Question

The value of $\underset{n\to \infty }{\lim }\left\{\sqrt[3]{n^2-n^3}+n\right\}$, is

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

Answer 1

Answer: (A)

We have,
$l=\underset{n\to \infty }{\lim }\left\{\sqrt[3]{n^2-n^3}+n\right\}$
$\Rightarrow l=\underset{n\to \infty }{\lim }\left\{n\sqrt[3]{\frac{1}{n}-1+n}\right\}$
$=\underset{n\to \infty }{\lim }n\left\{{\left(\frac{1}{n}-1\right)}^{1/3}+1^{1/3}\right\}$
$\Rightarrow l=\underset{n\to \infty }{\lim }n\left[\frac{\left(\frac{1}{n}-1\right)+1}{{\left(\frac{1}{n}-1\right)}^{2/3}-{\left(\frac{1}{n}-1\right)}^{1/3}+1}\right]$
$\left[\because a+b=\frac{a^3+b^3}{a^2-ab+b^2}\right]$
$\Rightarrow l=\underset{n\to \infty }{\lim }\left[\frac{1}{{\left(\frac{1}{n}-1\right)}^{2/3}-{\left(\frac{1}{n}-1\right)}^{1/3}+1}\right]=\frac{1}{3}$