The value of lim_{x to 0} left( frac{x^2 sin^ | vedclass

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(B) We use the Taylor series expansion for $\sin x$ near $x = 0$: $\sin x = x - \frac{x^3}{6} + O(x^5)$.Then, $\sin^2 x = (x - \frac{x^3}{6} + O(x^5))

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$3$

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(B) We use the Taylor series expansion for $\sin x$ near $x = 0$: $\sin x = x - \frac{x^3}{6} + O(x^5)$.Then, $\sin^2 x = (x - \frac{x^3}{6} + O(x^5))^2 = x^2 - 2(x)(\frac{x^3}{6}) + O(x^6) = x^2 - \frac{x^4}{3} + O(x^6)$.Substituting this into the denominator: $x^2 - \sin^2 x = x^2 - (x^2 - \frac{x^4}{3} + O(x^6)) = \frac{x^4}{3} + O(x^6)$.Now, substitute this into the limit expression: $\lim_{x 	o 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} = \lim_{x 	o 0} \frac{x^2 (x^2 + O(x^4))}{\frac{x^4}{3} + O(x^6)}$.$= \lim_{x 	o 0} \frac{x^4 + O(x^6)}{\frac{x^4}{3} + O(x^6)} = \frac{1}{1/3} = 3$.

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

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