Evaluate the limit
$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})$$
We know that :
$$\sin (\pi-x)=\sin x$$
So we have :
$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi-\pi\sqrt{4n^2+\sqrt{n}})$$
Now :
$$(\pi-\pi\sqrt{4n^2+\sqrt{n}})\cdot \frac{\pi+\pi\sqrt{4n^2+\sqrt{n}}}{\pi+\pi\sqrt{4n^2+\sqrt{n}}}=\frac{\pi^2-\pi^2(4n^2+\sqrt{n})}{\pi+\pi\sqrt{4n^2+\sqrt{n}}}$$
$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin \left(\frac{\pi^2-\pi^2(4n^2+\sqrt{n})}{\pi+\pi\sqrt{4n^2+\sqrt{n}}}\right)$$
Now what ?
I think the following is better. $$\lim_{n\rightarrow+\infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})=-\lim_{n\rightarrow+\infty}4\sqrt{n}\sin \left(2\pi n-\pi\sqrt{4n^2+\sqrt{n}}\right)=$$ $$=\lim_{n\rightarrow+\infty}4\sqrt{n}\sin\frac{\pi\sqrt{n}}{2 n+\sqrt{4n^2+\sqrt{n}}}=\lim_{n\rightarrow+\infty}4\sqrt{n}\sin\frac{\pi}{\sqrt{n}\left(2 +\sqrt{4+\frac{1}{n\sqrt{n}}}\right)}=$$ $$=\lim_{n\rightarrow+\infty}\left(\frac{\sin\frac{\pi}{\sqrt{n}\left(2 +\sqrt{4+\frac{1}{n\sqrt{n}}}\right)}}{\frac{\pi}{\sqrt{n}\left(2 +\sqrt{4+\frac{1}{n\sqrt{n}}}\right)}}\cdot\frac{4\pi}{2 +\sqrt{4+\frac{1}{n\sqrt{n}}}}\right)=\pi.$$