The limit you provided is:

$\lim_{x \to 0} \frac{(\sin 3x)^2}{x^2 \cos x}$

Steps to Solve:

1. Expression Simplification: Start by expressing $\sin 3x$ in a more convenient form near $x = 0$.

   $\sin 3x \approx 3x \quad \text{as} \quad x \to 0$

   So,

   $(\sin 3x)^2 \approx (3x)^2 = 9x^2$

2. Substitute in the Limit:

   Substitute this approximation into the original limit expression:

   $\lim_{x \to 0} \frac{9x^2}{x^2 \cos x}$

3. Simplification:

   Simplify the expression by canceling $x^2$ in the numerator and denominator:

   $\lim_{x \to 0} \frac{9}{\cos x}$

4. Evaluate the Limit:

   As $x \to 0$, $\cos x \to 1$. Therefore:

   $\lim_{x \to 0} \frac{9}{\cos x} = \frac{9}{1} = 9$

Final Answer:

$\lim_{x \to 0} \frac{(\sin 3x)^2}{x^2 \cos x} = 9$

Would you like more details or have any other questions?

Here are 5 related questions to further your understanding:

1. How would the limit change if the expression were $\lim_{x \to 0} \frac{(\sin 5x)^2}{x^2 \cos x}$?
2. What happens if $\cos x$ were replaced with $\cos(2x)$ in the denominator?
3. How does L'Hôpital's Rule apply to solving limits involving indeterminate forms like $\frac{0}{0}$?
4. What is the approximation for $\sin x$ near $x = 0$, and why is it useful?
5. Can you apply the small-angle approximation to $\cos x$ to verify the solution?

Tip: Small-angle approximations are useful in simplifying trigonometric functions near 0, especially in limit problems.