Gamma function approximation

Question

I found this approximation online:

When $h \to 0$, $\dfrac{\Gamma(h+\alpha)\cdot h}{\Gamma(h+1)}\sim \Gamma(\alpha)\cdot h$

How is this so, please? Thanks.

Answer 1

Since $ x \mapsto \Gamma(\cdot)$ is a continuous function, as $h \to 0$, one has $$ \Gamma(h+\alpha) \sim \Gamma(0+\alpha)=\Gamma(\alpha), \quad \Gamma(h+1)\sim \Gamma(1)=1 $$ giving $$ \dfrac{\Gamma(h+\alpha)\cdot h}{\Gamma(h+1)}\sim \dfrac{\Gamma(0+\alpha)\cdot h}{\Gamma(0+1)}=\Gamma(\alpha)\cdot h $$

Answer 2

$\displaystyle\lim_{h\to 0}\dfrac{\Gamma(h+\alpha) \;h}{\Gamma(h+1)\;\Gamma(\alpha)\; h}=\lim_{h\to 0}\dfrac{\Gamma(h+\alpha) }{\Gamma(h+1)\;\Gamma(\alpha)}\underbrace{=}_{\Gamma\text{ continuous}}\frac{\Gamma(\alpha)}{\Gamma(1)\Gamma(\alpha)}=1.$