \) for small \( x \), the limit becomes: \[ \lim_{x \to 0} \frac{\pi x^2}{x^2} = \pi \] Thus, the limit evaluates to 0.