How to find the following limit $\lim_{x\to+\infty}x^{\alpha} B(\alpha, x), \hspace{20pt} \alpha > 0 $

Question

How to find the following limit $$\lim_{x\to+\infty}x^{\alpha} B(\alpha, x), \hspace{20pt} \alpha > 0, \alpha, x \in \mathbb{R} $$

Answer 1

We use the following definition of gamma and beta functions: For positive reals $x, y$ $$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ $$\Gamma(x) = (x-1)\Gamma(x-1)$$ Observe that the ratio: $$\frac{\Gamma(x)}{\Gamma(x+y)} = \frac{1}{(x+y-1)(x+y-2)\ldots (x)}$$ With this, we have \begin{align*} x^\alpha B(\alpha, x) &= x^\alpha \left(\frac{\Gamma(x)}{\Gamma(x+\alpha)}\right)\Gamma(\alpha) \\ &= \left(\frac{x^\alpha}{(x+\alpha-1)(x+\alpha-2)\ldots (x)}\right)\Gamma(\alpha) \\ &= \frac{\Gamma(\alpha)}{(1+\frac{\alpha-1}{x})(1+\frac{\alpha-2}{x})\ldots (1)} \\ \end{align*}

Therefore, $$\lim_{x \rightarrow \infty} x^\alpha B(\alpha, x) = \Gamma(\alpha)$$