<s>[INST] Find the limit of $x_n = \sin(2\pi (n^3 − n^2 + 1)^{1/3} )$ Show that the sequence $(x_n)_{n\geq 1}$ defined by $x_n = \sin(2\pi (n^3 − n^2 + 1)^{1/3}
)$ is convergent and compute its limit.
What I tried:
Let us write $c_n = n^3 - n^2 + 1$. Then we can see
$$
(n-1)^3 < c_n < n^3
$$ for sufficiently large $n$. This means that
$$
(c_n)^{1/3} - (n-1) \in (0,1).
$$
By definition, $x_n = \sin(2\pi (c_n)^{1/3})$ and since the sine function is $2\pi$-periodic we see
$$
x_n = \sin(2\pi [(c_n)^{1/3} - (n-1)]).
$$
Now if $(c_n)^{1/3} - (n-1)$ could be shown to have a limit, I would be done. But I have no idea on how to proceed from here. Any help would be much appreciated!
 [/INST] Hint. As $n$ tends to $+\infty$, you may write
$$
\begin{align}
x_n &=\sin \left( 2\pi \sqrt[\large 3]{n^3-n^2+1}\right)\\\\
&=\sin \left( 2\pi n \:\sqrt[\large 3]{1-\frac{1}{n}+\frac1{n^3}}\right)\\\\
&=\sin \left( 2\pi n \:\left(1-\frac{1}{3n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)\right)\\\\
&=\sin \left( 2\pi n -\frac{2\pi}{3}+\mathcal{O}\left(\frac{1}{n}\right)\right)\\\\
&=-\sin \left(\frac{2\pi}{3}+\mathcal{O}\left(\frac{1}{n}\right)\right)\\\\
&\longrightarrow -\sin \left(\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}
\end{align}
$$
</s>