Question

If $f(x) = \dfrac{\log(1+ax) - \log(1-bx)}{x}$ for $x \neq 0$ and $f(0) = k$, and $f(x)$ is continuous at $x = 0$, then $k$ equals

• A. $a + b$
• B. $a - b$
• C. $a$
• D. $b$

Hint:

Standard limit: $\displaystyle\lim_{x \to 0}\frac{\ln(1+px)}{x} = p$. Apply this separately to each logarithmic term.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For $f$ to be continuous at $x=0$, we need $f(0) = \lim_{x \to 0} f(x)$.
Step 2: Detailed Explanation:
Using $\displaystyle\lim_{x \to 0}\frac{\log(1+px)}{x} = p$:
$k = \displaystyle\lim_{x\to0}\frac{\log(1+ax)-\log(1-bx)}{x} = a + b$.
Step 3: Final Answer:
$k = a + b$.