`Lt_(x to 0)(cos^(2)x-sin^(2)x-1)/(sqrt(x^(2)+4)-2)` is equal to
(i) 4
(ii) `-4`
(iii) 8
(iv) `-8`

Options

A. 4
B. `-4`
C. 8
D. `-8`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2 + 4} - 2} \), we will follow these steps:

### Step 1: Simplify the numerator
The numerator can be rewritten using the identity \( \cos^2 x - \sin^2 x = \cos(2x) \):
\[ \cos^2 x - \sin^2 x - 1 = \cos(2x) - 1 \]
Thus, the limit becomes:
\[ \lim_{x \to 0} \frac{\cos(2x) - 1}{\sqrt{x^2 + 4} - 2} \]

### Step 2: Use the identity for cosine
Using the identity \( \cos(2x) - 1 = -2\sin^2(x) \), we can rewrite the numerator:
\[ \lim_{x \to 0} \frac{-2\sin^2(x)}{\sqrt{x^2 + 4} - 2} \]

### Step 3: Rationalize the denominator
To simplify the denominator, we can rationalize it by multiplying the numerator and denominator by the conjugate of the denominator:
\[ \sqrt{x^2 + 4} + 2 \]
So we have:
\[ \lim_{x \to 0} \frac{-2\sin^2(x)(\sqrt{x^2 + 4} + 2)}{(\sqrt{x^2 + 4} - 2)(\sqrt{x^2 + 4} + 2)} \]
The denominator simplifies as follows:
\[ (\sqrt{x^2 + 4})^2 - 2^2 = x^2 + 4 - 4 = x^2 \]
Thus, the limit becomes:
\[ \lim_{x \to 0} \frac{-2\sin^2(x)(\sqrt{x^2 + 4} + 2)}{x^2} \]

### Step 4: Substitute the limit
Now we can separate the limit:
\[ \lim_{x \to 0} -2 \cdot \frac{\sin^2(x)}{x^2} \cdot (\sqrt{x^2 + 4} + 2) \]
Using the fact that \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \), we get:
\[ \lim_{x \to 0} -2 \cdot 1 \cdot (\sqrt{0^2 + 4} + 2) = -2 \cdot (2 + 2) = -2 \cdot 4 = -8 \]

### Final Answer
Thus, the limit is:
\[ \boxed{-8} \]