Question

$\underset{n\to \infty }{lim}\frac{2^3-1^3}{2^3+1^3}.\frac{3^3-1^3}{3^3+1^3}....\frac{n^3-1}{n^3+1}$ equals?

• A. $\frac{4}{5}$
• B. $\frac{2}{3}$
• C. $\frac{2}{5}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{2}{3}$

$\underset{n\to \infty }{lim}\frac{2^3-1^3}{2^3+1^3}.\frac{3^3-1^3}{3^3+1^3}....\frac{n^3-1}{n^3+1}=\underset{n\to \infty }{lim}\frac{(2-1)(2^2+2.1+1)}{(2+1)(2^2-2.1+1)}.\frac{(3-1)(3^2+3.1+1)}{(3+1)(3^2-3.1+1)}....\frac{(n-1)(n^2+n.1+1)}{(n+1)(n^2-n.1+1)}=\underset{n\to \infty }{lim}\frac{1.7}{3.3}.\frac{2.13}{4.7}.\frac{3.21}{5.13}....\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}=\underset{n\to \infty }{lim}\frac{2}{3}.\frac{n^2+n+1}{n(n+1)}=\frac{2}{3}.\frac{1}{1}=\frac{2}{3}$