<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] $\sqrt[3]{n^3 + n^2 + 2n} = n \sqrt[3]{1 + \frac{1}{n} + \frac{2}{n^2}}$
In the limit, this cube root $\rightarrow 1 + \frac{1}{3n}$, so this subexpression $\rightarrow n\pi + \pi/3$, so the argument to cosine goes to an integer multiple of $\pi$ plus $\pi/3$, and so the square of the cosine of such angles goes to $1/4$.
We use Newton's binomial formula.
\begin{align*}
&\lim_{n \rightarrow \infty} \cos^2\left(\pi \sqrt[3]{n^3 + n^2 + 2n}\right)  \\
&\qquad = \lim_{n \rightarrow \infty} \cos^2 \left(n \pi \sqrt[3]{1 + \frac{1}{n} + \frac{2}{n^2}}\right)  \\
&\qquad = \lim_{n \rightarrow \infty} \cos^2 \left( n \pi \sum_{k=0}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k   \right)  \\
&\qquad = \lim_{n \rightarrow \infty} \cos^2 \left( n \pi \left( \binom{1/3}{0} + \binom{1/3}{1} \left( \frac{1}{n} + \frac{2}{n^2} \right) + \sum_{k=2}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k \right) \right)  \\
&\qquad = \lim_{n \rightarrow \infty} \cos^2 \left( n \pi + \frac{\pi}{3} + \frac{2 \pi}{3n} + n \pi\sum_{k=2}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k \right)   \\
&\qquad = \lim_{n \rightarrow \infty} \left( \cos \left( n \pi + \frac{\pi}{3}\right) \cos \left( u \right)  -  \sin \left( n \pi + \frac{\pi}{3}\right) \sin \left( u \right)
\right)^2   \text{,}
\end{align*}
where $u = \frac{2 \pi}{3n} + n \pi\sum_{k=2}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k$.  By continuity, \begin{align*}
&\lim_{n \rightarrow \infty} \left( \cos \left( n \pi + \frac{\pi}{3}\right) \cos \left( u \right)  -  \sin \left( n \pi + \frac{\pi}{3}\right) \sin \left( u \right)
\right)^2  \\
&\qquad = \left( \left( \lim_{n \rightarrow \infty} \cos \left( n \pi + \frac{\pi}{3}\right) \right) \left( \lim_{n \rightarrow \infty}\cos \left( u \right) \right)  -  \left( \lim_{n \rightarrow \infty} \sin \left( n \pi + \frac{\pi}{3}\right) \right) \left( \lim_{n \rightarrow \infty} \sin \left( u \right) \right)
\right)^2  \\
&\qquad = \left( \left( \lim_{n \rightarrow \infty} \begin{cases} 1/2, & \text{$n$ even} \\ -1/2, & \text{$n$ odd} \end{cases} \right) \left( \lim_{n \rightarrow \infty}\cos \left( u \right) \right)  -  \left( \lim_{n \rightarrow \infty} \begin{cases} \sqrt{3}/2, & \text{$n$ even} \\ -\sqrt{3}/2, & \text{$n$ odd} \end{cases} \right) \left( \lim_{n \rightarrow \infty} \sin \left( u \right) \right) \right)^2  \text{.}
\end{align*}
So we examine
$$  \lim_{n \rightarrow \infty}\cos \left( u \right) 
= \lim_{n \rightarrow \infty}\cos \left( \frac{2 \pi}{3n} + n \pi\sum_{k=2}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k \right)
\text{.}  $$
First, $\left| \binom{1/3}{k} \right| \leq 1$ (in fact, we can tighten this to ${} \leq 1/3$).  Second, for $n > 2$, $\frac{1}{n} + \frac{2}{n^2} < \frac{2}{n}$, so the sum is bounded using the geometric series $\frac{1}{3} \sum_{k=2}^\infty \left( \frac{2}{n} \right)^k = \frac{4}{3n(n-2)}$, which goes to $0$ as $n$ goes to $\infty$, as does $\frac{2 \pi}{3n}$.  Consequently, again using continuity,
$$  \lim_{n \rightarrow \infty}\cos \left( u \right) 
=\cos \left( \lim_{n \rightarrow \infty} u \right) = \cos 0 = 1  $$
and also
$$  \lim_{n \rightarrow \infty}\sin \left( u \right) 
=\sin \left( \lim_{n \rightarrow \infty} u \right) = \sin 0 = 0  \text{.}  $$
Therefore,
\begin{align*}
&\lim_{n \rightarrow \infty} \cos^2\left(\pi \sqrt[3]{n^3 + n^2 + 2n}\right)  \\
&\qquad = \left( \left( \lim_{n \rightarrow \infty} \begin{cases} 1/2, & \text{$n$ even} \\ -1/2, & \text{$n$ odd} \end{cases} \right) \cdot 1  -  \left( \lim_{n \rightarrow \infty} \begin{cases} \sqrt{3}/2, & \text{$n$ even} \\ -\sqrt{3}/2, & \text{$n$ odd} \end{cases} \right) \cdot 0 \right)^2  \\
&\qquad = 1/4  \text{.}
\end{align*}
</s>