Is really :$\lim _{x\to -n } ( x+n)\Gamma(x)=0 $, with $n\in \mathbb{N}$?

Question

I have accrossed this limit :$\lim _{x\to -n } ( x+n)\Gamma(x)=0 $, with $n\in \mathbb{N}$ , in my textbook When i read some properties about Gamma function at negative values , really i don't succed to show how this limit is 0 as it were assumed by Wolfram alpha , then my question here is:

Question: Is really :$\lim _{x\to -n } ( x+n)\Gamma(x)=0 $, with $n\in \mathbb{N}$ ?

Answer 1

No. The limit is $\frac{(-1)^n}{n!}$. This follows from the fact that$$(z+n)\Gamma(z)=\frac{\Gamma(z+n+1)}{z(z+1)\ldots(z+n-1)}.$$