Question

The value of $f\left(0\right)$ so that the function
$f\left(x\right)=\frac{\log (1+\frac{x}{a})-\log \left(\begin{array}{c}1-\frac{x}{b}\end{array}\right)}{x},(x\ne 0)$ is continuous at $x=0$ is :

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

Answer 1

The correct option is A $\frac{a+b}{ab}$

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{\log (1+\frac{x}{a})-\log (1-\frac{x}{b})}{x}$

$\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1$

Using this in the above limit,

$\underset{x\to 0}{lim}\frac{\frac{x}{a}\cdot \frac{\log (1+\frac{x}{a})}{\frac{x}{a}}+\frac{x}{b}\cdot \frac{\log (1-\frac{x}{b})}{\frac{-x}{b}}}{x}$

$=\frac{x(\frac{1}{a}+\frac{1}{b})}{x}$

$=\frac{1}{a}+\frac{1}{b}$

$=\frac{a+b}{ab}$