Proof that these two expressions are equivalent I'm looking for algebraic proof that  

$$ y=\lim_{N \to \infty}\frac{\prod\limits_{n=0}^{N}x-n}{\prod\limits_{n=0}^{N-\lfloor x\rfloor}{x-\lfloor x\rfloor-n}}=\frac{\Gamma{(x+1)}}{\Gamma{(x-\lfloor x\rfloor}+1)}$$

I know that they are equivalent because when I graph $y$ vs. $x$ for both of these, the functions intersect at every point.

I can see how they are related when $x$ is an integer. Although the middle expression is undefined for those values (it requires dividing by zero), taking the limit as $x$ goes to $z$ ($\lim{x \to z}$, where $z \in \Bbb Z$) will make both functions equal to $z$ factorial. 
I'm not sure about values in which $x$ is not an integer though, which is why I'm asking this question. 

The reason I want to prove that this equation is true, is because I think that if you can find a good approximation for $\Gamma{(x-\lfloor x\rfloor+1)}$ ($ -\frac{1}{9}|\sin  \pi x \space|+1$ for example) then you could also approximate the gamma function pretty well.
$\small\text{Graph of the function}$


$\small\text{Graph of} \space \space \Gamma{(x-\lfloor x\rfloor+1)} \space\space\text{and} \space -\frac{1}{9}\left| \sin \left( \pi x \right) \right|+0.75$

 A: The proof is fairly straightforward.
We'll start by looking at one part of the limit you give, namely, $$\prod_{n=0}^{N}x-n$$ which, expanded, gives $$x(x-1)(x-2)\ldots(x-(N-1))(x-N)$$ 
assuming $N$ is a non-negative integer. This is equal to $$\frac{\Gamma\left(x+1\right)}{\Gamma\left(x-N\right)}$$ which can be shown using $\Gamma(a+1)=a\Gamma(a)$ for all $a$.
A similar manipulation can be done with the denominator of the limit you gave, substituting $N$ for $N-\lfloor x\rfloor$ and $x$ for $x - \lfloor x \rfloor$, giving $$\frac{\Gamma(x-\lfloor x\rfloor+1)}{\Gamma(x-N)}$$
after a small amount of simplification. This is only true if $N-\lfloor x\rfloor$ is a non-negative integer, as before.
Together these give $$\lim_{N\rightarrow \infty}\frac{\frac{\Gamma\left(x+1\right)}{\Gamma\left(x-N\right)}}{\frac{\Gamma(x-\lfloor x\rfloor+1)}{\Gamma(x-N)}} = \frac{\Gamma\left(x+1\right)}{\Gamma(x-\lfloor x\rfloor+1)}$$
The reason the limit cannot be dropped in the original expression—despite the cancellation here—is due to the fact that $N-\lfloor x\rfloor$ must be a non-negative integer for a product to be taken.
