If the function

(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

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

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

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

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

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

$\Rightarrow \frac{2-1}{2+1}=f(0)$

$\Rightarrow f(0)=\frac{1}{3}$