Evaluating $\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)$. (WolframAlpha says it doesn't exist; I get $0$.) I have tried to solve limit, which wolfram says that DNE, but according to my calculations it is equal to 0. Limit is given below
$$\begin{align}
\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)
&=\sin(\sqrt{n^2+1}\pi-n\pi+n\pi) \\
&=(-1)^n\sin(\sqrt{n^2+1}\pi-n\pi) \\
&=(-1)^n\sin\left(\frac{(\sqrt{n^2+1}\pi-n\pi)(\sqrt{n^2+1}\pi+n\pi)}{(\sqrt{n^2+1}\pi+n\pi)}\right) \\
&=(-1)^n\sin\left(\frac{n^2\pi^2+\pi^2-n^2\pi^2}{\sqrt{n^2+1}\pi+n\pi}\right) \\
&=(-1)^n\sin\left(\frac{\pi^2}{n\pi(\sqrt{1+\frac{1}{n^2}}+1}\right) \\
&=0 
\end{align}$$
It is because denominator of sin goes to infinity so everything inside sin goes to 0. as we know, sin of that would go to 0 too. And we know that $(-1)^n$ is bounded so we got that bounded * 0 has to be equal to 0.
Am I doing some mistake here ?

I don't think you need such a long computation. Once you write your sequence as
$$
a_n=(-1)^n\sin((\sqrt{n^2+1}-n)\pi)
$$
you can just observe that $\lim\,(\sqrt{n^2+1}-n)=0$ and, by continuity,
$$
\lim\, a_n=0.
$$
