The given expression involves evaluating the limit:

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

Step-by-step solution:

1. Simplify the numerator $1 - \cos^2 x$:

Using the trigonometric identity: $1 - \cos^2 x = \sin^2 x$

So, the expression simplifies to: $\lim_{x \to 0} \frac{\sin^2 x}{x^2 \cot\left( x - \frac{\pi}{3} \right)}$

2. Use a known limit for the sine function:

We know from standard limits that: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Thus, for small $x$, we can approximate: $\frac{\sin^2 x}{x^2} \approx 1$

3. Analyze the cotangent term:

We now need to evaluate the cotangent function near $x = 0$: $\cot\left( x - \frac{\pi}{3} \right) = \frac{1}{\tan\left( x - \frac{\pi}{3} \right)}$ As $x \to 0$, the argument of the tangent function approaches $-\frac{\pi}{3}$, and: $\tan\left( -\frac{\pi}{3} \right) = -\sqrt{3}$

Therefore: $\cot\left( x - \frac{\pi}{3} \right) \approx -\frac{1}{\sqrt{3}} \quad \text{as} \quad x \to 0$

4. Final evaluation:

Now substitute the approximations back into the limit expression: $\lim_{x \to 0} \frac{\sin^2 x}{x^2} \times \frac{1}{\cot\left( x - \frac{\pi}{3} \right)} \approx \frac{1}{-\frac{1}{\sqrt{3}}} = -\sqrt{3}$

Thus, the value of the limit is:

$\boxed{-\sqrt{3}}$

Would you like further clarification or a more detailed explanation? Here are some related questions you might find interesting:

1. How do you derive the small-angle approximation for trigonometric functions?
2. What are the common limit laws that help evaluate limits involving trigonometric functions?
3. How does L'Hopital's rule apply to limits with indeterminate forms like $\frac{0}{0}$?
4. What are some other key trigonometric identities useful for simplifying expressions?
5. How would this limit change if the argument inside the cotangent function were different?

Tip: Always check for trigonometric identities when simplifying complex expressions, especially when involving limits.