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Question:

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

Solution:

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

If f(x) is continuous at x = 0, then

$\lim _{x \rightarrow 0} f(x)=f(0)$

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

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

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

$\Rightarrow \frac{1}{a} \times 1-\left(-\frac{1}{b}\right) \times 1=f(0) \quad\left[\right.$ Using : $\left.\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=f(0)$

$\Rightarrow \frac{a+b}{a b}=f(0)$