The value of `lim_(xrarr(pi)/(3))(2-sqrt3sinx-cosx)/((3x-pi)^(2))` is equal to the reciprocal of the number

Answer

To solve the limit problem, we will follow these steps:

### Step 1: Write the limit expression
We start with the limit expression:
\[ \lim_{x \to \frac{\pi}{3}} \frac{2 - \sqrt{3} \sin x - \cos x}{(3x - \pi)^2} \]

### Step 2: Simplify the numerator
We can rewrite the numerator:
\[ 2 - \sqrt{3} \sin x - \cos x = 2 - 2 \left(\frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x\right) \]
Using the identity \( \cos(a - b) = \cos a \cos b + \sin a \sin b \), we can express:
\[ \frac{\sqrt{3}}{2} = \cos\left(\frac{\pi}{3}\right) \quad \text{and} \quad \frac{1}{2} = \sin\left(\frac{\pi}{3}\right) \]
Thus, we can write:
\[ 2 - 2 \left(\frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x\right) = 2(1 - \cos(x - \frac{\pi}{3})) \]

### Step 3: Substitute into the limit
Now substituting this back into our limit gives:
\[ \lim_{x \to \frac{\pi}{3}} \frac{2(1 - \cos(x - \frac{\pi}{3}))}{(3x - \pi)^2} \]

### Step 4: Change of variable
Let \( t = x - \frac{\pi}{3} \). Then as \( x \to \frac{\pi}{3} \), \( t \to 0 \). We also have:
\[ 3x - \pi = 3\left(t + \frac{\pi}{3}\right) - \pi = 3t \]
Thus, our limit becomes:
\[ \lim_{t \to 0} \frac{2(1 - \cos t)}{(3t)^2} = \lim_{t \to 0} \frac{2(1 - \cos t)}{9t^2} \]

### Step 5: Use the standard limit
We know that:
\[ \lim_{t \to 0} \frac{1 - \cos t}{t^2} = \frac{1}{2} \]
Therefore, we can substitute this into our limit:
\[ \lim_{t \to 0} \frac{2(1 - \cos t)}{9t^2} = \frac{2}{9} \cdot \frac{1}{2} = \frac{1}{9} \]

### Conclusion
Thus, we find:
\[ \lim_{x \to \frac{\pi}{3}} \frac{2 - \sqrt{3} \sin x - \cos x}{(3x - \pi)^2} = \frac{1}{9} \]
The problem states that this limit is equal to the reciprocal of a number, which means:
\[ \frac{1}{9} = \frac{1}{N} \implies N = 9 \]

### Final Answer
The value of the limit is equal to the reciprocal of the number \(9\).

---