The problem you've presented involves finding the limit:

$\lim_{{x \to 0}} \frac{\sin(3x^2)\cos(5x)}{2x^2}$

To solve this limit, we will use the small-angle approximations for sine and cosine functions and the limit properties.

Step 1: Use small-angle approximations

For small values of $x$, $\sin(x) \approx x$ and $\cos(x) \approx 1$. Therefore, for small values of $3x^2$ and $5x$, $\sin(3x^2) \approx 3x^2 \quad \text{and} \quad \cos(5x) \approx 1$

Step 2: Substitute and simplify

Substituting these approximations into the limit gives: $\lim_{{x \to 0}} \frac{3x^2 \cdot 1}{2x^2} = \lim_{{x \to 0}} \frac{3x^2}{2x^2} = \lim_{{x \to 0}} \frac{3}{2} = \frac{3}{2}$

Hence, the limit is $\frac{3}{2}$.

Do you have any further questions on this, or is there another problem you need help with? Here are a few related questions you might find interesting:

1. How would the solution change if the cosine term was $\cos(5x^2)$ instead?
2. What happens to the limit if the denominator was $x^3$ instead of $2x^2$?
3. Can you apply L'Hopital's Rule to solve this limit? If yes, how?
4. How does the limit behave if $x$ approaches a different value, say $\pi$?
5. What role do the coefficients 3 and 5 play in the behavior of the limit?

Tip: Always check if a function behaves nicely around the limit point (like being continuous and differentiable) before applying L'Hopital’s rule, as it requires both the numerator and denominator to be differentiable.