If `f(x)=(1+sinx-cosx)/(1-sinx-cosx), x ne 0` . The value of f(0) so that f is a continuous function is

Options

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

Answer

To find the value of \( f(0) \) such that the function \( f(x) = \frac{1 + \sin x - \cos x}{1 - \sin x - \cos x} \) is continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-step Solution:

1. **Identify the requirement for continuity**:
For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ \lim_{x \to 0} f(x) = f(0) \]

2. **Calculate the limit**:
We need to evaluate:
\[ \lim_{x \to 0} \frac{1 + \sin x - \cos x}{1 - \sin x - \cos x} \]

3. **Substituting the values**:
As \( x \) approaches 0, we know:
- \( \sin(0) = 0 \)
- \( \cos(0) = 1 \)

Substituting these values into the function:
\[ f(0) = \frac{1 + 0 - 1}{1 - 0 - 1} = \frac{0}{0} \]
This is an indeterminate form, so we need to apply L'Hôpital's Rule or simplify further.

4. **Simplifying the expression**:
We can rewrite \( 1 - \cos x \) using the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \) and \( \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \):
\[ \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \]
Thus, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{1 + 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) - (1 - 2 \sin^2\left(\frac{x}{2}\right))}{1 - 2 \sin^2\left(\frac{x}{2}\right) - (1 - 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right))} \]

5. **Further simplification**:
After simplifying the numerator and denominator, we can cancel out common factors:
\[ = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right) + 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}{2 \sin^2\left(\frac{x}{2}\right) - 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)} \]

6. **Taking the limit**:
As \( x \) approaches 0, \( \sin\left(\frac{x}{2}\right) \) approaches 0 and \( \cos\left(\frac{x}{2}\right) \) approaches 1:
\[ = \frac{0 + 0}{0 - 0} = \frac{0}{0} \]
We can apply L'Hôpital's Rule again or substitute small angle approximations.

7. **Final evaluation**:
After applying L'Hôpital's Rule or further simplification, we find:
\[ \lim_{x \to 0} f(x) = -1 \]

8. **Setting \( f(0) \)**:
Therefore, for \( f(x) \) to be continuous at \( x = 0 \):
\[ f(0) = -1 \]

### Conclusion:
The value of \( f(0) \) so that \( f \) is a continuous function is \( \boxed{-1} \).