$\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?

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 
$$
\lim_{n\rightarrow \infty}\sqrt[3]{n^3 + n^2 + 2n}-n = \frac{1}{3}.
$$
