The value of `lim_(xrarr(pi)/(6))(2cos(x+(pi)/(3)))/((1-sqrt3tanx))` is equal to

Options

A. `(1)/(2)`
B. `sqrt3`
C. `(sqrt3)/(4)`
D. `(sqrt3)/(2)`

Answer

To solve the limit problem \( \lim_{x \to \frac{\pi}{6}} \frac{2 \cos\left(x + \frac{\pi}{3}\right)}{1 - \sqrt{3} \tan x} \), we will follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{6} \)

First, we substitute \( x = \frac{\pi}{6} \) into the limit expression to check if we get an indeterminate form.

\[ \cos\left(\frac{\pi}{6} + \frac{\pi}{3}\right) = \cos\left(\frac{\pi}{6} + \frac{2\pi}{6}\right) = \cos\left(\frac{3\pi}{6}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

Now for the denominator:

\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \implies 1 - \sqrt{3} \tan\left(\frac{\pi}{6}\right) = 1 - \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 - 1 = 0 \]

Thus, we have the form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

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

**Denominator:**
\[ \frac{d}{dx}\left(1 - \sqrt{3} \tan x\right) = -\sqrt{3} \sec^2 x \]

### Step 3: Rewrite the limit

Now we can rewrite the limit using the derivatives we found:

\[ \lim_{x \to \frac{\pi}{6}} \frac{-2 \sin\left(x + \frac{\pi}{3}\right)}{-\sqrt{3} \sec^2 x} = \lim_{x \to \frac{\pi}{6}} \frac{2 \sin\left(x + \frac{\pi}{3}\right)}{\sqrt{3} \sec^2 x} \]

### Step 4: Substitute \( x = \frac{\pi}{6} \) again

Now we substitute \( x = \frac{\pi}{6} \) again:

\[ \sin\left(\frac{\pi}{6} + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{6} + \frac{2\pi}{6}\right) = \sin\left(\frac{3\pi}{6}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

For \( \sec^2\left(\frac{\pi}{6}\right) \):

\[ \sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \implies \sec^2\left(\frac{\pi}{6}\right) = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]

### Step 5: Calculate the limit

Now substituting these values into our limit expression:

\[ \lim_{x \to \frac{\pi}{6}} \frac{2 \cdot 1}{\sqrt{3} \cdot \frac{4}{3}} = \frac{2}{\sqrt{3} \cdot \frac{4}{3}} = \frac{2 \cdot 3}{4 \sqrt{3}} = \frac{6}{4 \sqrt{3}} = \frac{3}{2 \sqrt{3}} = \frac{3 \sqrt{3}}{6} = \frac{\sqrt{3}}{2} \]

### Final Answer

Thus, the value of the limit is:

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