$\lim\limits_{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{x^3}$ is equal to

1/2

$\lim\limits_{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{x^3}$                      [$\frac{0}{0}$ form]

$=\lim\limits_{x \rightarrow 0} \frac{\frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x^2}}{3 x^2}$              (L’ Hospital rule)

$=\lim\limits_{x \rightarrow 0}\left[\frac{\left(1+x^2\right)-\sqrt{1-x^2}}{3 x^2\left(1+x^2\right) \sqrt{1-x^2}} \times \frac{\left(1+x^2\right)+\sqrt{1-x^2}}{\left(1+x^2\right)+\sqrt{1-x^2}}\right]$

$=\lim\limits_{x \rightarrow 0}\left[\frac{\left(1+x^2\right)^2-\left(1-x^2\right)}{3 x^2\left((1+x^2)^2 \sqrt{1-x^2}+\left(1+x^2\right)\left(1-x^2\right)\right)}\right]$

$=\lim\limits_{x \rightarrow 0}\left[\frac{\left(3+x^2\right)}{3\left(\left(1+x^2\right)^2 \sqrt{1-x^2}+\left(1-x^4\right)\right)}\right]=\frac{1}{3}\left(\frac{3}{2}\right)=\frac{1}{2}$

Hence (4) is the correct answer.