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. 
 A: 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
$$
\lim_{x\to 0}\left(\frac{\sin(\arcsin(x)-\arctan(x))}{x^3}\right)\equiv\lim_{x\to 0}\left(\frac{x^3}{\sin(x^3)}\right)\frac{1}{(1+\sqrt{1-x^2})\sqrt{1+x^2}} = \frac{1}{2}
$$
A: Using Maclaurin series:
$$\arcsin x-\arctan x=x+\frac16x^3+\frac3{40}x^5+\ldots-\left(x-\frac13x^3+\frac15x^5-\ldots\right)=$$
$$=\frac12x^3+\mathcal O(x^5)$$
so
$$\lim_{x\to0}\frac{\arcsin x-\arctan x}{x^3}=\lim_{x\to0}\frac{\frac12x^3+\mathcal O(x^5)}{x^3}=\frac12$$
A: This is easy!
Lets start by noting that $$arcsin(x)=x+x^3/6+3x^5/40+...$$
Next $$arctan(x)=x-x^3/3+x^5/5-...$$
When this is substituted in you limit equation above we have for the numerator:
$$x^3/2-x^5/8 ...$$
When this is divided by $x^3$ and we pass to the limit we get the simple result of 1/2.
