<s>[INST] Limit of a Sequence involving cubic root I succeed in finding the following limit applying binomials and squeeze theorem:
$$\lim(\sqrt{n+1} - \sqrt{n}) = \lim\frac{1}{\sqrt{n+1} + \sqrt{n}} = 0$$ because $0 \leq \frac{1}{\sqrt{n+1} + \sqrt{n}} \leq \frac{1}{\sqrt{n}}$
But I need help because I'm not finding any way to simplify and solve the following limit:
$$\lim(\sqrt[3]{1-n^3} + n)$$
 [/INST] The point is that $\sqrt[3]{n^3-1}$ is awfully close to $n$ when $n$ is large. For a crude estimate, note that 
$$\left(n-\frac{1}{n}\right)^3=n^3-3n+\frac{3}{n}-\frac{1}{n^3}\lt  n^3 -1$$ if $n\ge 2$. It follows that if $n\ge 2$ then 
$$ 0\lt  n-\sqrt[3]{n^3-1} \lt n-\left(n-\frac{1}{n}\right)=\frac{1}{n},$$
and the rest follows by Squeezing.
</s>