Question

The value of \( \lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} \) is equal to:

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
This limit results in an indeterminate form of \( \frac{0}{0} \). To solve it efficiently, we can use Taylor series expansion for \( \sin x \), which states \( \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots \). This avoids multiple applications of L'Hôpital's Rule.
Step 2: Key Formula or Approach:
1. \( \sin x \approx x - \frac{x^3}{6} \) for \( x \to 0 \).
2. \( \sin^2 x \approx (x - \frac{x^3}{6})^2 = x^2 - \frac{x^4}{3} + \frac{x^6}{36} \).
Step 3: Detailed Explanation:
1. Expand the numerator: \[ x^2 \sin^2 x \approx x^2 (x^2 - \frac{x^4}{3}) = x^4 - \frac{x^6}{3} \] 2. Expand the denominator: \[ x^2 - \sin^2 x \approx x^2 - (x^2 - \frac{x^4}{3} + \frac{x^6}{36}) = \frac{x^4}{3} - \frac{x^6}{36} \] 3. Substitute these into the limit expression: \[ \lim_{x \to 0} \frac{x^4 - \frac{x^6}{3}}{\frac{x^4}{3} - \frac{x^6}{36}} \] 4. Divide the numerator and denominator by \( x^4 \): \[ \lim_{x \to 0} \frac{1 - \frac{x^2}{3}}{\frac{1}{3} - \frac{x^2}{36}} \] 5. As \( x \to 0 \), the \( x^2 \) terms vanish: \[ \frac{1}{\frac{1}{3}} = 3 \]
Step 4: Final Answer:
The value of the limit is 3.