An almost impossible limit The following limit appeared in a qualification exam: Find the limit of
$$\lim_{x \to 0} \left( \frac{\tan (\sin (x))-\sin (\tan (x))}{x^7} \right).$$
I ended up doing it in Mathematica, is there any other way?
Thanks in advance!

Recall that 
$$\sin(x) = x-x^3/6+x^5/120-x^7/5040+O(x^9)$$
$$\tan(x)=x+x^3/3+2x^5/15+17x^7/315+O(x^9)$$
thus
$$\begin{align}
\tan(\sin(x)) &= x-x^3/6+x^5/120+17x^7/315+O(x^9)\\
 &\;+\;(x-x^3/6+x^5/120+17x^7/315+O(x^9))^3/3\\
 &\;+\;2(x-x^3/6+x^5/120+17x^7/315+O(x^9))^5/15\\
 &\;-\;(x-x^3/6+x^5/120+17x^7/315+O(x^9))^7/5040+O(x^9)\\
&= x-x^3/6+x^5/120+17x^7/315+O(x^9)\\
 &\;+\;(x\cdot x \cdot x+3x\cdot x\cdot -x^3/6+3x\cdot (-x^3/6)^2+3x\cdot x\cdot x^5/120 + O(x^9))/3\\
 &\;+\;2(x\cdot x\cdot x\cdot x\cdot x + 5x\cdot x\cdot x\cdot x\cdot -x^3/6 + O(x^9))/15\\
 &\;-\;(x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x + O(x^9))/5040\\
&= x + x^3/6-x^5/40-107x^7/5040 + O(x^9)
\end{align}$$
which agrees with WolframAlpha's result. Similarly we can compute
$$\sin(\tan(x))=x + x^3/6-x^5/40-55x^7/1008 + O(x^9)$$
and so
$$\begin{align}
\lim_{x\to 0}\frac{\tan(\sin(x))-\sin(\tan(x))}{x^7}
&=\lim_{x\to 0}\frac{-107x^7/5040 + 55x^7/1008+O(x^9)}{x^7}\\
&=\frac{-107}{5040}+\frac{55}{1008}=\frac{1}{30}
\end{align}$$
