Evaluate the following limits :
`lim_(x to 0) (sin 3x)/(sin 2x)`

Answer

To evaluate the limit \( L = \lim_{x \to 0} \frac{\sin 3x}{\sin 2x} \), we can use the standard limit property that states \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

### Step-by-Step Solution:

1. **Rewrite the limit**:
\[ L = \lim_{x \to 0} \frac{\sin 3x}{\sin 2x} \]

2. **Multiply and divide by \( 3x \) and \( 2x \)**:
\[ L = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \frac{3x}{\sin 2x} \cdot \frac{2x}{2x} \]
This can be rearranged to:
\[ L = \lim_{x \to 0} \left( \frac{\sin 3x}{3x} \cdot \frac{3}{2} \cdot \frac{2x}{\sin 2x} \right) \]

3. **Separate the limits**:
\[ L = \frac{3}{2} \cdot \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \lim_{x \to 0} \frac{2x}{\sin 2x} \]

4. **Evaluate the limits**:
- From the standard limit property, we know:
\[ \lim_{x \to 0} \frac{\sin 3x}{3x} = 1 \]
- Also, using the same property:
\[ \lim_{x \to 0} \frac{2x}{\sin 2x} = 1 \]

5. **Combine the results**:
\[ L = \frac{3}{2} \cdot 1 \cdot 1 = \frac{3}{2} \]

### Final Answer:
\[ L = \frac{3}{2} \]