Limit of $\sin(\pi \sqrt{4n^2+n})$

Question

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.

Answer 1

Consider that

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

Answer 2

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 $\frac{\pi}{4} + o(1)\xrightarrow[n\to\infty]{}\frac{\pi}{4}$).