I want to find the relation between $\Gamma(\frac{1}{2}-x)$ and $\Gamma(-x)$ for small $x$.
For large $x$, we can use asymptotic expansion, i.e., $\Gamma(x+a) \sim \Gamma(x) x^{a}$, here I am considering small $x$ limit.
For integer difference i.e., $\Gamma(x+n)$ and $\Gamma(x)$, there is a nice relation
\begin{align}
\frac{\Gamma\left(x+n\right)}{\Gamma\left(x\right)}&=\prod_{k=0}^{n-1}{\left(x+k\right)}
\end{align}
which comes from the properties of $\Gamma(x+1)=x\Gamma(x)$.
Best Answer
For small values of $x$ $$\Gamma \left(\frac{1}{2}-x\right)=\sqrt{\pi }+\sqrt{\pi } (\gamma +2\log (2))x+O\left(x^2\right)\tag 1$$ $$\Gamma \left(-x\right)=-\frac{1}{x}-\gamma -\frac{6 \gamma ^2+\pi ^2}{12} x+O\left(x^2\right)\tag2 $$
Solve $(1)$ for $x$ and plug in $(2)$.