sequences and series - Finding $lim_{ntoinfty}frac{sinleft(frac{pi}{n}+frac{2 pi log(n)}{sqrt n}right)}{n log(n)sinleft(frac{pi}{n}right)}$

Is there a possible way to compute this limit ? I tried the L'Hôpital's rule, but it seems like the function become more complicated.

$$\lim_{n\to\infty} \frac{1}{n \log(n)} \frac{\sin\left(\frac{\pi}{n}+\frac{2 \pi \log(n)}{\sqrt n}\right)}{\sin\left(\frac{\pi}{n}\right)}$$

All suggestions are welcomed.

Answer

First, $$\lim_n\frac{\sin\frac\pi n}{\frac1n}=\pi,$$ so you might as well calculate
$$\lim_{n\to\infty} \dfrac{\sin(\frac{\pi}{n}+\frac{2 \pi \log(n)}{\sqrt n})}{\log n}.$$
You have
$$\left|\dfrac{\sin(\frac{\pi}{n}+\frac{2 \pi \log(n)}{\sqrt n})}{\log n}\right|\leq\frac1{\log n}\xrightarrow{\ \ \ \ \ }0,$$
so the limit is zero.