Question

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

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

Answer 1

The correct option is A $\frac{1}{3}$

$t=\underset{n\to \infty }{lim}[\sqrt[3]{3n^2-n^3}+n]\Rightarrow t=\underset{n\to 0}{lim}\sqrt[3]{3\frac{1}{n^2}-\frac{1}{n^3}}+\frac{1}{n}\Rightarrow t=\underset{n\to 0}{lim}\frac{\sqrt[3]{3n-1}}{n}+\frac{1}{n}\Rightarrow t=\underset{n\to 0}{lim}\frac{\sqrt[3]{3n-1}+1}{n}$

This is $\frac{0}{0}$ form so by using the L hospital rule differentiating numerator and denomirator

$\Rightarrow t={lim}_{n\to 0}\frac{\frac{1}{3}{(n-1)}^{-\frac{2}{3}}+0}{1}\Rightarrow t={lim}_{n\to 0}\frac{{(n-1)}^{-\frac{2}{3}}}{3}\Rightarrow t=\frac{{(0-1)}^{-\frac{2}{3}}}{3}=\frac{1}{3}$

So, option A is correct.