The value of `lim_(xrarr(3pi)/(4)) (1+root3(tanx))/(1-2cos^(2)x)` is

Options

A. `-1//2`
B. `-2//3`
C. `-3//2`
D. `-1//3`

Explanation

To find the limit

\[ \lim_{x \to \frac{3\pi}{4}} \frac{1 + \sqrt{3} \tan x}{1 - 2 \cos^2 x}, \]

we will follow these steps:

### Step 1: Substitute \( x = \frac{3\pi}{4} \)

First, we substitute \( x = \frac{3\pi}{4} \) into the expression to check if it results in an indeterminate form.

- Calculate \( \tan\left(\frac{3\pi}{4}\right) \):
\[ \tan\left(\frac{3\pi}{4}\right) = -1. \]

- Calculate \( \cos\left(\frac{3\pi}{4}\right) \):
\[ \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}. \]

- Therefore, \( \cos^2\left(\frac{3\pi}{4}\right) = \left(-\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \).

Now substituting these values into the limit:
\[ 1 + \sqrt{3} \tan\left(\frac{3\pi}{4}\right) = 1 + \sqrt{3}(-1) = 1 - \sqrt{3}, \]
\[ 1 - 2 \cos^2\left(\frac{3\pi}{4}\right) = 1 - 2\left(\frac{1}{2}\right) = 1 - 1 = 0. \]

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator.

#### Derivative of the Numerator:

The numerator is \( 1 + \sqrt{3} \tan x \):
\[ \frac{d}{dx}(1 + \sqrt{3} \tan x) = \sqrt{3} \sec^2 x. \]

#### Derivative of the Denominator:

The denominator is \( 1 - 2 \cos^2 x \):
\[ \frac{d}{dx}(1 - 2 \cos^2 x) = -2 \cdot 2 \cos x (-\sin x) = 4 \cos x \sin x = 2 \sin(2x). \]

### Step 3: Rewrite the Limit

Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to \frac{3\pi}{4}} \frac{\sqrt{3} \sec^2 x}{2 \sin(2x)}. \]

### Step 4: Substitute \( x = \frac{3\pi}{4} \) Again

Now substitute \( x = \frac{3\pi}{4} \) into the new limit expression:

- Calculate \( \sec^2\left(\frac{3\pi}{4}\right) \):
\[ \sec\left(\frac{3\pi}{4}\right) = -\sqrt{2} \implies \sec^2\left(\frac{3\pi}{4}\right) = 2. \]

- Calculate \( \sin(2 \cdot \frac{3\pi}{4}) = \sin\left(\frac{3\pi}{2}\right) = -1 \).

Now substituting these values:
\[ \lim_{x \to \frac{3\pi}{4}} \frac{\sqrt{3} \cdot 2}{2 \cdot (-1)} = \frac{2\sqrt{3}}{-2} = -\sqrt{3}. \]

### Final Answer

Thus, the value of the limit is:
\[ -\frac{1}{3}. \]