The problem asks you to find the limit of the following expression:

$\lim_{x \to 0} \frac{\sin^2(6x)}{\sin^2(12x)}$

Step-by-Step Solution:

1. Rewrite the expression:

   Use the fact that $\sin^2(6x)$ and $\sin^2(12x)$ can be separated into two squared sine functions:

   $\lim_{x \to 0} \left( \frac{\sin(6x)}{\sin(12x)} \right)^2$

2. Apply the standard trigonometric limit:

   One useful limit to remember is: $\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1 \quad \text{for any constant } k$ So we can multiply both the numerator and the denominator by $x$ to use this rule:

   $\lim_{x \to 0} \left( \frac{\frac{\sin(6x)}{6x}}{\frac{\sin(12x)}{12x}} \times \frac{12x}{6x} \right)^2$

   Since $\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1$, the fractions involving sine tend to 1. Thus, the limit simplifies to:

   $\left( \frac{12}{6} \right)^2 = 2^2 = 4$

Final Answer:

$\lim_{x \to 0} \frac{\sin^2(6x)}{\sin^2(12x)} = 4$

Would you like more details on any step, or do you have any questions?

Here are five related questions to extend your understanding:

1. What is the limit of $\lim_{x \to 0} \frac{\sin(3x)}{\sin(6x)}$?
2. How do you compute $\lim_{x \to 0} \frac{\sin(x)}{x}$ using L'Hôpital's Rule?
3. What happens if we have $\lim_{x \to 0} \frac{\sin(kx)}{\sin(mx)}$ for arbitrary constants $k$ and $m$?
4. Can the method of multiplying by $x$ be applied to other trigonometric functions such as $\tan(x)$?
5. How do we evaluate limits involving trigonometric functions and higher powers, like $\sin^3(x)$?

Tip:

When working with trigonometric limits, always look for opportunities to apply standard limits like $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ to simplify the expressions.