Question

Let $f\left(x\right)=\frac{\log \left(1+{\displaystyle \frac{x}{a}}\right)-\log \left(1-{\displaystyle \frac{x}{b}}\right)}{x}$, x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

Answer 1

Given: $f\left(x\right)=\frac{\log \left(1+{\displaystyle \frac{x}{a}}\right)-\log \left(1-\frac{x}{b}\right)}{x},x\ne 0$

If f(x) is continuous at x = 0, then

​$\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$
$\Rightarrow \underset{x\to 0}{\lim }\left(\frac{\log \left(1+{\displaystyle \frac{x}{a}}\right)-\log \left(1-\frac{x}{b}\right)}{x}\right)=f\left(0\right)$
$\Rightarrow \underset{x\to 0}{\lim }\left(\frac{\log \left(1+{\displaystyle \frac{x}{a}}\right)}{{\displaystyle \frac{ax}{a}}}-\frac{\log \left(1-\frac{x}{b}\right)}{{\displaystyle \frac{bx}{b}}}\right)=f\left(0\right)$
$\Rightarrow \frac{1}{a}\underset{x\to 0}{\lim }\left(\frac{\log \left(1+{\displaystyle \frac{x}{a}}\right)}{{\displaystyle \frac{x}{a}}}\right)-\left(-\frac{1}{b}\right)\underset{x\to 0}{\lim }\left(\frac{\log \left(1-\frac{x}{b}\right)}{{\displaystyle \frac{-x}{b}}}\right)=f\left(0\right)$
$$\begin{gathered} \Rightarrow \frac{1}{a}\times 1-\left(-\frac{1}{b}\right)\times 1=f\left(0\right)\left[\mathrm{Using}:\underset{x\to 0}{\lim }\frac{\log \left(1+x\right)}{x}=1\right] \\ \Rightarrow \frac{1}{a}+\frac{1}{b}=f\left(0\right) \\ \Rightarrow \frac{a+b}{ab}=f\left(0\right) \end{gathered}$$