We are asked to evaluate the following limit:

$\mathop {\lim }\limits_{h \to 0} \frac{{2\left( {\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)} \right)}}{{\sqrt 3 h(\sqrt 3 \cosh - \sinh )}}$

Step 1: Expand the trigonometric functions

We can use trigonometric sum formulas to simplify the expressions for $\sin\left(\frac{\pi}{6} + h\right)$ and $\cos\left(\frac{\pi}{6} + h\right)$.

$\sin\left(\frac{\pi}{6} + h\right) = \sin\frac{\pi}{6}\cos h + \cos\frac{\pi}{6}\sin h = \frac{1}{2}\cos h + \frac{\sqrt{3}}{2}\sin h$

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

Step 2: Substitute these into the numerator

Now substitute these expressions into the numerator:

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

Simplifying:

$= \frac{\sqrt{3}}{2}\cos h + \frac{3}{2}\sin h - \frac{\sqrt{3}}{2}\cos h + \frac{1}{2}\sin h$

$= \frac{4}{2}\sin h = 2\sin h$

Step 3: Simplify the denominator

The denominator contains $\sqrt{3}h(\sqrt{3}\cosh - \sinh)$. As $h \to 0$, the hyperbolic approximations can be used:

$\cosh h \approx 1 + \frac{h^2}{2}, \quad \sinh h \approx h$

Thus:

$\sqrt{3} h (\sqrt{3} \cosh h - \sinh h) \approx \sqrt{3} h (\sqrt{3}(1 + \frac{h^2}{2}) - h) = \sqrt{3} h (\sqrt{3} - h)$

Step 4: Evaluate the limit

The expression becomes:

$\lim_{h \to 0} \frac{2\sin h}{\sqrt{3}h(\sqrt{3} - h)}$

As $h \to 0$, $\sin h \approx h$, so the limit simplifies to:

$\lim_{h \to 0} \frac{2h}{\sqrt{3}h\sqrt{3}} = \frac{2}{3}$

Thus, the value of the limit is:

$\boxed{\frac{2}{3}}$

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Would you like further clarification or more details on any part of this? Here are 5 related questions you could explore:

1. What is the significance of small angle approximations in limit calculations?
2. How can hyperbolic functions be approximated for small values of $h$?
3. What are other techniques for evaluating trigonometric limits?
4. How does the sum and difference formula for trigonometric functions work?
5. Can this limit be solved using L'Hopital's Rule?

Tip: When evaluating limits with trigonometric functions, approximating them for small values (like $h \to 0$) can greatly simplify the process.