\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 )}} is equal to

\frac{4}{3}

–\frac{4}{3}

\frac{2}{3}

\frac{3}{4}

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:

\[
\boxed{\frac{2}{3}}
\]

---

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.

The query submitted by the user:\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 )}} is equal to \frac{4}{3}, -\frac{4}{3}, \frac{2}{3}, \frac{3}{4}.

Learn how to evaluate a limit involving trigonometric and hyperbolic functions as \( h \to 0 \). This problem demonstrates the use of trigonometric sum formulas, small angle approximations, and hyperbolic function expansions. Suitable for advanced high school students in Grades 11-12.

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