The given limit expression is:

$\lim_{x \to 0} \frac{x^2 \sin(x) \left(\frac{1}{2} \right) \sin(x) \sqrt{x}}{x^3}$

Let's simplify the expression step by step:

1. The expression in the numerator is: $x^2 \sin(x) \cdot \frac{1}{2} \cdot \sin(x) \sqrt{x}$ This can be rewritten as: $\frac{1}{2} x^2 \sin(x) \sin(x) \sqrt{x}$ Which simplifies to: $\frac{1}{2} x^2 \sin^2(x) \sqrt{x}$

2. The full expression for the limit becomes: $\lim_{x \to 0} \frac{\frac{1}{2} x^2 \sin^2(x) \sqrt{x}}{x^3}$

3. Simplify the fraction: $\frac{\frac{1}{2} x^2 \sin^2(x) \sqrt{x}}{x^3} = \frac{1}{2} \cdot \frac{\sin^2(x) \sqrt{x}}{x}$

4. Now, as $x \to 0$, we know that: $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ So, using this fact, we can approximate $\sin^2(x)$ near zero: $\frac{\sin^2(x)}{x^2} \approx 1$

5. Therefore, the limit becomes: $\lim_{x \to 0} \frac{1}{2} \cdot \frac{\sin^2(x) \sqrt{x}}{x} = \frac{1}{2} \cdot 0 = 0$

Thus, the value of the limit is:

$\boxed{0}$

Would you like more details on any step? Here are some related questions:

1. How do we apply limits involving trigonometric functions like $\sin(x)$?
2. Why does $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ hold?
3. What are other methods to simplify limits involving $\sin(x)$?
4. How does L'Hopital's Rule apply to this kind of limit problem?
5. Could Taylor series expansions help simplify this expression?

Tip: When working with limits involving trigonometric functions, knowing common limits like $\lim_{x \to 0} \frac{\sin(x)}{x}$ can save time!