Evaluating: $\lim_{n \to 0} \prod_{\substack{i=nk \\k \in \Bbb Z_{\geq 0}}}^{2-n} \left( 2-i \right) $ How would we evaluate:
$$\lim_{n \to 0} \prod_{\substack{i=nk \\k \in \Bbb Z_{\geq 0}}}^{2-n} \left( 2-i \right) $$
Is it possible to evaluate this manually? Or do we have to make a program to get an approximation? 
EDIT:
Since people are getting confused in the comments, here is an example:
If we take n=0.5:
$$\prod_{\substack{i=nk \\k \in \Bbb Z_{\geq 0}}}^{2-n} \left( 2-i \right) = \prod_{\substack{i=0.5k \\k \in \Bbb Z_{\geq 0}}}^{1.5} \left( 2-i \right)=(2-0)(2-0.5)(2-1)(2-1.5) $$
Hopes this clears up the misunderstanding.  
 A: The product can be written in an equivalent form as:
$$\Pi(n)=\prod_{k=0}^{\frac{2-n}{n}} (2-kn)=\prod_{k=0}^{\frac{2-n}{n}} -n(k-\frac{2}{n})=\prod_{k=0}^{\frac{2-n}{n}} -n\prod_{k=0}^{\frac{2-n}{n}} (k-\frac{2}{n})=$$
$$=2(-n)^{\frac{2-n}{n}} \left(\frac{n-2}{n}\right)_{\frac{2-n}{n}}$$
This may be helpful.
A: This is just a start, but I think you'll be able to carry it to a conclusion.
Let $$f(x)=\prod_{0\leq k<2/x}(2-nk)$$ and consider the sequence $$f\left(\frac12\right),f\left(\frac14\right),\dots,f\left(2^{-n}\right),\dots\tag{1}$$  We have $${f\left(2^{-n-1}\right)\over f\left(2^{-n}\right)}={1\cdot3\cdot5\cdots(2^{n+2}-1)\over(2^{n+1})^{2^{n+1}}}\tag{2}$$
This is because every other term in the product for $f(2^{-n-1})$ also occurs in the product for $f(2^{-n}),$ and the expression on the right-hand side of $(2)$ is what is left after cancellation.  The expression on the right-hand side of $(2)$ goes to $0$ as $n\to\infty,$ so the sequence in $(1)$ also goes to $0.$
I don't have time right now to try to extend this general $x.$  I would first try to show that if $2^{-n-1}<x<2^{-n}$ then $f\left(2^{-n-1}\right)<f(x)<f\left(2^{-n}\right).$  Of course, if you can show that, you are done.    
