Given, `lim_(xto0) (sin3x)/(sin7x)`.

Options

A. `3/5`
B. `3/7`
C. `1/5`
D. `7/3`

Explanation

To evaluate the limit \( \lim_{x \to 0} \frac{\sin 3x}{\sin 7x} \), we can use the standard limit property that states:

\[ \lim_{x \to 0} \frac{\sin kx}{kx} = 1 \]

for any constant \( k \).

### Step-by-Step Solution:

1. **Rewrite the limit**: We can express the limit in a form that allows us to apply the standard limit property.
\[ \lim_{x \to 0} \frac{\sin 3x}{\sin 7x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \frac{3x}{7x} \cdot \frac{7x}{\sin 7x} \]

2. **Separate the limit**: Now we can separate the limit into three parts:
\[ = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \lim_{x \to 0} \frac{3x}{7x} \cdot \lim_{x \to 0} \frac{7x}{\sin 7x} \]

3. **Evaluate each limit**:
- For the first limit:
\[ \lim_{x \to 0} \frac{\sin 3x}{3x} = 1 \]
- For the second limit:
\[ \lim_{x \to 0} \frac{3x}{7x} = \frac{3}{7} \]
- For the third limit:
\[ \lim_{x \to 0} \frac{7x}{\sin 7x} = 1 \]

4. **Combine the results**:
\[ \lim_{x \to 0} \frac{\sin 3x}{\sin 7x} = 1 \cdot \frac{3}{7} \cdot 1 = \frac{3}{7} \]

### Final Result:
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin 3x}{\sin 7x} = \frac{3}{7} \]