Prove $$\lim_{x \to \infty} \frac{\Gamma(x+1/2)}{\Gamma(x)~x^{1/2}}=1.$$
I got this problem from Probability and Statistics by Degroot & Schervish. There is a hint to use Stirling's formula which is stated as:
$$\lim_{x \to \infty}\frac{(2\pi)^\frac{1}{2}x^{x-1/2}e^{-x}}{\Gamma(x)}=1.$$
One way is to reduce the problem by using the hint to
$$\lim_{x \to \infty} \frac{\Gamma(x+1/2)~e^x}{(2\pi)^\frac{1}{2}x^{x}}=1.$$
But that did not helped me much. I though about doing the following, but I'm not sure that it is correct:
$$\begin{align}\lim_{x \to \infty} \frac{\Gamma(x+1/2)}{\Gamma(x)~x^{1/2}}&= \lim_{x \to \infty} \frac{\int_o^\infty u^{x-1/2}e^{-u}du}{\int_o^\infty u^{x-1}e^{-u}x^{1/2}du}\\ &=\lim_{x \to \infty} \left(\lim_{k \to \infty} \frac{\int_o^k u^{x-1/2}e^{-u}du}{\int_o^k u^{x-1}e^{-u}x^{1/2}du} \right)\\ &= \lim_{x \to \infty} \left(\lim_{k \to \infty} \frac{k^{x-1/2}e^{-k}}{k^{x-1}e^{-k}x^{1/2}} \right) \\ &=\lim_{x \to \infty} \left(\lim_{k \to \infty} \frac{k^{1/2}}{x^{1/2}} \right) \\ &=1.\end{align}$$
I basically tried to apply L'hopital's rule and to play with double limits but I'm really not sure whether this is so to speak 'permissible'. Thus, I would be grateful for any comments or corrections. Please note: the book is relatively at an introductory level so elementary methods should suffice. Thank you!
Wouldn't this be needlessly complicating things? Stirling formula tells you that $$\Gamma(x+1/2)\sim(2\pi)^{1/2}(x+1/2)^{x}e^{-x-1/2}$$ and that $$ x^{1/2}\,\Gamma(x)\sim(2\pi)^{1/2}x^{x}e^{-x} $$ hence $$ \frac{\Gamma(x+1/2)}{x^{1/2}\,\Gamma(x)}\sim\frac{(2\pi)^{1/2}(x+1/2)^{x}e^{-x-1/2}}{(2\pi)^{1/2}x^{x}e^{-x}}=\left(1+\frac1{2x}\right)^xe^{-1/2}\sim\ldots $$ To check that you understood what is going on, you might try to compute, for every fixed $a$, the limit of the ratio $$ \frac{\Gamma(x+a)}{x^a\,\Gamma(x)}. $$