I was playing around with $\binom{x}{k}$ for fixed $x$ as $k\to\infty$ and realized it could be compared with the Weierstrass factorization of the reciprocal $1/\Gamma(x)$ to get $\binom{x}{k}\sim k^{-(x+1)}/\Gamma(-x)$. (And apparently Euler's historical definition of the Gamma function was more along these lines.) It's natural to ask if we could extend this to an asymptotic series, though maybe it's more natural to ask for an asymptotic series of its logarithm.
From my numerical experiments, I suspect there is an asymptotic series
$$ \ln\binom{x}{k}~\sim\, -(x+1)\ln k\,-\,\ln\Gamma(-x)\,+\,\sum_{r=1}^\infty \frac{P_r(x+\tfrac{1}{2})}{k^r} $$
where $P_r$ is a polynomial of degree $r+1$ with the same parity as $r+1$ (that is, it is an even function if $r$ is odd, or vice-versa), when $P_1(x+\frac{1}{2})=\tfrac{1}{2}(x+\tfrac{1}{2})^2-\tfrac{1}{8}$ in particular. I don't have the computational power to guess what $P_2(x+\tfrac{1}{2})$ ought to be. Is this series known, or can someone find it with e.g. Euler-Maclaurin? (Perhaps a better version would replace the denominator $k^r$ with rising or falling factorials of $k$.)
Assume that $x$ is not a non-negative integer. Note that $$ \binom{x}{k} = ( - 1)^k \frac{{\Gamma (k - x)}}{{\Gamma ( - x)\Gamma (k + 1)}}. $$ Thus by the general asymptotic expansion for the gamma function, we find \begin{align*} \ln \left|\binom{x}{k} \right| & = - \ln \left| {\Gamma ( - x)} \right| + \ln \Gamma (k - x) - \ln \Gamma (k + 1) \\ & \sim - (x + 1)\ln k - \ln \left| {\Gamma ( - x)} \right| + \sum\limits_{j = 2}^\infty {\frac{{( - 1)^j (B_j ( - x) - B_j (1))}}{{j(j - 1)}}\frac{1}{{k^{j - 1} }}} \end{align*} as $k\to +\infty$, where $B_j(z)$ are the Bernoulli polynomials.
For the corresponding expansion(s) for $\binom{x}{k}$, see page 35 of Yudell L. Luke's book The Special Functions and Their Approximations: Volume 1.