Why do Mathematica and Wolfram|Alpha say $\Gamma(-\infty)=0$?

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According to Mathematica and Wolfram|Alpha, $\lim_{x\to -\infty}\Gamma(x)$ is equal to zero. See e.g https://www.wolframalpha.com/input/?i=gamma+function (at the bottom of the page), or try Limit[Gamma[x], x -> -Infinity] on Mathematica.

This contradicts the fact that $|\Gamma(x)|$ can be arbitrarily large when x is close enough to any negative integer.

Is Mathematica and Wolfram|Alpha wrong on this, or is there any other way to interpret this result?

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If $x$ is a continuous variable, then $\lim_{x \to -\infty} \Gamma(x) $ does not exist.

But if $-\infty$ is approached in a discrete manner, then the limit might exist.

For example, let's replace $x$ with $N+ \frac{1}{2}$, where $N$ is an integer.

The reflection formula for the gamma function states that $$\Gamma(x) = \frac{\pi \csc(\pi x)}{\Gamma(1-x)} $$ for all $x \notin \mathbb{Z}. $

Therefore, $$ \lim_{N \to - \infty}\Gamma \left(N + \frac{1}{2} \right) = \lim_{N \to -\infty} \frac{\pi \csc \left(\pi \left(N + \frac{1}{2} \right) \right)}{\Gamma \left(\frac{1}{2}-N \right)} = \lim_{N \to -\infty} \frac{\pi \sec \left(\pi N \right)}{\Gamma \left(\frac{1}{2}-N \right)} =0 $$ since $\sec (\pi N)$ is just bouncing between $1$ and $-1$, while $\Gamma \left(\frac{1}{2}-N \right)$ is going to infinity.

EDIT:

In fact, as long as you stay greater than some fixed distance away from the negative integers, the limit will seemingly be zero.