The infinite product in question is
$$\prod_{n=1}^{\infty}(1-\frac{x}{n\pi})$$
I see that this product is similar to that of $$\frac{\sin(x)}{x}= \prod_{n=1}^{\infty}(1-\frac{x^2}{n^2\pi^2})$$
The difference is that the Infinite Product in question is missing the positive products found in the $\frac{\sin(x)}{x}$
Thank you very much for your time and help!
Assuming $x>0$, consider the partial product $$P_k=\prod_{n=1}^{k}(1-\frac{x}{n\pi})=\frac{\left(1-\frac{x}{\pi }\right)_k}{k!}$$ where appear the Pochhammer symbol.
Taking logarithms, using Stirling approximation and continuing with Taylor expansion, we have $$\log(P_k)=-\frac{x \log \left({k}\right)}{\pi }+\log \left(\frac{\sqrt{2 \pi }}{\Gamma \left(1-\frac{x}{\pi }\right)}\right)-\frac{1}{2} \log (2 \pi )+\frac{x(x-\pi)}{2 \pi ^2 k}+O\left(\frac{1}{k^2}\right)$$ $$P_k=e^{\log(P_k)}=\left(\frac{1}{k}\right)^{\frac{x}{\pi }} \left(\frac{1}{\Gamma \left(1-\frac{x}{\pi }\right)}+\frac{\pi -x}{2 \pi \Gamma \left(-\frac{x}{\pi }\right) k}+O\left(\frac{1}{k^2}\right)\right)$$ that is to say $$P_k \sim \frac 1 {k^{\frac{x}{\pi }} \Gamma \left(1-\frac{x}{\pi }\right)}$$ So, the limit seems to be $0$.
Trying for a few values of $x$ and $k$ $$\left( \begin{array}{cccc} x & k & \text{approximation} & \text{exact} \\ 1 & 250 & 0.1298697696 & 0.1298134350 \\ 1 & 500 & 0.1041567084 & 0.1041341127 \\ 1 & 750 & 0.0915452169 & 0.0915319761 \\ 1 & 1000 & 0.0835346049 & 0.0835255429 \\ & & & \\ 2 & 250 & 0.0121465823 & 0.0121409627 \\ 2 & 500 & 0.0078129033 & 0.0078110960 \\ 2 & 750 & 0.0060354447 & 0.0060345139 \\ 2 & 1000 & 0.0050254019 & 0.0050248207 \\ & & & \\ 3 & 250 & 0.0002369272 & 0.0002369068 \\ 3 & 500 & 0.0001222229 & 0.0001222176 \\ 3 & 750 & 0.0000829846 & 0.0000829823 \\ 3 & 1000 & 0.0000630507 & 0.0000630494 \end{array} \right)$$