Let's say I am looking for a limit $z\rightarrow 0$ of a cumbersome expression such as
$P(z)=\frac{a_1 J_{\alpha}^2(z) + a_2 J_{\beta}^2(z)+ ...}{b_1 J_{\alpha}^2(z) + b_2 J_{\beta}^2(z)+...}$
and I take a limit of the Bessel as $J_n(z) \rightarrow \frac{z^n}{2^nn!}$ for a small argument and get something like
$P(z)=\frac{A_1 z^n + A_2 z^{n-2}+ ...+c1}{B_1 z^n + B_2 z^{n-2}+ ...+c2}$
where $a_i, A_i, b_i, B_i, c1$ and $c2$ are constants. Is it legit to take another limit and write
$P(z) = \frac{c1}{c2}$ ?
Sure it is. What you are doing is you are taking the expansion of the Bessel function around $z=0$. To be more precise, you could say
$$J_n(z)=\frac{z^n}{2^nn!}+O(z^{n\pm1})$$
where $O(\dots)$ refers to big-O-notation. You could think of the "O" as the maximum error between two expressions, one that decreases to $0$ as $x$ approaches some value, $0$ in this case.
Note the equal sign above, as the "O" takes care of the problem of using limits.
Using such notation could allow you to take your limit, so long as you are careful that there is no $\frac1z$ terms that could appear.
However, the result would not be $P(z)\to\frac{c_1}{c_2}$, but rather the numerator and denominator would be off by a constant amount at most. To show $P(z)\to\frac{c_1}{c_2}$, you would need to show the error/O's approach $0$ as $z\to0$. Only then will you have shown the limit is equal to this.