$\lim\limits_{n \to \infty}\sin(\pi\sqrt{n^2+1})$ 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!
 A: Using $\sqrt{n^2+1}=n+\dfrac{1}{n+\sqrt{n^2+1}}$, we can deduce that:
$$\sin (\pi \sqrt{n^2+1})=(-1)^n\sin \left(\frac{\pi}{n+\sqrt{n^2+1}}\right)$$
$(-1)^n$ is bounded and:
$$\lim_{n\to \infty} \sin \left(\frac{\pi}{n+\sqrt{n^2+1}}\right) = 0$$
so the final results is $0$.
A: One elementary way to do this is to "extract" the equivalent $n\pi$ like this :
$$\pi\sqrt{n^2+1}=n\pi\sqrt{1+\frac{1}{n^2}}$$
and observe the behaviour of the square root :
$$\epsilon_n = \sqrt{1+\frac{1}{n^2}}-1 = \frac{(\sqrt{1+\frac{1}{n^2}}-1)(\sqrt{1+\frac{1}{n^2}}+1)}{\sqrt{1+\frac{1}{n^2}}+1} = \frac{\frac{1}{n^2}}{\sqrt{1+\frac{1}{n^2}}+1}\le \frac{1}{2n^2}$$
so
$$\sin(\pi\sqrt{1+n^2}) = \sin(n\pi+n\epsilon_n) = (-1)^n\sin(n\epsilon_n)$$
and as $n\epsilon_n\xrightarrow[]{n\to\infty}0$, you can conclude.
Note that this can also be used to prove that the series $\sum\sin(\pi\sqrt{n^2+1})$ converges.
A: $$L=\lim_{n \rightarrow \infty} \sin (\pi\sqrt{n^2+1})=\lim_{n \rightarrow \infty} \sin (n\pi (1+\frac{1}{n^2})^{1/2})$$
$$L=\lim_{n \rightarrow \infty} \sin (n\pi(1+\frac{1}{2n^2})=\sin n\pi=0$$
A: $|\sin (π(n^2+1)^{1/2}-nπ)+nπ)|=$
$|\sin(π(n^2+1)^{1/2})\cos nπ+$
$ \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 
$\lim_{n \rightarrow \infty }|\sin (f(n))|=$
$|\sin (\lim_{n \rightarrow \infty}f(n))|= |\sin (0)|=0$.
