Why is $\lim\limits_{x \to \infty} \frac{(\sqrt{x})!}{(\sqrt{x+1})!}$ = 1 and not 0?

Question

Wouldn't we expect it to be equal to 0? I suspect that $\lim\limits_{x \to \infty} \frac{1}{\sqrt{x+1}}$ should be equal to $\lim\limits_{x \to \infty} \frac{(\sqrt{x})!}{(\sqrt{x+1})!}$. And the limit of the former is 0. Why is this not the case for the latter?

Answer 1

Note that $(\sqrt{x+1})!\neq \sqrt{(x+1)!}$. In the former case note that $$ \sqrt {x + 1} = \sqrt x \sqrt {1 + \frac{1}{x}} = \sqrt x + \frac{1}{{2\sqrt x }} + \mathcal{O}\!\left( {\frac{1}{x}} \right). $$ Now for fixed $a$ and $b$, $$ \frac{{(z + a)!}}{{(z + b)!}} = \frac{{\Gamma (z + 1 + a)}}{{\Gamma (z + 1 + b)}} \sim z^{a - b} $$ as $z\to +\infty$ (see here). Thus, $$ \frac{{(\sqrt x )!}}{{(\sqrt {x + 1} )!}} \!=\! \frac{{\Gamma (\sqrt x + 1)}}{{\Gamma (\sqrt {x + 1} + 1)}} \!=\! \frac{{\Gamma (\sqrt x + 1)}}{{\Gamma \left( {\sqrt x + 1 + \frac{1}{{2\sqrt x }} + \mathcal{O}\!\left( {\frac{1}{x}} \right)} \right)}} \sim (\sqrt x )^{ - \frac{1}{{2\sqrt x }}} \!=\! \mathrm{e}^{ - \frac{1}{4}\frac{{\log x}}{{\sqrt x }}} \to 1 $$ as $x\to +\infty$.