How can we show the convergence of $x_n = \sin(2\pi (n^3-n^2+1)^{\frac{1}{3}})$? Show that the sequence $(x_n)_{n\geq 1}$ defined by $$x_n = \sin(2\pi (n^3-n^2+1)^{\frac{1}{3}})$$ converges and compute its limit.

It smells like an easy high school question
$$\lim_{n\to\infty}\sin(2\pi (\sqrt[3]{n^3-n^2+1}-n)+2\pi n)=$$
$$\lim_{n\to\infty}\sin(2\pi (\sqrt[3]{n^3-n^2+1}-n))=-\frac{\sqrt{3}}{2}$$
because
$$\lim_{x\to0} \frac{\sqrt[3]{x^3-x+1}-1}{x}=\lim_{x\to0} \frac{3x^2-1}{3(x^3-x+1)^{2/3}}=-\frac{1}{3}$$
by the cute  l'Hôpital's rule, and then
$$\lim_{n\to\infty} (\sqrt[3]{n^3-n^2+1}-n)=-\frac{1}{3}$$ 
$$\sin \left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}$$
Chris.
