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

$$\sqrt[3]{n^3+1}-\sqrt{n^2-1}=\sqrt[3]{n^3+1}-n-\sqrt{n^2-1}+n
\\=\frac1{(n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2}+\frac1{\sqrt{n^2-1}+n}$$
and both terms are decreasing. The supremum is with $n=1$,
$$x_1=\sqrt[3]2.$$
