Limit of $\frac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}$ as $x \rightarrow 0$ Find $\lim_{x \to 0} \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}$
I came across this limit a long time ago and could easily obtain a straightforward solution by finding the asymptotic expansion. But since the limit turns out to be nice despite the messy coefficients, I'm just curious if there is some reason other than just coincidental coefficients.
 A: 
Proposition. Consider two odd functions $f$ and $g$ defined in a neighborhood
  $(-\epsilon,\epsilon)$, of $0$. Suppose that $f$ and $g$ have the
  following asymptotic expansions $$ \eqalign{f(x)&=x+\alpha x^3+\beta
x^5+\gamma x^7+{\cal O}(x^9)\cr g(x)&=x+\alpha' x^3+\beta' x^5+\gamma'  x^7+{\cal O}(x^9) } $$ Then, if$~~ $
  $\color{red}{3\alpha\alpha'(\alpha'-\alpha)\ne 2(\beta'\alpha-\beta\alpha')}$,$~$ we have 
  $$\color{blue}{ \lim_{x\to 0}\frac{f\circ
g(x)-g\circ f(x)}{f^{-1}\circ g^{-1}(x)-g^{-1}\circ f^{-1}(x)}=1.} $$

Proof. 
$\qquad$ Note that the assumptions on $f$ and $g$ insure that they define invertible functions in a neighborhood of $0$, (may be smaller than a $(-\epsilon,\epsilon)$). Moreover, $f^{-1}$ and $g^{-1}$ have asymptotic expansion in the neighborhood of $0$ to the same order as $f$ and $g$ respectively. This expansion, can be calculated using the the method of undetermined coefficients. We find
$$ \eqalign{f^{-1}(x)&=x-\alpha x^3+(3\alpha^2-\beta)
x^5+(-12\alpha^3+8\alpha\beta-\gamma) x^7+{\cal O}(x^9)\cr 
g^{-1}(x)&=x-\alpha' x^3+(3\alpha'^2-\beta')
x^5+(-12\alpha'^3+8\alpha'\beta'-\gamma') x^7+{\cal O}(x^9) }$$
Now, it is easy to find
that
$$ f\circ g(x) =x+(\alpha+\alpha') x^3+(3\alpha\alpha'+\beta+\beta')
x^5+(5\alpha'\beta+3\alpha(\alpha'^2+\beta')+\gamma+\gamma') x^7+{\cal O}(x^9)
$$
So subtracting the similar formula for $g\circ f$ we obtain
$$
f\circ g(x)-g\circ f(x)=(3\alpha\alpha'(\alpha'-\alpha)- 2(\beta'\alpha-\beta\alpha'))x^7
+{\cal O}(x^9)\tag{1}
$$
Similarly, (but harder), we find
$$ f^{-1}\circ g^{-1}(x) =x-(\alpha+\alpha') x^3+(3(\alpha^2+\alpha\alpha'+\alpha'^2)-\beta-\beta')
x^5+(-12\alpha^3-12\alpha'^3-15\alpha^2\alpha'-12\alpha\alpha'^2+5\alpha'\beta+3\alpha\beta'+8\alpha\beta+8\alpha'\beta'-\gamma-\gamma') x^7+{\cal O}(x^9)
$$
So subtracting the similar formula for $g^{-1}\circ f^{-1}$ we obtain
$$
f^{-1}\circ g^{-1}(x)-g^{-1}\circ f^{-1}(x)=(3\alpha\alpha'(\alpha'-\alpha)- 2(\beta'\alpha-\beta\alpha'))x^7
+{\cal O}(x^9)\tag{2}
$$
Now the result follows by comparing $(1)$ and $(2)$.$\qquad\square$
A: (this is the answer I gave in the comments below the question. I guess a nicer solution exists.)
If $f,g$ are say analytic functions of one variable such that $f(0)=g(0)=0$ and $f'(0)=g'(0)=1$, and $f\neq g^{-1}$ (the Taylor series of $f^{-1}$ and $g$ are not equal)  then
$$\lim_{x\to 0}\frac{f^{-1}(x)-g(x)}{g^{-1}(x)-f(x)}=1,$$
hence the answer is $1$.
To see it (the best way is with pictures, but formulas will do):
Since $f(x)=x+O(x^2)$ we have
$$\lim_{x\to 0}\frac{f^{-1}(x)-g(x)}{f^{-1}(f(x))-g(f(x))}=1$$ as the Taylor series of $f^{-1}(x)-g(x)$ and $f^{-1}(f(x))-g(f(x))$ have the same leading term $ax^n$ (whatever $a$ and $n$ are).
On the other hand
$$\lim_{x\to 0}\frac{g(g^{-1}(x))-g(f(x))}{g^{-1}(x)-f(x)}=1,$$
because $g'(0)=1$ and $g^{-1}(x),f(x)\to 0$ as $x\to 0$.
Now just use $f^{-1}(f(x))=g(g^{-1}(x))=x$.
A: I suppose and think that this is really related to the coincidental coefficients. We can show it building for each piece a Taylor expansion built at $x=0$. As results, we have 
$$\tan ^{-1}\left(\sin ^{-1}(x)\right)=x-\frac{x^3}{6}+\frac{13 x^5}{120}-\frac{173 x^7}{5040}+O\left(x^{9}\right)$$
$$\sin ^{-1}\left(\tan ^{-1}(x)\right)=x-\frac{x^3}{6}+\frac{13 x^5}{120}-\frac{341 x^7}{5040}+O\left(x^{9}\right)$$
$$\tan (\sin (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+O\left(x^{9}\right)$$
$$\sin (\tan (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+O\left(x^{9}\right)$$ As a result $$ \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))} \approx 1$$ In my opinion, it is quite important to notice for the terms appearing in numerator and denominator the identity of the first three terms (this is why a fourth term had to be introduced in order to avoid a $\frac{0}{0}$ situation).
Pushing the expansion to much higher orders would give
$$ \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}=1-\frac{5 x^2}{3}+\frac{3937 x^4}{1890}-\frac{24779 x^6}{11907}+O\left(x^8\right)$$ If we modify the problem to $$ \dfrac{\tan^{-1}(\sin^{-1}(a x))-\sin^{-1}(\tan^{-1}(a x))}{\tan(\sin(b x))-\sin(\tan(b x))}$$ the result of the expansion would be $$\frac{a^7}{b^7}-\frac{5 x^2 \left(13 a^9+29 a^7 b^2\right)}{126
   b^7}+O\left(x^4\right)$$ For a completely general formulation such as  $$ \dfrac{\tan^{-1}(\sin^{-1}(a x))-\sin^{-1}(\tan^{-1}(b x))}{\tan(\sin(c x))-\sin(\tan(d x))}$$ ($a,b,c,d$ being different) the result of the expansion would be $$\frac{a-b}{c-d}-\frac{x^2 \left((a-b) \left(a^2+a b+b^2+c^2+c d+d^2\right)\right)}{6
   (c-d)}+O\left(x^4\right)$$
