What is $\lim_{t \rightarrow0} \frac{\Gamma(\alpha t)}{\Gamma(t)} (\Gamma$ is the Gamma function)? What is $\lim_{t \rightarrow0} \frac{\Gamma(\alpha t)}{\Gamma(t)} (\Gamma$ is the Gamma function)?
I took the numerator, used the change of coordinates $u = \alpha t$ and got that the limit was $\frac{1}{\alpha}$..This seems wrong but I'm not sure why. Any help would be appreciated!
 A: $$\frac{\Gamma(\alpha t)}{\Gamma(t)}  = \frac1{\alpha} \frac{\alpha t \Gamma(\alpha t)}{t \Gamma(t)} = \frac1{\alpha} \frac{\Gamma(1+\alpha t)}{ \Gamma(1+t)}$$
Now take the limit as $t \to 0$...
A: For arbitrary $a\neq 0$ as
\begin{equation*}
\lim_{z \to 0}\dfrac{\Gamma(a\,z)}{\Gamma(z)}=\lim_{z \to 0}\dfrac{\pi}{\Gamma(1-a\,z)\sin(a\pi\,z)}\times \frac{\Gamma(1-z)\sin(\pi\,z)}{\pi} 
=\frac{1}{a}\lim_{z \to 0}\dfrac{\pi\,a\,z}{\sin(a\pi\,z)}\times\lim_{z \to 0} \frac{\sin(\pi\,z)}{\pi\,z}=\frac1a,
\end{equation*} 
and if you want to avoid 
$$\lim_{t\to 0} \Gamma(1-\alpha t)=\Gamma(1-\lim_{t\to 0} \alpha t)
$$
write
\begin{align*}
\lim_{z \to 0}\dfrac{\Gamma(a\,z)}{\Gamma(z)}&=\lim_{z \to 0}\dfrac{\pi}{\Gamma(1-a\,z)\sin(a\pi\,z)}\times \frac{\Gamma(1-z)\sin(\pi\,z)}{\pi} \\
&=\frac1a \lim_{z \to 0}\frac{\Gamma(-z)}{\Gamma(-az)}
\lim_{z \to 0}\dfrac{\pi}{\sin(a\pi\,z)}\times \frac{\sin(\pi\,z)}{\pi}\\
&=\frac1a \lim_{-z \to 0}\frac{\Gamma(-z)}{\Gamma(-az)}
\lim_{z \to 0}\dfrac{\pi}{\sin(a\pi\,z)}\times \frac{\sin(\pi\,z)}{\pi}\\
\end{align*} 
or
\begin{align*}
\left(\lim_{z \to 0}\dfrac{\Gamma(a\,z)}{\Gamma(z)}\right)^2=\left(\frac1a\right)^2.
\end{align*}
