Let L=lim x→0(tan (sin x)-sin (tan x))/(x^7. Find 1/L.

Expand with Maclaurin series. $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}$ and $\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}$.
Composing, both $\tan(\sin x)$ and $\sin(\tan x)$ equal $x+\frac{x^3}{6}-\frac{x^5}{40}+\dots$ up to the $x^5$ term, so their difference starts at $x^7$. Carrying the composition to order $7$ gives
$$\tan(\sin x)=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107x^7}{5040}+\dots,\qquad \sin(\tan x)=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55x^7}{1008}+\dots$$
Subtracting, the coefficient of $x^7$ is $-\frac{107}{5040}+\frac{275}{5040}=\frac{168}{5040}=\frac{1}{30}$ (confirmed by computer algebra). Since higher terms vanish in the limit, $L=\frac1{30}$ and $\frac1L=30$. Using L'Hopital seven times would be far longer, which is why the series route is preferred here.