Solving the limit of $\lim_{x\to 1-n} (\exp(2 \pi i x)-1)\Gamma(x)$

Question

I am reading the book Introduction to Cyclotomic Fields. I found the following limit on page number 33. I have no idea how to obtain the following limit.

$$\lim_{x\to 1-n} (e^{(2 \pi i x)}-1)\Gamma(x)=\frac{(2\pi i)(-1)^{(n-1)}}{(n-1)!}.$$

Do we need to use L'Hospital's rule for this? Is there any connection with the residue of Gamma function at negative values?

The residue at $z=k$ is given by $\operatorname{Re}s_{z=k} \Gamma(z)=\frac{(-1)^{k}}{k!}.$

Please help me to understand this.

Answer 1

Start:

Note that if $f(x)=e^{2\pi i x}$ then $f(1-n)=1$ when $n$ is an integer, and you can write your limit is $$\lim_{x\to 1-n} \frac{f(x)-f(1-n)}{x-(1-n)} \cdot (x-(1-n)) \Gamma(x)=f'(1-n) \cdot \lim_{n\to 1-n}(x-(1-n))\Gamma(x)$$

But $f'(1-n)=2\pi i$, so you need to show:

$$\lim_{x\to 1-n} (x-(1-n))\Gamma(x)=\frac{(-1)^{n-1}}{(n-1)!}$$

This follows from your residue result (corrected to be at $z=-k$,) if you can show that the pole at $z=1-n$ is of degree $1.$