Limit of $\sin(\pi \sqrt{4n^2+n})$ We can notice that $\sqrt{4n^2 + n} = \sqrt{4n^2(1+ \frac{1}{4n})} = 2n\sqrt{1+\frac{1}{4n}}$. Therefore
$$\lim_{n \to \infty} \sin (2n\pi \sqrt{1+ \frac{1}{4n}}) \text{ will be an even number}$$
Because the square root becomes $1$ and we end up with an even number: $\sin(\text{even number})$
And sine of an even number is $0$. But apparently that is not the right answer.

Consider that
$$\sin(\pi\sqrt{4n^2+n})=\sin(\pi\sqrt{4n^2+n}-2\pi n)=\sin\left(\frac{\pi n}{\sqrt{4n^2+n}+2n}\right)\to\sin\frac\pi4.$$
