<s>[INST] $\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right) $, where $n \in \mathbb{N}$ $$
\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right)
$$
where $n \in \mathbb{N}$.
In this question, what I thought was, since $n \to \infty$ and $\cos ^2x$ is periodic , all I need is actually the fractional part of this. And it easy to say that $n+1> \sqrt[3]{n^3+n^2+2n}> n$.
But evaluation of $\sqrt[3]{n^3+n^2+2n} - n$ , is getting tricky. I'm sure there must be a short way to solve it. Can someone help me with it?
 [/INST] $\lim _{n \to \infty} \cos^2(\pi \sqrt[3]{n^3+n^2+2n}) = ?
$
$(n+1/3)^3
=n^3+3(1/3)n^2+3(1/9)n+1/27
=n^3+n^2+n/3+1/27
$
so
$n^3+n^2+2n
\gt (n+1/3)^3
$.
$(n+1/3+a/n)^3
=n^3+n^2+n(3a+1/3)+O(1)
\gt n^3+n^2+2n
$
if
$3a+1/3 > 2$
or
$a > 5/9
$.
Choosing
$a = 2/3$,
$(n+1/3+2/(3n))^3
\lt n^3+n^2+2n
$
so that
$n+1/3
\lt \sqrt[3]{n^3+n^2+2n}
\lt n+1/3+2/(3n)
$
for all large enough $n$.
From this you can proceed
as some of the other answers.
</s>