We are tasked with finding the limit:

$\lim_{x \to 0} \frac{3 - 3\cos^2 x}{x^2}$

Step 1: Factor out the constant

We can factor out the constant $3$ from the numerator:

$\lim_{x \to 0} \frac{3(1 - \cos^2 x)}{x^2}$

Step 2: Use a trigonometric identity

Recall that $1 - \cos^2 x = \sin^2 x$ (from the Pythagorean identity). Substituting this into the expression:

$\lim_{x \to 0} \frac{3\sin^2 x}{x^2}$

Step 3: Simplify the limit

We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$, so we can rewrite the expression:

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

Thus, the value of the limit is:

$\boxed{3}$

Would you like further clarification on any step?

Here are five related questions you can explore:

1. What is the limit of $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$?
2. How can we derive $\lim_{x \to 0} \frac{\sin x}{x}$ using L'Hopital's Rule?
3. What are some common trigonometric limits and how are they derived?
4. How does L'Hopital's Rule apply to limits involving indeterminate forms?
5. What is the Maclaurin series for $\sin x$ and $\cos x$, and how can it be used in limits?

Tip: For small angles, trigonometric functions like $\sin x$ and $\cos x$ can often be approximated using Taylor or Maclaurin series expansions.