Question

The value of $\displaystyle\lim _{n \rightarrow \infty}\left[\sqrt[3]{n^{2}-n^{3}}+n\right]$ is

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $-\frac{2}{3}$

Answer 1

Answer: (A)

$\displaystyle\lim _{n \rightarrow \infty}\left[\sqrt[3]{n^{2}-n^{3}}+n\right]$
$=\displaystyle\lim _{n \rightarrow \infty} n\left[\left(-1+\frac{1}{n}\right)^{1 / 3}+1^{1 / 3}\right]$
$=\displaystyle\lim _{n \rightarrow \infty} n \cdot \frac{\left(\frac{1}{n}-1\right)+1}{\left(\frac{1}{n}-1\right)^{2 / 3}+1-\left(\frac{1}{n}-1\right)^{1 / 3}}$
$\left(\right.$ Using $\left.a+b=\frac{a^{3}+b^{3}}{a^{2}-a b+b^{2}}\right)$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{\left(\frac{1}{n}-1\right)^{2 / 3}+1-\left(\frac{1}{n}-1\right)^{1 / 3}}$
$=\frac{1}{1+1+1}=\frac{1}{3}$