The problem presented is to evaluate the limit:

$\lim_{{x \to \frac{\pi}{3}}} \frac{\sqrt{3} \cos x - \sin x}{x - \frac{\pi}{3}}$

To solve this, we can proceed by substituting values or by using L'Hôpital's Rule because the direct substitution results in an indeterminate form of $\frac{0}{0}$. Here's the step-by-step solution:

1. Substitute $x = \frac{\pi}{3}$: $\sqrt{3} \cos \left(\frac{\pi}{3}\right) - \sin \left(\frac{\pi}{3}\right) = \sqrt{3} \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} = 0$ $x - \frac{\pi}{3} = 0$ So, this substitution yields $\frac{0}{0}$, an indeterminate form.

2. Apply L'Hôpital's Rule: Differentiate the numerator and the denominator with respect to $x$:

   • Numerator: $\frac{d}{dx}(\sqrt{3} \cos x - \sin x) = -\sqrt{3} \sin x - \cos x$
   • Denominator: $\frac{d}{dx}(x - \frac{\pi}{3}) = 1$

   So, the limit now becomes: $\lim_{{x \to \frac{\pi}{3}}} -\sqrt{3} \sin x - \cos x$

3. Substitute $x = \frac{\pi}{3}$ into the new expression: $-\sqrt{3} \sin \left(\frac{\pi}{3}\right) - \cos \left(\frac{\pi}{3}\right) = -\sqrt{3} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2}$ $= -\frac{3}{2} - \frac{1}{2} = -2$

Therefore, the value of the limit is:

$\boxed{-2}$

Would you like further details on any of the steps or additional information on limit calculations?

Here are five related questions for further exploration:

1. What is L'Hôpital's Rule and when can it be applied?
2. How do you identify an indeterminate form in limits?
3. What are other methods to solve limits besides L'Hôpital's Rule?
4. How do trigonometric limits differ from polynomial limits?
5. How do you handle limits approaching infinity in trigonometric functions?

Tip: When solving limits, always check for an indeterminate form before applying L'Hôpital's Rule.