$\displaystyle \begin{align*} \left( \frac{\sqrt{\frac{\sin{(2nx)}}{1+\cos{(nx)}}}}{(x-\frac{\pi}{2n})(x+\frac{\pi}{2n})4n^2}\right)^2 &= \frac{\frac{\sin{(\pi + 2nx - \pi)}}{1+\cos{(nx)}}}{(2nx-\pi)^2(2nx+\pi)^2} \\ &= \frac{\frac{\sin{(\pi)}\cos{(2nx - \pi)} + \cos{(\pi)}\sin{(2nx - \pi)}}{1+\cos{(nx)}}}{(2nx-\pi)^2(2nx+\pi)^2} \\ &= \frac{\frac{-\sin{(2nx - \pi)}}{1+\cos{(nx)}}}{(2nx-\pi)^2(2nx+\pi)^2} \\ &= - \frac{1}{2nx-\pi} \cdot \frac{\sin{(2nx-\pi)}}{2nx-\pi} \cdot \frac{1}{1+\cos{(nx)}} \cdot \frac{1}{(2n+x)^2} \end{align*}$
So I think we are heading to $-\infty$.
(There's an error in here somewhere, I think, but my conclusion remains).
Approaching from the right (with $n$ positive), it says that the square would approach negative infinity. The limit takes the square root of negative sign(n). Not sure how it came to that conclusion, though. I thought it went to positive infinity, but I did not feel like looking into it.
one thing you have to be careful with when using WA or Mathematica is that it makes no assumptions on the types of numbers being dealt with.
In particular here it assumes both $x$ and $n$ are complex numbers when I believe it's meant that $x$ is real, and $n$ is at least an integer if not an non-negative one.
There are ways in Mathematica to tell it what types of numbers are being dealt with. WA may have this functionality as well, I don't know.