If the value of `lim_(xrarr(pi)/(6))(cos(x+(pi)/(3)))/((1-sqrt3tanx))` is equal to `lambda`, then the value of `120lambda^(2)` is equal to

Answer

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

### Step 1: Identify the limit
We need to evaluate the limit:
\[ \lim_{x \to \frac{\pi}{6}} \frac{\cos\left(x + \frac{\pi}{3}\right)}{1 - \sqrt{3} \tan x} \]

### Step 2: Substitute \(x = \frac{\pi}{6}\)
Substituting \(x = \frac{\pi}{6}\):
- The numerator becomes:
\[ \cos\left(\frac{\pi}{6} + \frac{\pi}{3}\right) = \cos\left(\frac{\pi}{6} + \frac{2\pi}{6}\right) = \cos\left(\frac{3\pi}{6}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]
- The denominator becomes:
\[ 1 - \sqrt{3} \tan\left(\frac{\pi}{6}\right) = 1 - \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 - 1 = 0 \]

Since both the numerator and denominator approach 0, we have an indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limit on the right side exists.

#### Differentiate the numerator:
\[ f(x) = \cos\left(x + \frac{\pi}{3}\right) \]
Using the chain rule:
\[ f'(x) = -\sin\left(x + \frac{\pi}{3}\right) \cdot \frac{d}{dx}\left(x + \frac{\pi}{3}\right) = -\sin\left(x + \frac{\pi}{3}\right) \]

#### Differentiate the denominator:
\[ g(x) = 1 - \sqrt{3} \tan x \]
Differentiating:
\[ g'(x) = -\sqrt{3} \sec^2 x \]

### Step 4: Rewrite the limit using derivatives
Now we rewrite the limit:
\[ \lim_{x \to \frac{\pi}{6}} \frac{-\sin\left(x + \frac{\pi}{3}\right)}{-\sqrt{3} \sec^2 x} = \lim_{x \to \frac{\pi}{6}} \frac{\sin\left(x + \frac{\pi}{3}\right)}{\sqrt{3} \sec^2 x} \]

### Step 5: Substitute \(x = \frac{\pi}{6}\) again
Substituting \(x = \frac{\pi}{6}\):
- The numerator becomes:
\[ \sin\left(\frac{\pi}{6} + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]
- The denominator becomes:
\[ \sqrt{3} \sec^2\left(\frac{\pi}{6}\right) = \sqrt{3} \cdot \left(\frac{2}{\sqrt{3}}\right)^2 = \sqrt{3} \cdot \frac{4}{3} = \frac{4\sqrt{3}}{3} \]

### Step 6: Evaluate the limit
Thus, we have:
\[ \lim_{x \to \frac{\pi}{6}} \frac{1}{\frac{4\sqrt{3}}{3}} = \frac{3}{4\sqrt{3}} = \frac{\sqrt{3}}{4} \]
Let \( \lambda = \frac{\sqrt{3}}{4} \).

### Step 7: Calculate \(120\lambda^2\)
Now we calculate:
\[ \lambda^2 = \left(\frac{\sqrt{3}}{4}\right)^2 = \frac{3}{16} \]
Then,
\[ 120 \lambda^2 = 120 \cdot \frac{3}{16} = \frac{360}{16} = 22.5 \]

### Final Answer
Thus, the value of \(120\lambda^2\) is \(\boxed{22.5}\).