Evaluate the limit $\lim_{n\to\infty}\prod_{r=1}^n\frac{4r}{4r+3}$

Evaluate the limit $$\lim_{n\to\infty}\prod_{r=1}^n\frac{4r}{4r+3}$$

My Attempt:

I am trying to do by Squeeze theorem and then idea of telescopic product but not able to get anywhere.

Using $A.M\geq G.M$ we get $$\frac{2r+1+2r+2}{2}\geq \sqrt{(2r+1)(2r+2)}$$$$\frac{4r+3}{4r}\geq \frac{\sqrt{(2r+1)(2r+2)}}{2r}$$$$\frac{4r}{4r+3}\leq \frac{2r}{\sqrt{(2r+1)(2r+2)}}<\frac{2r+1}{\sqrt{(2r+1)(2r+2)}}$$$$\frac{4r}{4r+3}< \sqrt{\frac{2r+1}{2r+2}}$$

Some clever manipulation or trick is required here

There is something here though Limit as $n\to+\infty$ of $\prod_{k=1}^{n} \frac{2k}{2k+1}$but no general method as such