For which $(a,b,n) \in \mathbb{Z}^3$ satisfying $a+b=n$ does $\frac{\Gamma(z+1)}{\Gamma(x+1)\Gamma(y+1)}$ approach a limit as $(x,y,z) \rightarrow (a,b,n)$ in $\mathbb{C}^3$, and what is that limit? (The case where $a$, $b$, and $n$ are all negative is the problematic one.) What if we restrict to $\{(x,y,z) \in \mathbb{C}^3 \ | \ x+y=z\}$?


I'm guessing that where the limit exists it agrees with the "windmill" shown in <https://i.stack.imgur.com/osFsj.png> (taken from page 197 of Hilton, Holton, and Pedersen's "Mathematical Reflections: In a Room with Many Mirrors").



Extended binomial coefficients and the gamma function
There's nothing special about the gamma function; the failure of the limit to exist when $a$, $b$, and $n$ are negative is exactly the same as the failure of
$$\lim\_{(x,y,z)\to(0,0,0)} \frac{xy}{z}$$ to exist.


I will renormalize in a way that I think makes clearer what's going on. First let's look at
$$\lim\_{(x,y,z)\to(0,0,0)} \frac{\Gamma(z)}{\Gamma(x)\Gamma(y)}.$$
We have
$$\frac{\Gamma(z)}{\Gamma(x)\Gamma(y)} = \frac{xy}{z}\frac{\Gamma(z+1)}{\Gamma(x+1)\Gamma(y+1)}.$$
Since $\Gamma(z+1)/\Gamma(x+1)\Gamma(y+1)$ is analytic and nonzero near $(0,0,0)$, $\Gamma(z)/\Gamma(x)\Gamma(y)$ behaves just like $xy/z$ near $(0,0,0)$. In particular, if you restrict to $x+y=z$, this is $xy/(x+y)$, for which the limit as $(x,y)\to(0,0)$ does not exist.


Now let's consider
$$\lim\_{(x,y,z)\to(0,0,0)} \frac{\Gamma(z-a)}{\Gamma(x-b)\Gamma(y-c)}$$ where $a$, $b$, and $c$ are nonnegative integers.
We have
$$\frac{\Gamma(z-a)}{\Gamma(x-b)\Gamma(y-c)} = \frac{\Gamma(z)}{\Gamma(x)\Gamma(y)}\cdot
\frac{(x-b)\cdots (x-1)(y-c)\cdots (y-1)}{(z-a)\cdots(z-1)}.$$
The second factor on the right is analytic and nonzero near $(0,0,0)$, so $\Gamma(z-a)/\Gamma(x-b)\Gamma(y-c)$ behaves just like $\Gamma(z)/\Gamma(x)\Gamma(y)$ (and thus like $xy/z$) near $(0,0,0)$.


