Find value of $r$ and the limit For some $r \in \mathbb Q$, the limit
$$\lim_{x \rightarrow \infty}x^r.\frac{1}2.\frac{3}4.\frac{5}6......\frac{2x-1}{2x}$$   exists and is non zero
What is that value of $r$ and what is that limit equal to?   

I rewrote the product $\frac{1}2.\frac{3}4.\frac{5}6......\frac{2x-1}{2x}$ = $\frac{(2x)!}{2^{2x}(x!)^2}$ but it didn't help.


Let $x^r.\frac{1}{2}.\frac{3}{4} \dots \frac{2x-1}{2x} = A \tag{1}$.
Clearly $A < x^r. \frac{2}{3} \tag{2}. \frac{4}{5}.\frac{6}{7} \dots \frac{2x}{2x+1}$ and $A > x^r. \frac{1}{2}.\frac{2}{3}.\frac{4}{5} \dots \frac{2x-2}{2x-1} \tag{3}$
Multiplying $(1)$ and $(3)$,
$A^2 > x^{2r}.\frac{1}{2(2x)} \text{ or }, A > \frac{x^r}{2 \sqrt{x}}$. The limit does not exist if $r >0.5$. Now we need to check if the limit exists for $r \leq 0.5$
Multiplying $(1)$ and $(2)$, we gwt
$A < \frac{x^r}{\sqrt{2x+1}}$ and its clear that $A$ is finite and nonzero for $r=0.5$ and $\lim_{x \to \infty} A \leq \frac{1}{\sqrt{2}}$. 
I am unable to go further and give the exct value of the limit. 
