$\underset{x→0}{\lim}\frac{x(1-\sqrt{1-x^2})}{\sqrt{1-x}.(sin^{-1}x)^3}$ is equal to

$\frac{1}{2}$

$\underset{x→0}{\lim}\frac{x(1-\sqrt{1-x^2})}{\sqrt{1-x}.(sin^{-1}x)^3}=\underset{x→0}{\lim}\frac{x(1-\sqrt{1-x^2})}{x^3\sqrt{1-x}}.(\frac{x}{sin^{-1}x})^3$

$=\underset{x→0}{\lim}\frac{1}{\sqrt{1-x}}.\underset{x→0}{\lim}\frac{1-\sqrt{1-x^2}}{x^2}=1$.$\underset{x→0}{\lim}\frac{1-(1-\frac{x^2}{2}+\frac{x^4}{8})}{x^2}=\frac{1}{2}$.

Hence (A) is the correct answer.