We need to find the limit of the expression $\frac{x \sin(3x)}{\sin^2(5x)}$ as $x$ approaches 0. That is, we need to evaluate $A = \lim_{x \to 0} \frac{x \sin(3x)}{\sin^2(5x)}$.

1. Problem Description

We need to find the limit of the expression $\frac{x \sin(3x)}{\sin^2(5x)}$ as $x$ approaches

0. That is, we need to evaluate $A = \lim_{x \to 0} \frac{x \sin(3x)}{\sin^2(5x)}$.

2. Solution Steps

We can rewrite the expression as:

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

We know that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, which also implies $\lim_{x \to 0} \frac{x}{\sin(x)} = 1$.

Then, $\lim_{x \to 0} \frac{\sin(ax)}{ax} = 1$ implies $\lim_{x \to 0} \frac{ax}{\sin(ax)} = 1$.

We can rewrite the given limit as:

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

$A = \lim_{x \to 0} \frac{x}{\sin(5x)} \cdot \lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)}$

Let's evaluate the first limit:

$\lim_{x \to 0} \frac{x}{\sin(5x)} = \lim_{x \to 0} \frac{5x}{5\sin(5x)} = \frac{1}{5} \lim_{x \to 0} \frac{5x}{\sin(5x)} = \frac{1}{5} \cdot 1 = \frac{1}{5}$

Now, let's evaluate the second limit:

$\lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{5x}{\sin(5x)} \cdot \frac{3x}{5x} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \lim_{x \to 0} \frac{5x}{\sin(5x)} \cdot \lim_{x \to 0} \frac{3x}{5x} = 1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5}$

Therefore,

$A = \frac{1}{5} \cdot \frac{3}{5} = \frac{3}{25}$

3. Final Answer

3/25