Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. $None$

Answer 1

The correct option is C $0$

We know that $a^3-b^3=(a-b)(a^2+b^2+ab)$

So, $(a-b)=\frac{a^3-b^3}{(a^2+b^2+ab)}$

So, $(n+1{)}^{2/3}-(n-1{)}^{2/3}$ $=\frac{(n+1{)}^2-(n-1{)}^2}{\left(\right(n+1{)}^4+(n-1{)}^4+(n+1{)}^2(n-1{)}^2}=\frac{2\times 2n}{3n^4+10n^2+3}=L$

So, ${lim}_{\to \infty }L=0$