Tangent of sine minus sine of tangent

Idea. For small $x$, $\tan(\sin x)$ and $\sin(\tan x)$ are both close to $x$, and, surprisingly, they agree up to order $5$. The limit measures how they differ at order $7$. This is a challenge in careful bookkeeping.

Step 1: the building blocks. As $u\to0$,
$$\sin u=u-\frac{u^3}6+\frac{u^5}{120}-\frac{u^7}{5040}+o(u^7),$$
and
$$\tan u=u+\frac{u^3}3+\frac{2u^5}{15}+\frac{17u^7}{315}+o(u^7).$$

Step 2: powers of $\sin x$ and $\tan x$, up to order $7$. Using $\sin x=x-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}$ and $\tan x=x+\frac{x^3}3+\frac{2x^5}{15}+\frac{17x^7}{315}$ and expanding:

For $u=\sin x$:
- $u^3=x^3-\frac{x^5}2+\frac{13x^7}{120}$,
- $u^5=x^5-\frac{5x^7}6$,
- $u^7=x^7$.

For $u=\tan x$:
- $u^3=x^3+x^5+\frac{11x^7}{15}$,
- $u^5=x^5+\frac{5x^7}3$,
- $u^7=x^7$.

Step 3: $\tan(\sin x)$. Put $u=\sin x$ in $u+\frac{u^3}3+\frac{2u^5}{15}+\frac{17u^7}{315}$ and collect:
- order $3$: $-\frac16+\frac13=\frac16$,
- order $5$: $\frac1{120}-\frac16+\frac2{15}=-\frac1{40}$,
- order $7$: $-\frac1{5040}+\frac{13}{360}-\frac19+\frac{17}{315}=-\frac{107}{5040}$.

So $\tan(\sin x)=x+\frac{x^3}6-\frac{x^5}{40}-\frac{107\,x^7}{5040}+o(x^7)$.

Step 4: $\sin(\tan x)$. Put $u=\tan x$ in $u-\frac{u^3}6+\frac{u^5}{120}-\frac{u^7}{5040}$:
- order $3$: $\frac13-\frac16=\frac16$,
- order $5$: $\frac2{15}-\frac16+\frac1{120}=-\frac1{40}$,
- order $7$: $\frac{17}{315}-\frac{11}{90}+\frac1{72}-\frac1{5040}=-\frac{275}{5040}$.

So $\sin(\tan x)=x+\frac{x^3}6-\frac{x^5}{40}-\frac{275\,x^7}{5040}+o(x^7)$.

Step 5: the difference. The terms up to order $5$ cancel, and
$$\begin{aligned} \tan(\sin x)-\sin(\tan x)&=\frac{-107+275}{5040}\,x^7+o(x^7) \\&=\frac{168}{5040}\,x^7+o(x^7)\\&=\frac{x^7}{30}+o(x^7). \end{aligned}$$

Conclusion. The limit is $\dfrac1{30}$.

Remark. This example is famous because a naive approach ("both are $\approx x$") or even a computation up to order $5$ gives $0$. It was used by Arnold to illustrate how deep the order of contact between two functions can be.