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.

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 $$\frac{a-b}{c-d}-\frac{x^2 \left((a-b) \left(a^2+a b+b^2+c^2+c d+d^2\right)\right)}{6
   (c-d)}+O\left(x^4\right)$$
