Evaluate `lim_(hto0) (2[sqrt(3)sin((pi)/(6)+h)-cos((pi)/(6)+h)])/(sqrt(3)h(sqrt(3)cosh-sinh)).`

Explanation

To evaluate the limit

\[ \lim_{h \to 0} \frac{2\left(\sqrt{3}\sin\left(\frac{\pi}{6}+h\right) - \cos\left(\frac{\pi}{6}+h\right)\right)}{\sqrt{3}h\left(\sqrt{3}\cos h - \sin h\right)}, \]

we will proceed step by step.

### Step 1: Substitute Known Values
First, we know that:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}. \]
Thus, we can rewrite the limit as:
\[ \lim_{h \to 0} \frac{2\left(\sqrt{3}\left(\frac{1}{2} + \cos h\right) - \frac{\sqrt{3}}{2} - \sin h\right)}{\sqrt{3}h\left(\sqrt{3}\cos h - \sin h\right)}. \]

### Step 2: Simplify the Numerator
Using the angle addition formulas:
\[ \sin\left(\frac{\pi}{6}+h\right) = \sin\left(\frac{\pi}{6}\right)\cos h + \cos\left(\frac{\pi}{6}\right)\sin h = \frac{1}{2}\cos h + \frac{\sqrt{3}}{2}\sin h, \]
\[ \cos\left(\frac{\pi}{6}+h\right) = \cos\left(\frac{\pi}{6}\right)\cos h - \sin\left(\frac{\pi}{6}\right)\sin h = \frac{\sqrt{3}}{2}\cos h - \frac{1}{2}\sin h. \]
Substituting these into the numerator gives:
\[ \sqrt{3}\left(\frac{1}{2}\cos h + \frac{\sqrt{3}}{2}\sin h\right) - \left(\frac{\sqrt{3}}{2}\cos h - \frac{1}{2}\sin h\right). \]

### Step 3: Combine Terms
This simplifies to:
\[ \frac{\sqrt{3}}{2}\cos h + \frac{3}{2}\sin h - \frac{\sqrt{3}}{2}\cos h + \frac{1}{2}\sin h = 2\sin h. \]
Thus, the limit now becomes:
\[ \lim_{h \to 0} \frac{2 \cdot 2\sin h}{\sqrt{3}h\left(\sqrt{3}\cos h - \sin h\right)} = \lim_{h \to 0} \frac{4\sin h}{\sqrt{3}h\left(\sqrt{3}\cos h - \sin h\right)}. \]

### Step 4: Use the Limit Property
We know that:
\[ \lim_{h \to 0} \frac{\sin h}{h} = 1. \]
Thus, we can rewrite the limit as:
\[ \lim_{h \to 0} \frac{4}{\sqrt{3}\left(\sqrt{3}\cos h - \sin h\right)}. \]

### Step 5: Evaluate the Remaining Limit
As \( h \to 0 \), we have:
\[ \cos h \to 1 \quad \text{and} \quad \sin h \to 0. \]
So,
\[ \sqrt{3}\cos h - \sin h \to \sqrt{3} \cdot 1 - 0 = \sqrt{3}. \]
Thus, the limit becomes:
\[ \frac{4}{\sqrt{3} \cdot \sqrt{3}} = \frac{4}{3}. \]

### Final Answer
Therefore, the limit is:
\[ \lim_{h \to 0} \frac{2\left(\sqrt{3}\sin\left(\frac{\pi}{6}+h\right) - \cos\left(\frac{\pi}{6}+h\right)\right)}{\sqrt{3}h\left(\sqrt{3}\cos h - \sin h\right)} = \frac{4}{3}. \]