To solve the limit $\lim_{x \to 0} \frac{x^2}{\sin(2x^2)}$, we need to analyze the behavior of the function as $x$ approaches 0.

First, we recall the limit property: $\lim_{u \to 0} \frac{\sin u}{u} = 1$ We can rewrite our given limit in a similar form by making a substitution $u = 2x^2$. As $x \to 0$, $u \to 0$ as well.

Therefore, $\lim_{x \to 0} \frac{x^2}{\sin(2x^2)} = \lim_{x \to 0} \frac{x^2}{\sin u} \quad \text{where} \quad u = 2x^2$

Rewriting the limit using this substitution gives: $\lim_{u \to 0} \frac{\frac{u}{2}}{\sin u} = \frac{1}{2} \lim_{u \to 0} \frac{u}{\sin u}$

Since we know that: $\lim_{u \to 0} \frac{u}{\sin u} = 1$

Thus, $\lim_{x \to 0} \frac{x^2}{\sin(2x^2)} = \frac{1}{2} \cdot 1 = \frac{1}{2}$

Therefore, the correct answer is $\boxed{\frac{1}{2}}$.

Would you like more details or have any questions about this solution?

Here are 5 more questions you might consider asking:

1. How do you handle limits involving indeterminate forms?
2. Can you explain the Squeeze Theorem for limits?
3. How do you find the limit of a trigonometric function as it approaches infinity?
4. What are the different techniques for evaluating limits in calculus?
5. Can you provide examples of limits that require L'Hôpital's rule?

Tip: When dealing with trigonometric limits, try to express the problem in a form where you can use standard limit properties, such as $\lim_{u \to 0} \frac{\sin u}{u} = 1$.