$\lim\limits_{n \to \infty}\sin(\pi\sqrt{n^2+1})$ I have a question regarding my method of finding a limit for my analysis class.
$\lim\limits_{n \to \infty}\sin(\pi\sqrt{n^2+1})$. 
The method I used was I noticed that $\lim\limits_{n \to \infty}n = \lim\limits_{n \to \infty}(n^2+1)$ I think, so then that means $\lim\limits_{n \to \infty}\sin(\pi\sqrt{n^2+1}) = \lim\limits_{n \to \infty}\sin(n\pi)$ then for all $n \in \Bbb N, \sin(n\pi) = 0$, thus $\lim\limits_{n \to \infty}\sin(\pi\sqrt{n^2+1}) = 0$. Is this a good way to do this?
If not, why, and what would be a good way to do this problem.
Also, what is a good way to show $\lim\limits_{n \to \infty}n = \lim\limits_{n \to \infty}(n^2+1)$. I reached the conclusion using intuition (and a C++ script), but I don't think thats rigorous enough.
Thanks in advance!

$|\sin (π(n^2+1)^{1/2}-nπ)+nπ)|=$
$|\sin(π(n^2+1)^{1/2})\cos nπ+$
$ \cos (π(n^2+1)^{1/2}-nπ)\sin nπ)|=$
$|\sin (π(n^2+1)^{1/2}-nπ)\ cos (nπ)|=$
$|\sin (π(n^2+1)^{1/2}-nπ)||\cos (nπ)|$
$=|\sin (π(n^2+1)^{1/2}-nπ)|\cdot 1.$
$f(n)=π(n^2+1)^{1/2}-nπ=$
$π\dfrac{1}{(n^2+1)^{1/2}+n};$
$\lim_{n \rightarrow \infty}f(n)=0$;
Finally 
$\lim_{n \rightarrow \infty }|\sin (f(n))|=$
$|\sin (\lim_{n \rightarrow \infty}f(n))|= |\sin (0)|=0$.
