Question

The value of $\underset{n\to \infty }{lim}[\sqrt[3]{3(n+1{)}^2}-\sqrt[3]{3(n-1{)}^2}]$ is ................

• A. $0$
• B. $-1$
• C. $\infty$
• D. $1$

Answer 1

The correct option is A $0$

$\underset{n\to \infty }{lim}[\sqrt[3]{3(n+1{)}^2}-\sqrt[3]{3(n-1{)}^2}]$
$=\underset{n\to \infty }{lim}{n}^{2/3}[{(1+\frac{1}{n})}^{2/3}-{(1-\frac{1}{n})}^{2/3}]$
$=\underset{n\to \infty }{lim}{n}^{2/3}[(1+\frac{2}{3}\cdot \frac{1}{n}+\frac{\frac{2}{3}(\frac{2}{3}-1)}{2!}\frac{1}{n^2}\cdot \cdot \cdot )-(1-\frac{2}{3}\cdot \frac{1}{n}+\frac{\frac{2}{3}(\frac{2}{3}-1)}{2!}\frac{1}{n^2}\cdot \cdot \cdot )]$
$=\underset{n\to \infty }{lim}{n}^{2/3}[\frac{4}{3}\cdot \frac{1}{n}+\frac{8}{81}\cdot \frac{1}{n^3}+\cdot \cdot \cdot ]$
$=\underset{n\to \infty }{lim}[\frac{4}{3}\cdot \frac{1}{n^{1/3}}+\frac{8}{81}\cdot \frac{1}{n^{7/3}}+\cdot \cdot \cdot ]=0$
Hence, option 'A' is correct.