How find this strange limit find this follow strange limit

$$\lim_{x\to 0}\dfrac{\arcsin{\arctan{\sin{\tan{\arcsin{\arctan{x}}}}}}-\sin{\tan{\arcsin{\arctan{\sin{\tan{x}}}}}}}{\arctan{\arcsin{\tan{\sin{\arctan{\arcsin{x}}}}}}-\tan{\sin{\arctan{\arcsin{\tan{\sin{x}}}}}}}\cdots (1)$$

and I think this limit is $1$,But I can't have proof.
so I have found this follow  same limit(But this is not hard)

$$\lim_{x\to0}\dfrac{\sin{\tan{x}}-\tan{\sin{x}}}{\arcsin{\arctan{x}}-\arctan{\arcsin{x}}}=1$$

and this limit is very famous,and this problem have two methods(maybe have more methods)
this proof [1] can see:http://math.berkeley.edu/~giventh/lim.pdf
and the proof [2] can see http://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%202e/upfiles/instructor/ess_wp_0807a_inst.pdf
and the proof [3]   can see: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=360183
Now  $(1)$  How prove it.Thank you .
 A: It suffices to know that $$\begin{align}\sin(x)&=x-\tfrac16x^3+O(x^4)\\ \tan(x) &= x+\tfrac13x^3+O(x^4)\\\arcsin(x)&=x+\tfrac16x^3+O(x^4)\\\arctan(x) &= x-\tfrac13x^3+O(x^4)\end{align}$$ and observe that $f(x)=x+ax^3+O(x^4)$, $g(x)=x+bx^3+O(x^4)$ implies that $f(g(x))=x+(a+b)x^3+O(x^4)$. [This also shows why the inverse functions have the coefficient of $x^3$ negated.]
Especially, when working only modulo $O(x^4)$ composition becomes commutative and most of the $\sin$ and $\tan$ cancel against $\arcsin$ and $\arctan$, respectively.
This way, the numerator becomes $$ x+(\tfrac16-\tfrac13)x^3 - x- (\tfrac13-\tfrac16)x^3+O(x^4)=-\tfrac13x^3+O(x^4)$$
[we only need that it is not $0+O(x^4)$ accidentally] and the denominator the same (by commutativity!), hence the quotient is 
$$ 1+O(x)$$
and hence the limit as $x\to 0$ is $1$.
The result does not change when you add more complextity by adding additional rounds in the same pattern.

Remark: It is not necessariy for finding the limit, but more precisely the quotient is $1-\tfrac 12x^6+O(x^8)$.
In a more abstract formulation, I used this: The set of germs of functions $f$ analytic at $0$ with $f(0)=0$, $f'(0)=1$, $f''(0)=0$ form a group under composition and $f\mapsto f'''(0)$ is a group homomorphism to $\mathbb R$. A similar statement does not hold if we increase precision in order to handle more than one nontrivial coefficient.
