Analysis question limit sin function as n goes to infinity can you help me with the following:
$\lim_{n \rightarrow \infty} \sin^{2} \pi \sqrt{n^2 + n}$
Thanks a lot!
 A: Notice that
$$(\sin(\pi(\sqrt{n^{2}+n}-n)))^{2}=((-1)^{n}\sin(\pi\sqrt{n^{2}+n}))^{2}=(\sin(\pi\sqrt{n^{2}+n}))^{2}$$
and that
$$\sqrt{n^{2}+n}-n=\frac{n}{\sqrt{n^{2}+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\to\frac{1}{2}$$ 
as $n\to\infty$. So 
$$\lim_{n\to\infty}(\sin(\pi\sqrt{n^{2}+n}))^{2}=\lim_{n\to\infty}(\sin(\pi(\sqrt{n^{2}+n}-n)))^{2}=\lim_{n\to\infty}\bigg(\sin\bigg(\pi\frac{1}{\sqrt{1+\frac{1}{n}}+1}\bigg)\bigg)^{2}$$
$$=(\sin(\frac{\pi}{2}))^{2}=1$$
A: For the limit $\lim_{n\to+\infty}\sin^2\sqrt{n^2+n}$, since $\{e^{in}\}_{n\in\mathbb{N}}$ is dense in the unit circle (it is a consequence of the irrationality of $\pi$ and the Dirichlet box principle), the same holds for the sequence $\{e^{i(n+1/2)}\}_{n\in\mathbb{N}}$ (a translation of the unit circle preserves density).
By taking imaginary parts (the projection preserves density too), we have that the sequence $\{\sin(n+1/2)\}_{n\in\mathbb{N}}$ is dense in the $[-1,1]$ interval.
Since the sine is a Lipschitz function and the distance between $\sqrt{n^2+n}$ and $n+1/2$ is bounded by $\frac{1}{4n}$, the sequence $\{\sin\sqrt{n^2+n}\}_{n\in\mathbb{N}}$ is dense in the interval $[-1,1]$, so your limit does not exist.

For the modified limit,
$$\lim_{n\to +\infty}\sin^2 \pi\sqrt{n^2+n},$$
the fact that the distance between $\sqrt{n^2+n}$ and $n+\frac{1}{2}$ is bounded by $\frac{1}{4n}$, together with the fact that the sine is a Lipschitz function, ensures that the limit is just $1$ (the sine function equals $\pm 1$ in the odd multiples of $\frac{\pi}{2}$).
A: The sine function doesn't have a limit if $x \rightarrow \infty$. The sine function goes up and down between -1 and 1 infinitely many times . 
A: Hint:
Evaluate your function for the sequence:
$$n_k = \frac{\sqrt{1+4k^2\pi^2}-1}{2}$$
Then do the same for this sequence:
$$n_k = \frac{\sqrt{1+4\pi^2(k^2+k+1/4)}-1}{2}$$
