Question

Evaluate the limit \( \lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} \)

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The limit is in the indeterminate form \( \frac{0}{0} \). To evaluate it, we can use the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) or expand the functions using Taylor series for higher precision in the denominator.

Step 2: Key Formula or Approach:
1. Divide numerator and denominator by \( x^4 \). 2. Use Taylor expansion: \( \sin x = x - \frac{x^3}{6} + O(x^5) \).

Step 3: Detailed Explanation:
1. Rewrite the expression: \[ L = \lim_{x \to 0} \frac{x^2 \sin^2 x}{(x - \sin x)(x + \sin x)} \] 2. Divide numerator and denominator by \( x^4 \): \[ L = \lim_{x \to 0} \frac{(\frac{\sin x}{x})^2}{\frac{x^2 - \sin^2 x}{x^4}} \] 3. Focus on the denominator: \( x^2 - \sin^2 x = (x - \sin x)(x + \sin x) \). - \( x - \sin x \approx x - (x - \frac{x^3}{6}) = \frac{x^3}{6} \) - \( x + \sin x \approx x + x = 2x \) 4. Denominator \( \approx \frac{x^3}{6} \cdot 2x = \frac{x^4}{3} \). 5. Substitute back: \[ L = \frac{1^2}{1/3} = 3 \]

Step 4: Final Answer:
The value of the limit is 3.