Evaluate Limit of Trigonometric Function as x Approaches Zero

To evaluate the limit $L = \lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x}$

We can use the Taylor series expansion for $\sin x$ near $x = 0$: $\sin x = x - \frac{x^3}{6} + O(x^5)$

Squaring both sides gives: $\sin^2 x = \left( x - \frac{x^3}{6} + O(x^5) \right)^2 = x^2 - \frac{x^4}{3} + O(x^6)$

Now, substitute this expansion into the numerator and denominator of the limit expression:

1. Numerator: $x^2 \sin^2 x = x^2 \left( x^2 - \frac{x^4}{3} + O(x^6) \right) = x^4 - \frac{x^6}{3} + O(x^8)$

2. Denominator: $x^2 - \sin^2 x = x^2 - \left( x^2 - \frac{x^4}{3} + O(x^6) \right) = \frac{x^4}{3} - O(x^6)$

Substitute these back into the limit: $L = \lim_{x \to 0} \frac{x^4 - \frac{x^6}{3} + O(x^8)}{\frac{x^4}{3} - O(x^6)}$

Divide the numerator and the denominator by $x^4$: $L = \lim_{x \to 0} \frac{1 - \frac{x^2}{3} + O(x^4)}{\frac{1}{3} - O(x^2)}$

Taking the limit as $x \to 0$: $L = \frac{1}{\frac{1}{3}} = 3$

Thus, the value of the limit is $3$, which corresponds to Option B.