Extract a constant term from a Laurent product

Use $(4-x)^{-1/2}=\tfrac12(1-x/4)^{-1/2}$. The binomial terms give

$\begin{aligned}&(4-x)^{-1/2}\\ & =\frac12\left(1+\frac{x}{8}+\frac{3x^2}{128}+\frac{5x^3}{1024}+\cdots\right)\\ & =\frac12+\frac{x}{16}+\frac{3x^2}{256}\\ &{}+\frac{5x^3}{2048}+\cdots.\end{aligned}$

The printed standard open binomial interval is $|x/4|<1$, or $-4<x<4$.

Multiply by $1+a/x$. The constant terms are $1\cdot(1/2)$ and $(a/x)(x/16)$, so $1/2+a/16=5$ and $a=72$. This is a Laurent coefficient, not evaluation at $x=0$: the product is undefined there. Within the printed open interval its domain is $0<|x|<4$.