Limit in previous version of WM and DiscreteLimit

Question

How can I find this limit (where $n$ is an integer number) in WM v 12.2: $\lim_{n\to+\infty} \sin^2(\pi\,\sqrt{n^2+n})$?

I mean, it's clear that it equals to one. Versions 9 and below easily produced the correct answer. Now in v12.2 I've a lot of problems finding the same limit. Any idea? Thank you

Answer 1

This can be done in 12.2 as follows.

DiscreteAsymptotic[Sin[Pi*Sqrt[n^2 + n]]^2, n -> Infinity]

1

Answer 2

I suppose it is not equal to one if it is over Reals:

Asymptotics`ClassicLimit[Sin[Pi*Sqrt[n^2 + n]]^2, n -> Infinity]

prints Interval[{0, 1}]. While

Asymptotics`ClassicLimit[Sin[Pi*Sqrt[n^2 + n]]^2, n -> Infinity, 
 Assumptions -> n \[Element] Integers]

does indeed print one. That is what versions 11.2 and below were using.

I suppose that is like one of those examples in DiscreteLimit[] documentation, after all "integers only" is far from the same as "all reals" limit.

(We cannot use modern Limit to solve this, since it does not allow any Assume[] or Assumptions if it is on the variable we are limiting.)

Edit:

In 13.2 it finally works!

DiscreteLimit[Sin[Pi*Sqrt[n^2 + n]]^2, n -> Infinity] 

prints 1.