#### Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 26

The correct option is (b)

Hint:

If a function f  is continuous at x = a, then

$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$

$\text { (i) } \lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} x}{x}\right)=1$

$\text { (ii) } \lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x}{x}\right)=1$

Given:

the function

$f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$

is continuous at each point of its domain

Solution:

Step 1: Understand that according to question, f(x) is continuous at x = 0

\begin{aligned} &\therefore \lim f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}=f(0) \end{aligned}

$\lim _{x \rightarrow \infty} \frac{x\left(2-\frac{\sin ^{-1} x}{x}\right)}{x\left(2+\frac{\tan ^{-1} x}{x}\right)}=f(0)$

Simplify further,

$\lim _{x \rightarrow 0} \frac{2-\frac{\sin ^{-1} x}{x}}{2+\frac{\tan ^{-1} x}{x}}=f(0)$

$\frac{2-\lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} x}{x}\right)}{2+\lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x}{x}\right)}=f(0)$

Apply formula (i) and formula (ii)

\begin{aligned} &\frac{2-1}{2+1}=f(0) \\ &f(0)=\frac{1}{3} \end{aligned}

Hence, the correct answer is option (b)