Let `f(x)=(In(1-ax)-In(1-bx))/x`,`x!=0`. If `f(x)` is continuous at` x=0`, then` f(0)`=

Options

A. a-b
B. a+b
C. b-a
D. In a-In b

Answer

To find \( f(0) \) for the function

\[ f(x) = \frac{\ln(1 - ax) - \ln(1 - bx)}{x}, \quad x \neq 0, \]

we need to ensure that \( f(x) \) is continuous at \( x = 0 \). This means we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Rewrite the function using properties of logarithms
We can rewrite the function using the property of logarithms:

\[ f(x) = \frac{\ln\left(\frac{1 - ax}{1 - bx}\right)}{x}. \]

### Step 2: Find the limit as \( x \) approaches 0
We need to find:

\[ f(0) = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\ln\left(\frac{1 - ax}{1 - bx}\right)}{x}. \]

### Step 3: Apply L'Hôpital's Rule
Since both the numerator and denominator approach 0 as \( x \to 0 \), we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator:

- The derivative of the numerator \( \ln\left(1 - ax\right) - \ln\left(1 - bx\right) \) is:

\[ \frac{-a}{1 - ax} - \frac{-b}{1 - bx} = \frac{-a(1 - bx) + b(1 - ax)}{(1 - ax)(1 - bx)} = \frac{b - a - (b - a)x}{(1 - ax)(1 - bx)}. \]

- The derivative of the denominator \( x \) is \( 1 \).

Thus, we have:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\frac{b - a - (b - a)x}{(1 - ax)(1 - bx)}}{1} = \frac{b - a}{(1 - 0)(1 - 0)} = b - a. \]

### Step 4: Conclusion
Since \( f(x) \) is continuous at \( x = 0 \), we find that:

\[ f(0) = b - a. \]

### Final Answer
Thus, the value of \( f(0) \) is:

\[ \boxed{b - a}. \]