# Limit of Gamma function

Compute $$\lim_{z\to 0}z^3\Gamma(z)\Gamma(z-1)\Gamma(z-2)$$.

Idea: Note that by the functional relation of $$\Gamma$$ we have that $$\Gamma(z+1)=z\Gamma(z)$$. Hence, $$\Gamma(z)=(z-1)(z-2)\Gamma(z-2), \Gamma(z-1)=(z-2)\Gamma(z-2)$$ and so plugging into our original expression we get:

$$$$\lim_{z\to 0}z^3(z-1)(z-2)^2\Gamma(z-2).$$$$

I'm stuck here - I know that there are simple poles at the negative integers for $$\Gamma(z)$$, so we have simple poles for the negative integers, and $$0,1,2$$ for $$\Gamma(z-2)$$. Even still, I don't know how to get around this problem- we cannot simply plug in $$z=0$$.

Use the functional equation $$z \Gamma(z) = \Gamma(z+1)$$ in reverse: \eqalign{\Gamma(z) &= \frac{\Gamma(z+1)}{z}\cr \Gamma(z-1) &= \frac{\Gamma(z)}{z-1} = \frac{\Gamma(z+1)}{z(z-1)} \cr \Gamma(z-2) &= \frac{\Gamma(z-1)}{z-2} = \frac{\Gamma(z+1)}{z(z-1)(z-2)}} so $$z^3 \Gamma(z) \Gamma(z-1) \Gamma(z-2) = \frac{\Gamma(z+1)^3}{(z-1)^2(z-2)}$$ Now take the limit as $$z \to 0$$, knowing $$\Gamma$$ is continuous at $$1$$ with $$\Gamma(1) = 1$$.