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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:

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

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 $z=0$.

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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 $\Gamma(1) = 1$.

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