Find the limit?

${\lim}_{n \rightarrow \infty} {x}_{n}$ where ${x}_{n} = \sqrt[3]{\left({n}^{2}\right) - \left({n}^{3}\right)} + n$

Mar 23, 2018

$\frac{1}{3}$

Explanation:

Assuming the limit as

${\lim}_{n \to \infty} - \sqrt[3]{\left\mid {n}^{2} - {n}^{3} \right\mid} + n$ we have

$- \sqrt[3]{\left\mid {n}^{2} - {n}^{3} \right\mid} + n = n \left(- \sqrt[3]{\left\mid \frac{1}{n} - 1 \right\mid} + 1\right) =$

$= \frac{- \sqrt[3]{1 - \frac{1}{n}} + \sqrt[3]{1}}{\frac{1}{n}} = \frac{\sqrt[3]{1 - \frac{1}{n}} + \sqrt[3]{1}}{- \frac{1}{n}}$

now calling $h = - \frac{1}{n}$ we have

${\lim}_{n \to \infty} - \sqrt[3]{\left\mid {n}^{2} - {n}^{3} \right\mid} + n = {\lim}_{h \to 0} \frac{\sqrt[3]{1 + h} + \sqrt[3]{1}}{h} = \frac{1}{3}$

Mar 23, 2018

Explanation:

Use $\left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right) = {a}^{3} + {b}^{3}$ to remove the third root in the numerator.

$\sqrt[3]{{n}^{2} - {n}^{3}} + n = \frac{\left(\sqrt[3]{{n}^{2} - {n}^{3}} + n\right)}{1} \cdot \frac{\left({\left(\sqrt[3]{{n}^{2} - {n}^{3}}\right)}^{2} - n \sqrt[3]{{n}^{2} - {n}^{3}} + {n}^{2}\right)}{\left({\left(\sqrt[3]{{n}^{2} - {n}^{3}}\right)}^{2} - n \sqrt[3]{{n}^{2} - {n}^{3}} + {n}^{2}\right)}$

$= \frac{{n}^{2} - {n}^{3} + {n}^{3}}{{\left(\sqrt[3]{{n}^{2} - {n}^{3}}\right)}^{2} - n \sqrt[3]{{n}^{2} - {n}^{3}} + {n}^{2}}$

$= {n}^{2} / \left({n}^{2} {\left(\sqrt[3]{\frac{1}{n} - 1}\right)}^{2} - {n}^{2} \sqrt[3]{\frac{1}{n} - 1} + {n}^{2}\right)$

$= \frac{1}{{\left(\sqrt[3]{\frac{1}{n} - 1}\right)}^{2} - \sqrt[3]{\frac{1}{n} - 1} + 1}$

So,

${\lim}_{n \rightarrow \infty} \left(\sqrt[3]{{n}^{2} - {n}^{3}} + n\right) = {\lim}_{n \rightarrow \infty} \frac{1}{{\left(\sqrt[3]{\frac{1}{n} - 1}\right)}^{2} - \sqrt[3]{\frac{1}{n} - 1} + 1}$

$= \frac{1}{{\left(\sqrt[3]{0 - 1}\right)}^{2} - \sqrt[3]{0 - 1} + 1}$

$= \frac{1}{{\left(- 1\right)}^{2} - \left(- 1\right) + 1}$

$= \frac{1}{3}$