(B) We know that $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x} = 1$. Given limit is $\lim _{x \ri

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(B) We know that $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x} = 1$. Given limit is $\lim _{x \rightarrow 0} \frac{\sin x \sin ^{-1} x}{x^2}$. This can be rewritten as $\lim _{x \rightarrow 0} \left(\frac{\sin x}{x}\right) \left(\frac{\sin ^{-1} x}{x}\right)$. Applying the limit properties,we get $1 \times 1 = 1$.