$\lim _{x \rightarrow 0} \frac{4 \sin x \sin ^{2} \frac{x}{2}}{x^{3}}$

Step 1 ：Find the derivatives of the numerator and the denominator: \(\frac{d}{dx}(4 \sin x \sin^2(\frac{x}{2})) = 4\sin(\frac{x}{2})^2\cos(x) + 4\sin(\frac{x}{2})\sin(x)\cos(\frac{x}{2})\) and \(\frac{d}{dx}(x^3) = 3x^2\)

Step 2 ：Apply L'Hopital's rule and find the limit as x approaches 0: \(\lim_{x \to 0} \frac{4\sin(\frac{x}{2})^2\cos(x) + 4\sin(\frac{x}{2})\sin(x)\cos(\frac{x}{2})}{3x^2} = \boxed{1}\)