Evaluate the following limits:
`lim_(xrarr0)(sin5x)/(sin8x)`

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{\sin 5x}{\sin 8x} \), we can use the property that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

### Step-by-Step Solution:

1. **Rewrite the limit**:
We can express the limit in a form that allows us to apply the known limit property. We can rewrite the expression as follows:
\[ \lim_{x \to 0} \frac{\sin 5x}{\sin 8x} = \lim_{x \to 0} \frac{\sin 5x}{5x} \cdot \frac{5x}{8x} \cdot \frac{8x}{\sin 8x} \]

2. **Separate the limit**:
Now we can separate the limit into three parts:
\[ = \lim_{x \to 0} \frac{\sin 5x}{5x} \cdot \lim_{x \to 0} \frac{5x}{8x} \cdot \lim_{x \to 0} \frac{8x}{\sin 8x} \]

3. **Evaluate each limit**:
- The first limit \( \lim_{x \to 0} \frac{\sin 5x}{5x} = 1 \) (using the property of sine).
- The second limit \( \lim_{x \to 0} \frac{5x}{8x} = \frac{5}{8} \).
- The third limit \( \lim_{x \to 0} \frac{8x}{\sin 8x} = 1 \) (again using the property of sine).

4. **Combine the results**:
Now we can combine the results of the limits:
\[ = 1 \cdot \frac{5}{8} \cdot 1 = \frac{5}{8} \]

5. **Final Answer**:
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{\sin 5x}{\sin 8x} = \frac{5}{8} \]