do this in min 4 different ways

It's a challenge....

my solutions...

#1 using expansion of tan^-1x and sin^-1x and changing sin³x to x³

#2 using LHR

i can't get any thing more.....:(

Or the elementary limits lim x → 0 sinx/x = 1 etc. This is very lengthy though. But still can be done.

\hspace{-16}\mathbf{\lim_{x\rightarrow 0}\left(\frac{\tan^{-1}(x)-\sin^{-1}(x)}{\sin^3 (x)}\right)}\\\\\\ \mathbf{\lim_{x\rightarrow 0}\left(\frac{\tan^{-1}(x)-\sin^{-1}(x)}{x^3 }\right).\lim_{x\rightarrow 0}\left(\frac{x^3}{\sin^3 (x)}\right)}$\\\\\\ $\mathbf{\lim_{x\rightarrow 0}\left\{\left(\frac{\tan^{-1}x-x}{x^3}\right)-\left(\frac{\sin^{-1}x-x}{x^3}\right)\right\}}$\\\\\\ $\mathbf{\lim_{x\rightarrow 0}\left(\frac{\tan^{-1}x-x}{x^3}\right)-\lim_{x\rightarrow 0}\left(\frac{\sin^{-1}x-x}{x^3}\right)}$\\\\\\ Now Put $\mathbf{\tan^{-1}(x)=t\Leftrightarrow x=\tan (t)}$ and Limit $t\rightarrow 0$\\\\ and Put $\mathbf{\sin^{-1}(x)=u\Leftrightarrow x=\sin (u)}$ and Limit $u\rightarrow 0$\\\\\\ $\mathbf{\lim_{t\rightarrow 0}\left(\frac{t-\tan t}{\tan^3 t}\right)-\lim_{u\rightarrow 0}\left(\frac{u-\sin u}{\sin^3 u}\right)=\lim_{t\rightarrow 0}\left(\frac{t-\tan t}{t^3 }\right)-\lim_{u\rightarrow 0}\left(\frac{u-\sin u}{u^3 }\right)}$\\\\\\

\hspace{-16}\mathbf{\lim_{t\rightarrow 0}\left(\frac{t-\tan t}{t^3}\right)-\lim_{t\rightarrow 0}\left(\frac{t-\sin t}{t^3}\right)}$\\\\\\ $\mathbf{\lim_{t\rightarrow 0}\left(\frac{t.\cos t-\sin t}{\cos t.t^3}\right)-\lim_{t\rightarrow 0}\left(\frac{t-\sin t}{t^3}\right)}$\\\\\\ $\mathbf{\lim_{t\rightarrow 0}\left(\frac{t.\cos t-\sin t}{t^3}\right)-\lim_{t\rightarrow 0}\left(\frac{t-\sin t}{t^3}\right)}$\\\\\\ $\mathbf{\lim_{t\rightarrow 0} \frac{t\cos t -t}{t^3}=\lim_{t\rightarrow 0} \frac{\cos t -1}{t^2}=-\frac{1}{2}}$

SIMPLEST :

\lim_{x\rightarrow 0}\frac{\tan^{-1} x-\sin^{-1} x}{x^{3}} = \lim_{x\rightarrow 0}\frac{\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-... \right)-\left(x+\frac{1^{2}}{3!}x^{3}+\frac{1^{2}3^{2}}{5!}x^{5}+... \right)}{x^{3}}

= -\frac{1}{2}