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

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

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    Actually, It is not full solution, because x and y are not natural number. Thus third line cannot be true. – 2017-02-22
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    @marka_17 This relations also holds for real numbers. It is only problematic if one of the factors in the denominator should happen to be zero, which can be excluded since we are only interested in large values of x. – 2017-02-22