Find the value of f(0) so that the function `f(x)=1/8 (1-cos^2 x+sin^2 x)/[sqrt(x^2+1)-1], x ne 0` is continuous .

Answer

To find the value of \( f(0) \) such that the function

\[ f(x) = \frac{1}{8} \frac{1 - \cos^2 x + \sin^2 x}{\sqrt{x^2 + 1} - 1}, \quad x \neq 0 \]

is continuous at \( x = 0 \), we need to ensure that

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

### Step-by-Step Solution:

1. **Simplify the Function**:
We start with the expression for \( f(x) \):

\[ f(x) = \frac{1}{8} \frac{1 - \cos^2 x + \sin^2 x}{\sqrt{x^2 + 1} - 1}. \]

Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can simplify \( 1 - \cos^2 x + \sin^2 x \) to:

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

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

\[ f(x) = \frac{1}{8} \frac{2 \sin^2 x}{\sqrt{x^2 + 1} - 1} = \frac{1}{4} \frac{\sin^2 x}{\sqrt{x^2 + 1} - 1}. \]

2. **Evaluate the Limit**:
We need to find \( \lim_{x \to 0} f(x) \):

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{4} \frac{\sin^2 x}{\sqrt{x^2 + 1} - 1}. \]

The expression \( \sqrt{x^2 + 1} - 1 \) approaches 0 as \( x \) approaches 0, so we can apply L'Hôpital's Rule or simplify further.

3. **Rationalizing the Denominator**:
To simplify \( \sqrt{x^2 + 1} - 1 \), we multiply the numerator and denominator by \( \sqrt{x^2 + 1} + 1 \):

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

Thus, we can rewrite the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{4} \frac{\sin^2 x}{\frac{x^2}{\sqrt{x^2 + 1} + 1}} = \lim_{x \to 0} \frac{1}{4} \cdot \frac{\sin^2 x (\sqrt{x^2 + 1} + 1)}{x^2}. \]

4. **Using the Limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)**:
We know:

\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} = \left( \lim_{x \to 0} \frac{\sin x}{x} \right)^2 = 1. \]

Therefore, we can evaluate:

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

5. **Setting \( f(0) \)**:
To make \( f(x) \) continuous at \( x = 0 \), we set:

\[ f(0) = \lim_{x \to 0} f(x) = \frac{1}{2}. \]

### Final Answer:
Thus, the value of \( f(0) \) that makes the function continuous is

\[
\boxed{\frac{1}{2}}.