The function `f(x) = (log(1+ax)-log(1-bx))/(x)` is not defined at x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0, is

Options

A. `a - b`
B. `a + b`
C. `log a + log b`
D. `log a - log b`

Answer

To find the value that should be assigned to \( f(0) \) so that the function \( f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x} \) 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 Limit**: We need to find \( \lim_{x \to 0} f(x) \):
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\log(1 + ax) - \log(1 - bx)}{x} \]

2. **Apply Logarithmic Properties**: Use the property of logarithms that states \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \):
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\log\left(\frac{1 + ax}{1 - bx}\right)}{x} \]

3. **Evaluate the Limit**: As \( x \) approaches 0, both the numerator and denominator approach 0, resulting in an indeterminate form \( \frac{0}{0} \). Thus, we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{\log\left(\frac{1 + ax}{1 - bx}\right)}{x} \]

4. **Differentiate the Numerator and Denominator**: Differentiate the numerator and denominator:
- The derivative of the numerator \( \log\left(\frac{1 + ax}{1 - bx}\right) \) using the chain rule:
\[ \frac{d}{dx} \log\left(\frac{1 + ax}{1 - bx}\right) = \frac{1}{\frac{1 + ax}{1 - bx}} \cdot \left(\frac{(1 - bx)(a) - (1 + ax)(-b)}{(1 - bx)^2}\right) \]
- The derivative of the denominator \( x \) is simply \( 1 \).

5. **Simplify the Derivative**: After differentiating, we substitute \( x = 0 \):
\[ \lim_{x \to 0} \frac{(1 - 0)(a) + (1)(b)}{(1 - 0)^2} = a + b \]

6. **Assign the Value to \( f(0) \)**: For \( f(x) \) to be continuous at \( x = 0 \), we need to assign:
\[ f(0) = a + b \]

### Conclusion:
The value that should be assigned to \( f(0) \) for continuity at \( x = 0 \) is \( a + b \).