The limit expression in the image is:

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

To solve this, we can first rewrite it as:

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

Now, let's solve step by step:

Step 1: Use the standard limit identity

One of the standard limit identities is:

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

To apply this, observe that we need the denominator to match the argument inside the sine function. In this case, we have $\sin(6x)$, but the denominator is $x$. We can fix this by multiplying and dividing by 6:

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

Step 2: Apply the limit identity

Since $\lim_{{x \to 0}} \frac{{\sin(6x)}}{6x} = 1$, the expression simplifies to:

$\lim_{{x \to 0}} \left( 6 \cdot 1 \right)^2 = 36$

Final Answer:

$\lim_{{x \to 0}} \frac{{\sin^2(6x)}}{x} = 36$

Would you like further clarification or details?

Here are 5 related questions to explore further:

1. How do you derive the standard limit $\lim_{{x \to 0}} \frac{{\sin(x)}}{x} = 1$?
2. What is L'Hopital's rule, and how can it be used to solve limits like this?
3. Can we apply the squeeze theorem to confirm this result?
4. How does this limit change if the argument inside the sine function is a different multiple of $x$?
5. What is the importance of limits in calculus and how do they connect to continuity?

Tip: When dealing with trigonometric limits, always look for ways to match the argument of the sine function with the denominator for easier simplifications.