$\underset{x\to 0}{\lim }\frac{{\sin }^{-1}x - {\tan }^{-1}x}{x^3}=$

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }\frac{{\sin }^{-1}\mathrm{x} - {\tan }^{-1}\mathrm{x}}{{\mathrm{x}}^3} \left(\frac{0}{0} \mathrm{form}\right) \\ \underset{\mathrm{x}\to 0}{\lim } \frac{{\displaystyle \frac{1}{\sqrt{1-{\mathrm{x}}^2}}}-{\displaystyle \frac{1}{1+{\mathrm{x}}^2}}}{3{\mathrm{x}}^2} \\ \underset{\mathrm{x}\to 0}{\lim } \frac{{(1-{\mathrm{x}}^2)}^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-{(1+{\mathrm{x}}^2)}^{-1}}{3{\mathrm{x}}^2} \left(\frac{0}{0} \mathrm{form}\right) \\ \left(1\right) \underset{\mathrm{x}\to 0}{\lim } \frac{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} {(1-{\mathrm{x}}^2)}^{{\displaystyle \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}(-2\mathrm{x})+{(1+{\mathrm{x}}^2)}^{{\displaystyle -2}}\left(2\mathrm{x}\right)}{6\mathrm{x}} \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{+{(1-{\mathrm{x}}^2)}^{{\displaystyle \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+2{(1+{\mathrm{x}}^2)}^{{\displaystyle -2}}}{6} \\ =\frac{1+2}{6}=\frac{1}{2} \end{gathered}$$