<s>[INST] Evaluation of $\lim _{ x\rightarrow 0 }{ \frac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } }{ { x }^{ 3 } } } \quad $ without using L'Hospital rule My aim is to evaluate the following limit

$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x }  }{ { x }^{ 3 } }  } \quad$ 

which evaluates to $\frac{1}{2}$ after using L'Hospital rule for rather $5$ steps, which of course is a tedious task. Could anyone suggest some shorter and elegant methods for evaluating this limit?
Thanks. 
 [/INST] We have
$$
\sin(\arcsin(x)-\arctan(x))=\frac{x(1-\sqrt{1-x^2})}{\sqrt{1+x^2}}
$$
and
$$
\lim_{x\to 0}\left(\frac{\arcsin(x)-\arctan(x)}{x^3}\right)\equiv \lim_{x\to 0}\left(\frac{\sin(\arcsin(x)-\arctan(x))}{\sin(x^3)}\right)
$$
then
$$
\frac{\sin(\arcsin(x)-\arctan(x))}{\sin(x^3)} = \frac{x^3(1-\sqrt{1-x^2})}{x^2\sqrt{1+x^2}\sin(x^3)} = \left(\frac{x^3}{\sin(x^3)}\right)\frac{x^2}{x^2(1+\sqrt{1-x^2})\sqrt{1+x^2}}
$$
then
$$
\lim_{x\to 0}\left(\frac{\sin(\arcsin(x)-\arctan(x))}{x^3}\right)\equiv\lim_{x\to 0}\left(\frac{x^3}{\sin(x^3)}\right)\frac{1}{(1+\sqrt{1-x^2})\sqrt{1+x^2}} = \frac{1}{2}
$$
</s>