Proof verification of $x_n = \sqrt[3]{n^3 + 1} – \sqrt{n^2 – 1}$ is bounded

Question

Let $n \in \mathbb N$ and:
$$
x_n = \sqrt[^3]{n^3 + 1} – \sqrt{n^2 – 1}
$$
Prove $x_n$ is bounded sequence.

Start with $x_n$:

$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} – \sqrt{n^2 – 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} – \sqrt{1 – {1\over n^2}}\right)
\end{align}
$$

From here:

$$
\sqrt[^3]{1 + {1\over n^3}} \gt 1 \\
\sqrt{1 – {1\over n^2}} \lt 1
$$

Therefore:

$$
\sqrt[^3]{1 + {1\over n^3}} – \sqrt{1 – {1\over n^2}} \gt 0
$$

Which means $x_n \gt 0$.

Consider the following inequality:

$$
\sqrt[^3]{n^3 + 1} \le \sqrt{n^2 + 1} \implies \\
\implies x_n < \sqrt{n^2 + 1} – \sqrt{n^2 – 1}
$$
Or:
$$
x_n < \frac{(\sqrt{n^2 + 1} – \sqrt{n^2 – 1})(\sqrt{n^2 + 1} + \sqrt{n^2 – 1})}{\sqrt{n^2 + 1} + \sqrt{n^2 – 1}} = \\
= \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 – 1}} <2
$$

Also $x_n \gt0$ so finally:

$$
0 < x_n <2
$$

Have i done it the right way?

Answer 1

$$x_n=\frac{(n^3+1)^2-(n^2-1)^3}{\sqrt[3]{(n^3+1)^6}+\sqrt[3]{(n^3+1)^5}\sqrt{n^2-1}+...}\rightarrow0,$$ which says that $\{x_n\}$ is bounded.