The value of $$\lim_{h \to 0} \left\{\frac{\sqrt{3}\sin\left(\frac{\pi}{6} + h\right) - \cos\left(\frac{\pi}{6} + h\right)}{\sqrt{3}h(\sqrt{3}\cos h - \sin h)}\right\}$$ is:

We simplify the numerator $$\sqrt{3}\sin\left(\frac{\pi}{6} + h\right) - \cos\left(\frac{\pi}{6} + h\right)$$. Writing this as $$2\left(\frac{\sqrt{3}}{2}\sin\left(\frac{\pi}{6}+h\right) - \frac{1}{2}\cos\left(\frac{\pi}{6}+h\right)\right) = 2\sin\left(\frac{\pi}{6}+h - \frac{\pi}{6}\right) = 2\sin h$$.

Similarly, the factor $$\sqrt{3}\cos h - \sin h = 2\left(\frac{\sqrt{3}}{2}\cos h - \frac{1}{2}\sin h\right) = 2\cos\left(h + \frac{\pi}{6}\right)$$.

Substituting these into the expression gives $$\frac{2\sin h}{\sqrt{3} \cdot h \cdot 2\cos\left(h + \frac{\pi}{6}\right)} = \frac{\sin h}{h} \cdot \frac{1}{\sqrt{3}\cos\left(h + \frac{\pi}{6}\right)}$$.

Taking the limit as $$h \to 0$$: $$\lim_{h \to 0}\frac{\sin h}{h} = 1$$ and $$\cos\left(0 + \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$. Therefore the limit equals $$\frac{1}{\sqrt{3} \cdot \frac{\sqrt{3}}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3}$$.