I am trying to evaluate a complicated limit, for which I have tried all the standard rules but to no avail (L'Hopitals, etc.). From numerical implementation, I think that the limit should be equal to $0$, but I have no idea how to approach the problem:
$$ \lim_{x \to \infty} \frac{\alpha^x e^{-\alpha x}(\beta x)^x}{\Gamma(x+1)}, $$ where $\alpha>0$ and $0 \leq\beta\leq 1$.
Thanks in advance for any tips.
On
Using Stirling's formulas for $\Gamma(x)$ large values of $x$ gives
$$ \lim_{x \to \infty} \frac{\alpha^x e^{-\alpha x}(\beta x)^x}{\Gamma(x+1)}=\lim_{x\rightarrow\infty}\frac{e^{x(\log(\alpha\beta) -\alpha +1)}}{\sqrt{2\pi x}} $$
All then depend on the reactions sign of $\log(\alpha\beta) -\alpha +1)$. This seems to at most zero (but I have not write down the details on paper yet) in which case the limit as you suspected is $0$.
Here is a picture to convince yourself that the exponent in the last limit is indeed at most $0$
By Stirling,
$$\frac{(\alpha\beta n)^ne^{-n\alpha}}{n!}\sim(\alpha\beta n)^ne^{-n\alpha}\frac{e^n}{\sqrt{2\pi n}n^n}=\frac{e^{n(1-\alpha+\log(\alpha\beta))}}{\sqrt{2\pi n}}.$$
The limit can be $0$ or $\infty$ depending on the sign of $1-\alpha+\log(\alpha\beta)$, which you have to discuss (this is relatively easy).