How can we prove $\lim_{n\to +\infty} \sin(\sqrt{n^2+1}\pi)$ exists, in another way? I was reading about this question just now Evaluating $\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)$. (WolframAlpha says it doesn't exist; I get $0$.) and I immediately got intrigued for I would have never came up with such a trick to evaluate the limit.
So I thought about: how could we prove that limit exists, without using such a trick?
For example, I studied about the Heine method, for limits with successions. This for example rapidly can prove the limit $\lim_{n\to +\infty} \sin(n)$ does not exist. Indeed I can choose a sequence $a_n = \pi n$ such that $a_n \to +\infty$ as $n\to +\infty$ or another sequence $b_n = \frac{2n+1}{2} \pi$, and if the limits exists then it does not depend upon the chosen sequence. In this case, I would get $\sin(a_n) = 0$ whereas $\sin(b_n) = \pm 1$, so the limit does not exist.
Can we use a similar argument to prove the existence of the other limit (the one in the linked question)?
$$\lim_{n\to +\infty} \sin(\sqrt{n^2+1}\pi)$$
Thank you so much.
 A: We have
$$ \sqrt{n^2+1}=n\sqrt{1+\frac{1}{n^2}}=n\left(1+\mathcal{O}\left(\frac{1}{n^2}\right)\right)=n+o(1) $$
therefore $\sin(\pi\sqrt{n^2+1})=\sin(n\pi+o(1))$. If you mean that $n$ takes integer values, then the limit exists and is $0$. If you want $n$ to be real (as mentioned in the comments) then the limit does not exist, taking $a_n=n$ and $b_n=2n+1/2$ gives you two different limits, namely $0$ and $1$ respectively.
A: Assuming that $n$ is an integer and using $b_n=\sqrt{n^2+1}-n$
$$a_n= \sin \left(\pi  \sqrt{n^2+1}\right)=\sin (n\pi +b_n\pi )=\color{red}{(-1)^n}\,\sin(b_n\pi)+\cos(b_n\pi)$$
Now, by Taylor
$$b_n=\sqrt{n^2+1}-n=\frac{1}{2 n}-\frac{1}{8
   n^3}+O\left(\frac{1}{n^5}\right)$$
$$\sin(b_n\pi)=\frac{\pi }{2 n}-\frac{\pi  \left(6+\pi ^2\right)}{48
   n^3}+O\left(\frac{1}{n^5}\right)$$
$$\cos(b_n\pi)=1-\frac{\pi ^2}{8 n^2}+\frac{\pi ^2 \left(24+\pi ^2\right)}{384
   n^4}+O\left(\frac{1}{n^6}\right)$$
So,
$$a_{2n}=\frac{\pi }{4 n}-\frac{\pi  \left(6+\pi ^2\right)}{384
   n^3}+O\left(\frac{1}{n^5}\right)\quad \to ~0^+$$
$$a_{2n+1}=-\frac{\pi }{4 n}+\frac{\pi }{8 n^2}+\frac{\pi  \left(\pi
   ^2-18\right)}{384 n^3}-\frac{\pi  \left(\pi ^2-2\right)}{256
   n^4}+O\left(\frac{1}{n^5}\right)\quad \to ~0^-$$
