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.