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.

 A: Taking your rewritten expression which is actually quite useful, I would use Stirling's approximation on it, where for large $x$
$\large x! \sim \sqrt{2\pi x}(\frac{x}{e})^x$
Thus we have, for $x \rightarrow \infty$ (skipping out some intermediate steps)
$\large x^r\frac{(2x)!}{2^{2x}(x!)^2} \sim \frac{x^r\sqrt{2\pi(2x)}(\frac{2x}{e})^{2x}}{2^{2x}(\sqrt{2\pi x}(\frac{x}{e})^x)^2}=\frac{x^{r-0.5}}{\sqrt{\pi}}$
The value of $r$ for which the expression converges to a finite nonzero value is $0.5$, as the power of $x$ will be $0$, resulting in $x^{r-0.5}=1$. 
The value to which the limit converges is $\large \frac{1}{\sqrt{\pi}}$
A: 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. 
