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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!

Evaluate the limit as x approaches 0 for the expression (x^2 * sin(x) * (1/2) * sin(x) * sqrt(x)) / x^3.

Explore the detailed computation of a trigonometric limit where sin(x) and x^2 are key components. This guide illustrates how to simplify complex limits using fundamental trigonometric limits and approximations. Suitable for university-level calculus courses.

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