Question

$f\left(x\right)=\frac{\log (1+ax)-\log (1-bx)}{x}$ 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-

• A. $a-b$
• B. $a+b$
• C. $\log a+\log b$
• D. $\frac{a+b}{2}$

Answer 1

Answer: (B)

$a+b$

Given $f\left(x\right)=\frac{\log (1+ax)-\log (1-bx)}{x}$ for $x\ne 0$
Since, $f\left(x\right)$ is continuous at $x=0$
$LHL=RHL=f\left(0\right)$
Now, $\underset{x\to 0}{lim}f\left(x\right)$
$=\underset{x\to 0}{lim}\frac{\log (1+ax)-\log (1-bx)}{x}$
$=\underset{x\to 0}{lim}\frac{\log (1+ax)}{x}-\underset{x\to 0}{lim}\frac{\log (1-bx)}{x}$
$=\underset{x\to 0}{lim}a\frac{\log (1+ax)}{ax}-\underset{x\to 0}{lim}(-b)\frac{\log (1-bx)}{-bx}$
$=a+b$ ($\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1$)
$\Rightarrow f\left(0\right)=a+b$