Solve $\lim_{x→0} \frac{\sin^{-1}x-\tan^{-1}x}{x^3}$ Prove that $$\lim_{x→0} \frac{\sin^{-1}x-\tan^{-1}x}{x^3}=\frac{1}{2}$$


$$
\lim_{x→0} \frac{\sin^{-1}x-\tan^{-1}x}{x^3}=\lim_{x→0} \frac{\tan^{-1}\frac{x}{\sqrt{1-x^2}}-\tan^{-1}x}{x^3}
\\
=\lim_{x→0} \frac{\tan^{-1}\bigg[\frac{\frac{x}{\sqrt{1-x^2}}-x}{1+\frac{x^2}{\sqrt{1-x^2}}}\bigg]}{x^3}
=\lim_{x→0} \frac{\tan^{-1}\bigg[\frac{x(1-\sqrt{1-x^2})}{\sqrt{1-x^2}+x^2}\bigg]}{x^3}\\
\text{since, }\frac{x^2}{\sqrt{1-x^2}}&gt;-1
$$

I think I am getting stuck here, do not really see how to proceed further with the steps.

Note: I don't want to use L'Hospital's rule as I 'd like to solve it without using the derivatives of $\sin^{-1}x$ and $\tan^{-1}x$.
\begin{align*}
&amp;\lim_{x\rightarrow 0}\dfrac{\sin^{-1}x-\tan^{-1}x}{x^{3}}\\
&amp;=\lim_{x\rightarrow 0}\dfrac{\sin(\sin^{-1}x-\tan^{-1}x)}{x^{3}}\cdot\dfrac{\sin^{-1}x-\tan^{-1}x}{\sin(\sin^{-1}x-\tan^{-1}x)}\\
&amp;=\lim_{x\rightarrow 0}\dfrac{\sin(\sin^{-1}x-\tan^{-1}x)}{x^{3}}\cdot 1\\
&amp;=\lim_{x\rightarrow 0}\dfrac{\sin(\sin^{-1}x)\cos(\tan^{-1}x)-\cos(\sin^{-1}x)\sin(\tan^{-1}x)}{x^{3}}\\
&amp;=\lim_{x\rightarrow 0}\dfrac{x\cdot\dfrac{1}{\sqrt{1+x^{2}}}-\sqrt{1-x^{2}}\cdot\dfrac{x}{\sqrt{1+x^{2}}}}{x^{3}}\\
&amp;=\lim_{x\rightarrow 0}\dfrac{1}{\sqrt{1+x^{2}}}\cdot\dfrac{1-\sqrt{1-x^{2}}}{x^{2}}\\
&amp;=\lim_{x\rightarrow 0}\dfrac{1-(1-x^{2})}{x^{2}}\cdot\dfrac{1}{1+\sqrt{1-x^{2}}}\\
&amp;=\lim_{x\rightarrow 0}\dfrac{1}{1+\sqrt{1-x^{2}}}\\
&amp;=\dfrac{1}{2}.
\end{align*}
hint

Sorry, i understood that $\sin^{-1}(x)$ means $\frac{1}{\sin(x)}$ instead of $\arcsin(x)$.