To solve the limit \(\lim_{x \to 0} \frac{\sin 2x}{\sin 6x}\), we'll use the small angle approximation:

\[
\sin x \approx x \quad \text{as} \quad x \to 0
\]

### Step-by-Step Solution:

1. **Apply the small angle approximation**:
   \[
   \sin 2x \approx 2x \quad \text{and} \quad \sin 6x \approx 6x \quad \text{as} \quad x \to 0
   \]

2. **Substitute these approximations into the limit**:
   \[
   \lim_{x \to 0} \frac{\sin 2x}{\sin 6x} \approx \lim_{x \to 0} \frac{2x}{6x}
   \]

3. **Simplify the expression**:
   \[
   \lim_{x \to 0} \frac{2x}{6x} = \lim_{x \to 0} \frac{2}{6} = \frac{1}{3}
   \]

### Final Answer:
\[
\lim_{x \to 0} \frac{\sin 2x}{\sin 6x} = \frac{1}{3}
\]

Would you like more details or have any questions about this solution? 

Here are 5 related questions to deepen your understanding:

1. How does the small angle approximation work for other trigonometric functions like \(\cos x\) and \(\tan x\)?
2. Can you solve \(\lim_{x \to 0} \frac{\sin(3x)}{\sin(9x)}\) using a similar approach?
3. How does L'Hôpital's rule apply to this limit problem?
4. What would be the result of \(\lim_{x \to 0} \frac{\sin^2(2x)}{\sin^2(6x)}\)?
5. What is the general approach for evaluating limits involving trigonometric functions as \(x\) approaches 0?

**Tip:** When evaluating limits, particularly trigonometric ones, the small angle approximation is a powerful tool to simplify the problem.

Explore how to solve the limit \(\lim_{x \to 0} \frac{\sin 2x}{\sin 6x}\) using the small angle approximation method. This problem involves trigonometric functions and their approximations, suitable for college-level students and those interested in calculus.

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