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!
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 $\frac{\pi}{2}$).
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 .