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 need to evaluate:

\[ \lim_{x \to \frac{\pi}{3}} \frac{2 - \sqrt{3} \sin x - \cos x}{(3x - \pi)^2} \]

### Step 1: Substitute \(x = \frac{\pi}{3}\)

First, we substitute \(x = \frac{\pi}{3}\) into the numerator and denominator to check if we get an indeterminate form.

\[ \text{Numerator: } 2 - \sqrt{3} \sin\left(\frac{\pi}{3}\right) - \cos\left(\frac{\pi}{3}\right) \]
\[ = 2 - \sqrt{3} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} \]
\[ = 2 - \frac{3}{2} - \frac{1}{2} = 2 - 2 = 0 \]

\[ \text{Denominator: } (3 \cdot \frac{\pi}{3} - \pi)^2 = (0)^2 = 0 \]

Since both the numerator and denominator approach 0, we have a \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

We differentiate the numerator and denominator:

**Numerator:**
\[ \frac{d}{dx}(2 - \sqrt{3} \sin x - \cos x) = -\sqrt{3} \cos x + \sin x \]

**Denominator:**
\[ \frac{d}{dx}((3x - \pi)^2) = 2(3x - \pi)(3) = 6(3x - \pi) \]

Now we can rewrite the limit:

\[ \lim_{x \to \frac{\pi}{3}} \frac{-\sqrt{3} \cos x + \sin x}{6(3x - \pi)} \]

### Step 3: Substitute \(x = \frac{\pi}{3}\) again

Substituting \(x = \frac{\pi}{3}\):

\[ \text{Numerator: } -\sqrt{3} \cos\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right) \]
\[ = -\sqrt{3} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = 0 \]

\[ \text{Denominator: } 6(3 \cdot \frac{\pi}{3} - \pi) = 6(0) = 0 \]

Again, we have a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 4: Differentiate again

**Numerator:**
\[ \frac{d}{dx}(-\sqrt{3} \cos x + \sin x) = \sqrt{3} \sin x + \cos x \]

**Denominator:**
\[ \frac{d}{dx}(6(3x - \pi)) = 18 \]

Now we can rewrite the limit:

\[ \lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} \sin x + \cos x}{18} \]

### Step 5: Substitute \(x = \frac{\pi}{3}\) again

Substituting \(x = \frac{\pi}{3}\):

\[ \text{Numerator: } \sqrt{3} \sin\left(\frac{\pi}{3}\right) + \cos\left(\frac{\pi}{3}\right) \]
\[ = \sqrt{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{3}{2} + \frac{1}{2} = 2 \]

Now substituting into the limit:

\[ \lim_{x \to \frac{\pi}{3}} \frac{2}{18} = \frac{1}{9} \]

### Conclusion

The value of the limit is:

\[ \frac{1}{9} \]

According to the problem statement, this is equal to the reciprocal of the number 9.