How rapidly does $\Gamma(x)$ diverge as $x$ approaches $0$?

Question

Notoriously $$\lim\limits_{x\to0^{\pm}}\Gamma(x)=\pm\infty,$$ but can we be more precise (tightly) bounding from above $\left\lvert \Gamma(x) \right\rvert$ when $x$ is close to $0$? I could not find anything related, thank you.

Answer 1

According to for instance Wolfram Alpha, one has that when $x$ approaches $0$ $$ \Gamma(x) = \frac{1}{x} - \gamma + o(1) $$ where $\gamma$ is Euler's constant. That is, $\lvert \Gamma(x)\rvert$ behaves asymptotically like $\frac{1}{\lvert x\rvert}$ around $0$.

You can also derive this from the Taylor expansion of the reciprocal Gamma function at $0$.