$\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?

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
$$\frac{n^2+2n}{\left(\sqrt[3]{n^3+n^2+2n}\right)^2+n\sqrt[3]{n^3+n^2+2n}+n^2}=\\
\frac{1+\frac{2}{n}}{\left(\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}\right)^2+\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}+1} \to \frac{1}{3}$$
