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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.

Find the limit of (sin(3x^2)cos(5x))/(2x^2) as x approaches 0.

Learn how to find the limit of (sin(3x^2)cos(5x))/(2x^2) as x approaches 0 using small-angle approximations for sine and cosine functions. This problem is suitable for high school and college students studying calculus.

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