If the function
$f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$
is continuous at each point of its domain, then the value of f (0) is
(a) 2
(b) $\frac{1}{3}$
(c) $-\frac{1}{3}$
(d) $\frac{2}{3}$
(b) $\frac{1}{3}$
Given: $f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$
If $f(x)$ is continuous at $x=0$, then
$\lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{x\left(2-\frac{\sin ^{-1} x}{x}\right)}{x\left(2+\frac{\tan ^{-1} x}{x}\right)}=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\left(2-\frac{\sin ^{-1} x}{x}\right)}{\left(2+\frac{\tan -1}{x}\right)}=f(0)$
$\Rightarrow \frac{2-\lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} x}{x}\right)}{2+\lim _{x \rightarrow 0}\left(\frac{\operatorname{tax}-1}{x}\right)}=f(0)$
$\Rightarrow \frac{2-1}{2+1}=f(0)$
$\Rightarrow f(0)=\frac{1}{3}$