$$ A=\dfrac{\sin(\frac{\pi}{n}+\frac{2 \pi \log(n)}{\sqrt n})}{\sin(\frac{\pi}{n})}=\frac{\sin \left(\frac{\pi }{n}+\frac{2 \pi \log (n)}{\sqrt{n}}\right)}{\frac{\pi
}{n}+\frac{2 \pi \log (n)}{\sqrt{n}}}\times\dfrac{\frac\pi n}{{\sin(\frac{\pi}{n})}}\times\frac{\frac{\pi }{n}+\frac{2 \pi \log (n)}{\sqrt{n}} }{\frac \pi n}$$ Since $n$ is large, you face twice the limit of $\frac{\sin(x)}x=1$ when $x\to 0$ and so, for large values of $n$ $$A\simeq \frac{\frac{\pi }{n}+\frac{2 \pi \log (n)}{\sqrt{n}} }{\frac \pi n}=2 \sqrt{n} \log (n)+1$$ So, $$\frac A{n\log(n)}\simeq \frac{2 \sqrt{n} \log (n)+1 }{n\log(n)}=\frac{2}{\sqrt n}+\frac 1{n\log(n)}$$ which shows not only the limit but also how it is approached.
Edit
We could even go further using composition of series and show that $$\frac A{n\log(n)}=\frac{2}{\sqrt n}+\frac 1{n\log(n)}-\frac{4 \pi ^2 \log ^2(n)}{3 n^{3/2}}+O\left(\frac{1}{n^2}\right)$$