To solve the limit

$\lim_{{x \to 0}} \frac{\sin(2x)}{\sin(6x)}$

we can make use of the standard limit result:

$\lim_{{x \to 0}} \frac{\sin(kx)}{kx} = 1$

for any constant $k$.

Step 1: Rewrite the expression

We can rewrite the limit expression by factoring out the constants $2$ and $6$ in the sine functions:

$\lim_{{x \to 0}} \frac{\sin(2x)}{\sin(6x)} = \lim_{{x \to 0}} \frac{\frac{\sin(2x)}{2x}}{\frac{\sin(6x)}{6x}} \times \frac{2}{6}$

Step 2: Apply the limit property

From the standard limit property, we know:

$\lim_{{x \to 0}} \frac{\sin(2x)}{2x} = 1 \quad \text{and} \quad \lim_{{x \to 0}} \frac{\sin(6x)}{6x} = 1$

Thus, we can simplify the limit expression:

$\lim_{{x \to 0}} \frac{\sin(2x)}{\sin(6x)} = \frac{1}{1} \times \frac{2}{6} = \frac{2}{6} = \frac{1}{3}$

Final Answer:

The limit is

$\boxed{\frac{1}{3}}$

Would you like more details or have further questions? Here are some related questions to explore:

1. How does the standard limit $\lim_{{x \to 0}} \frac{\sin(kx)}{kx} = 1$ apply to other trigonometric functions like cosine?
2. What would happen if we had $\lim_{{x \to 0}} \frac{\sin(2x)}{x}$ instead?
3. How can we use L'Hôpital's Rule to evaluate limits of indeterminate forms like $\frac{0}{0}$?
4. What are some other common limits used in calculus for evaluating trigonometric expressions?
5. Can this approach be extended to limits involving more complicated trigonometric functions, such as $\tan(x)$?

Tip: When dealing with limits involving trigonometric functions as $x \to 0$, remember the key result $\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1$, which is crucial in simplifying many such expressions.