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!


$$\frac{\Gamma(\alpha t)}{\Gamma(t)} = \frac1{\alpha} \frac{\alpha t \Gamma(\alpha t)}{t \Gamma(t)} = \frac1{\alpha} \frac{\Gamma(1+\alpha t)}{ \Gamma(1+t)}$$

Now take the limit as $t \to 0$...

  • 1
    $\begingroup$ Ah man I knew this had to be simple. I was way over-thinking this one. I was about to get really fancy in an answer, but this is way more elegant than what I had in mind. $\endgroup$ – Cameron Williams Jun 28 '16 at 18:27
  • $\begingroup$ @CameronWilliams: simple limit problems with simple answers typically have simple solutions. $\endgroup$ – Ron Gordon Jun 28 '16 at 18:28
  • $\begingroup$ Very true! $\,\,$ $\endgroup$ – Cameron Williams Jun 28 '16 at 18:28

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