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.

When $n\to\infty$, we have
$$
\sqrt{4n^2+n} = 2n\sqrt{1+\frac{1}{4n}}
= 2n\left(1+\frac{1}{8n} + o\!\left(\frac{1}{n}\right)\right)
= 2n + \frac{1}{4} + o(1) \tag{1}
$$
using the Taylor expansion of $\sqrt{1+x}$ for $x$ around $0$. Recalling that $$\forall a,b,\qquad \sin(a+b)=\sin a \cos b + \cos a \sin b \tag{2}$$ we get
$$\begin{align*}
\sin(\pi\sqrt{4n^2+n}) &=\sin\left(2n\pi + \frac{\pi}{4} + o(1)\right) \\
&= \sin(2n\pi) \cos\left(\frac{\pi}{4} + o(1)\right)
+ \cos(2n\pi) \sin\left(\frac{\pi}{4} + o(1)\right) \\
&= 0 + \sin\left(\frac{\pi}{4} + o(1)\right) \\
&\xrightarrow[n\to\infty]{} \sin\frac{\pi}{4} = \boxed{\frac{1}{\sqrt{2}}}
\end{align*}$$
using continuity of $\sin$ (as $\frac{\pi}{4} + o(1)\xrightarrow[n\to\infty]{}\frac{\pi}{4}$).
