If the function `f(x) = (cos^(2)x - sin^(2)x-1)/(sqrt(x^(2)+1)-1), x != 0`, is continuous at x = 0, then f(0) is equal to

Options

A. -2
B. -1
C. 0
D. -4

Answer

To determine the value of \( f(0) \) for the function

\[ f(x) = \frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2 + 1} - 1}, \quad x \neq 0, \]

and ensure that it is continuous at \( x = 0 \), we will follow these steps:

### Step 1: Evaluate the limit as \( x \) approaches 0
We need to find

\[ \lim_{x \to 0} f(x). \]

Substituting \( x = 0 \) directly into the function gives us:

\[ f(0) = \frac{\cos^2(0) - \sin^2(0) - 1}{\sqrt{0^2 + 1} - 1} = \frac{1 - 0 - 1}{1 - 1} = \frac{0}{0}, \]

which is an indeterminate form.

### Step 2: Simplify the expression
To resolve the \( \frac{0}{0} \) form, we can manipulate the expression. We will multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2 + 1} - 1} \cdot \frac{\sqrt{x^2 + 1} + 1}{\sqrt{x^2 + 1} + 1}. \]

This gives us:

\[ = \frac{(\cos^2 x - \sin^2 x - 1)(\sqrt{x^2 + 1} + 1)}{(\sqrt{x^2 + 1})^2 - 1^2}. \]

The denominator simplifies to:

\[ x^2 + 1 - 1 = x^2. \]

So we have:

\[ f(x) = \frac{(\cos^2 x - \sin^2 x - 1)(\sqrt{x^2 + 1} + 1)}{x^2}. \]

### Step 3: Simplify the numerator
Recall that

\[ \cos^2 x - \sin^2 x = \cos(2x), \]

thus:

\[ \cos^2 x - \sin^2 x - 1 = \cos(2x) - 1 = -2\sin^2(x). \]

So, we can rewrite \( f(x) \):

\[ f(x) = \frac{-2\sin^2 x (\sqrt{x^2 + 1} + 1)}{x^2}. \]

### Step 4: Evaluate the limit
Now we can rewrite \( \sin^2 x \) in terms of \( x \):

\[ \sin^2 x = \left(\frac{\sin x}{x}\right)^2 x^2. \]

Thus, we have:

\[ f(x) = -2 \left(\frac{\sin x}{x}\right)^2 (\sqrt{x^2 + 1} + 1). \]

Taking the limit as \( x \to 0 \):

\[ \lim_{x \to 0} f(x) = -2 \cdot 1 \cdot (1 + 1) = -2 \cdot 2 = -4. \]

### Conclusion
Since \( f(x) \) is continuous at \( x = 0 \), we find that:

\[ f(0) = \lim_{x \to 0} f(x) = -4. \]

Thus, the value of \( f(0) \) is

\[ \boxed{-4}. \]