#### Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 30 Maths Textbook Solution.

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

Hint:

$f\left ( x \right )$  must be defined. The limit of the  $f\left ( x \right )$  approaches the value  $x$ must exist.

Given:

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

If $f\left ( x \right )$ is continuous at $x=0$ , then

\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0}\left(\frac{\log \left(1+\frac{x}{a}\right)-\log \left(1-\frac{x}{b}\right)}{x}\right)=f(0) \end{aligned}

$\lim _{x \rightarrow 0}\left(\frac{\log \left(1+\frac{x}{a}\right)}{\frac{a x}{a}}-\frac{\log \left(1-\frac{x}{b}\right)}{\frac{b x}{b}}\right)=f(0)$                                        [Multiplying the denominator of first term by $\frac{a}{a}$ ]

[Multiplying the denominator of second term by $\frac{b}{b}$ ]

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

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

\begin{aligned} \\\\ &\frac{1}{a}+\frac{1}{b}=f(0) \\\\ &\frac{a+b}{a b}=f(0) \end{aligned}