Question

If $f(x)=\frac{\log (1+ax)-\log (1-bx)}{x}$ for $x\ne 0$ and $f(0)=k$ and $f(x)$ is continuous at $x=0$ then k is equal to:

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

Answer 1

Answer: (A)

$f(x)=\frac{\log (1+ax)-\log (1-bx)}{x}$ $f(x)$ is continuous at $x=k$ and $f(0)=k$ . $\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\frac{\log (1+ax)-\log (1-bx)}{x}$ $\left(\frac{0}{0}form\right)$ $=\underset{x\to 0}{\lim }\left(\frac{1}{1+ax}.a+\frac{b}{1-bx}\right)$ $=\underset{x\to 0}{\lim }\left(\frac{a}{1+ax}+\frac{b}{1-bx}\right)=a+b$ $\therefore$ $a+b=f(0)=k$