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For this question, perhaps it is possible to prove rigorously prove that the answer is no, with Liouville-Ritt elementary functions. But so far I was unable to prove that $W_j$ is non-elementary as a function of, say $a_1$, when $n\geq 3$. (When $n\leq 3$ it is elementary, and it was known that the mean motion has an elementary expression when $n\leq 3$ (Bohl).

For this question, perhaps it is possible to prove rigorously that the answer is no, with Liouville-Ritt elementary functions. But so far I was unable to prove that $W_j$ is non-elementary as a function of, say $a_1$, when $n\geq 3$. (When $n\leq 3$ it is elementary, and it was known that the mean motion has an elementary expression when $n\leq 3$ (Bohl).

Let $f(z,a)$ be an elementary function of two variables. Is it true that $g(a)=\lim_{z\to z_0} f(z,a)$ is an elementary function of $a$, provided that the limit exists for all $a$ on some interval?

With "purely real elementary functions", in the spirit of your definition, the answer is still no, if we allow the $\arcsin{}$ in your definition to have the domain $(-1,1)$. Then all elementary functions are analytic (as compositions of analytic functions), but we have a non-analytic limit $$\lim_{k\to\infty}\arctan(kx)=\frac{\pi}{2}{\mathrm{sgn}}(x),\quad \lim_{n\to\infty}(1+x^n)^{1/n}.$$ If you allow $\arcsin$ on $[-1,1]$ then elementary functions can be discontinuous like $\arcsin(\sin x)$, and the above counterexample does not work.

Let $f(x,a)$ be an elementary function of two variables. Is it true that $g(a)=\lim_{x\to x_0} f(x,a)$ is an elementary function of $a$, provided that the limit exists for all $a$ on some interval?

With "purely real elementary functions", in the spirit of your definition, the answer is still no, if we allow the $\arcsin{}$ in your definition to have the domain $(-1,1)$. Then all elementary functions are analytic (as compositions of analytic functions), but we have a non-analytic limit: ($x\to+\infty$ along the positive ray) $$\lim_{x\to+\infty}\arctan(ax)=\frac{\pi}{2}{\mathrm{sgn}}(a),\quad \lim_{x\to+\infty}(1+a^x)^{1/x}.$$ If you allow $\arcsin$ on $[-1,1]$ then elementary functions can be discontinuous like $\arcsin(\sin x)$, and the above counterexample does not work.

With "purely real elementary functions", in the spirit of your definition, the answer is still no, if we allow the $\arcsin{}$ in your definition to have the domain $(-1,1)$. Then all elementary functions are analytic (as compositions of analytic functions), but we have a non-analytic limit $$\lim_{k\to\infty}\arctan(kx)=\frac{\pi}{2}{\mathrm{sgn}}(x).$$ If you allow $\arcsin$ on $[-1,1]$ then elementary functions can be discontinuous like $\arcsin(\sin x)$, and the above counterexample does not work.

With "purely real elementary functions", in the spirit of your definition, the answer is still no, if we allow the $\arcsin{}$ in your definition to have the domain $(-1,1)$. Then all elementary functions are analytic (as compositions of analytic functions), but we have a non-analytic limit $$\lim_{k\to\infty}\arctan(kx)=\frac{\pi}{2}{\mathrm{sgn}}(x),\quad \lim_{n\to\infty}(1+x^n)^{1/n}.$$ If you allow $\arcsin$ on $[-1,1]$ then elementary functions can be discontinuous like $\arcsin(\sin x)$, and the above counterexample does not work.