$\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?
 A: Use the binomial theorem on $(1+x)^{1/3}$ to get
$$
\sqrt[3]{n^3 + n^2 + 2n} = n\sqrt[3]{1 + n^{-1} + 2n^{-2}}= n\left[1 + \frac{n^{-1}+2n^{-2}}{3} - \frac{(n^{-1}+2n^{-2})^2}{9}+...\right] \\= n + \frac{1}{3} + \frac{5}{9n} + O(n^{-2})
$$
So 
$$
\lim_{n\rightarrow \infty}\sqrt[3]{n^3 + n^2 + 2n}-n = \frac{1}{3}.
$$
A: Try with $$\cos(a+b)=\cos(a) \cos(b) - \sin(a) \sin(b) \tag{1}$$
In this case 
$$\cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right)=\cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}-\pi n+\pi n\right)=\\
\left(\cos\left(\pi \sqrt[3]{n^3+n^2+2n}-\pi n\right) \cdot \color{red}{\cos(\pi n)}-\sin\left(\pi \sqrt[3]{n^3+n^2+2n}-\pi n\right)\cdot \color{red}{\sin(\pi n)}\right)^2=
\left(\cos\left(\pi \sqrt[3]{n^3+n^2+2n}-\pi n\right)\right)^2=\\
\left(\cos\left(\pi \frac{n^2+2n}{\left(\sqrt[3]{n^3+n^2+2n}\right)^2+n\sqrt[3]{n^3+n^2+2n}+n^2}\right)\right)^2$$
from $$a^3-b^3=(a-b)(a^2+ab+b^2) \tag{2}$$
leading to
$$\color{red}{\cos^2\left(\frac{\pi}{3}\right)}$$
because
$$\frac{n^2+2n}{\left(\sqrt[3]{n^3+n^2+2n}\right)^2+n\sqrt[3]{n^3+n^2+2n}+n^2}=\\
\frac{1+\frac{2}{n}}{\left(\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}\right)^2+\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}+1} \to \frac{1}{3}$$
A: $\sqrt[3]{n^3 + n^2 + 2n} = n \sqrt[3]{1 + \frac{1}{n} + \frac{2}{n^2}}$
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*}
A: $\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.
