Question

The function $f(x)=\frac{\ln (1+a x)-\ln (1-b x)}{x}$ 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

• A. $a -b$
• B. $a +b$
• C. $\ln a+\ln b$
• D. None of these

Answer 1

Answer: (B)

For $f(x)$ to be continuous, we must have
$f(0)=\displaystyle\lim _{x \rightarrow 0} f(x)$
$\therefore \displaystyle\lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{a \log (1+a x)}{a x}+\frac{b \log (1-b x)}{-b x}$
$=a \cdot 1+b \cdot 1 \left[\text { using, } \displaystyle\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$ $=a +b$
$\therefore f(0)=a +b$