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)

$\underset{n\to \infty }{\lim }\left[\sqrt[3]{n^2-n^3}+n\right]$
$=\underset{n\to \infty }{\lim }n\left[{\left(-1+\frac{1}{n}\right)}^{1/3}+1^{1/3}\right]$
$=\underset{n\to \infty }{\lim }n\cdot \frac{\left(\frac{1}{n}-1\right)+1}{{\left(\frac{1}{n}-1\right)}^{2/3}+1-{\left(\frac{1}{n}-1\right)}^{1/3}}$
$($ Using $a+b=\frac{a^3+b^3}{a^2-ab+b^2})$
$=\underset{n\to \infty }{\lim }\frac{1}{{\left(\frac{1}{n}-1\right)}^{2/3}+1-{\left(\frac{1}{n}-1\right)}^{1/3}}$
$=\frac{1}{1+1+1}=\frac{1}{3}$