Answer
$\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {\sqrt[3]{{{n^3} + 1}} - n} \right) = \frac{1}{3}$
Work Step by Step
We have
$\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {\sqrt[3]{{{n^3} + 1}} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt[3]{{{n^3} + 1}} - n}}{{\frac{1}{{{n^2}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n\left( {\sqrt[3]{{1 + \frac{1}{{{n^3}}}}} - 1} \right)}}{{\frac{1}{{{n^2}}}}}$
$\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {\sqrt[3]{{{n^3} + 1}} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt[3]{{1 + \frac{1}{{{n^3}}}}} - 1}}{{\frac{1}{{{n^3}}}}}$
Using L'Hôpital's Rule on the right-hand side, we get
$\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {\sqrt[3]{{{n^3} + 1}} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{3}{{\left( {1 + \frac{1}{{{n^3}}}} \right)}^{ - 2/3}}\left( { - \frac{3}{{{n^4}}}} \right)}}{{ - \frac{3}{{{n^4}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{3}{\left( {1 + \frac{1}{{{n^3}}}} \right)^{ - 2/3}}$
So,
$\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {\sqrt[3]{{{n^3} + 1}} - n} \right) = \frac{1}{3}\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{{\left( {1 + \frac{1}{{{n^3}}}} \right)}^{2/3}}}} = \frac{1}{3}$
