<s>[INST] How find this limit $\lim\limits_{x\to 0^{+}}\frac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$ 
Find the limit
$$\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$$

My attempt: Since
$$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+o(x^7)$$
$$\tan{x}=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\dfrac{1}{63}x^7+o(x^3)$$
So
$$\sin{(\tan{x})}=\tan{x}-\dfrac{1}{3!}(\tan{x})^3+\dfrac{1}{5!}(\tan{x})^5-\dfrac{1}{7!}(\tan{x})^7+o(x^7)$$
Though this method might solve, I think this problem has nicer methods. Thanks.
 [/INST] The interesting thing (which I cannot explain) is that if you have two odd functions
$$f(x)=x+a_3x^3+a_5 x^5+a_7 x^7+?x^9,\quad g(x)=x+b_3x^3+b_5 x^5+b_7 x^7+?x^9$$
with $f'(0)=g'(0)=1$ (or $=-1$) then they "commute up to order 5", i.e.,
$$f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)= ?x^7\qquad(x\to0)\ .$$
To prove this we do the computation for $f\bigl(g(x)\bigr)$:
$$\eqalign{f\bigl(g(x)\bigr)&=x+(a_3+b_3)x^3+(a_5+3a_3b_3+b_5)x^5 + \cr &\qquad\qquad(a_7+5a_5b_3+3a_3b_3^2+3a_3b_5+b_7)x^7\ +\ ?x^9\ .\cr}\tag{1}$$
We see that the coefficients of  $x^3$ and $x^5$ both are symmetric in $a$ and $b$, and in addition  $a_7+b_7$ will cancel when forming $f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)$. From inspection of $(1)$ we therefore can deduce that
$$f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)=\bigl(2(a_5 b_3-a_3 b_5)+3(a_3b_3^2-a_3^2 b_3)\bigr)x^7\ +\ ?x^9\ .$$
Inserting here the known coefficients for $\sin$ and $\tan$ we find that the limit in question is $-{1\over30}$.
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