Question

$li{m}_{n\to \infty }\frac{2^3-1^3}{2^3+1^3}.\frac{3^3-1^3}{3^3+1^3}....\frac{n^3-1}{n^3+1}$ equal

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

Answer 1

Answer: (B)

$\frac{2}{3}$

Now,

$\underset{x\to \infty }{lim}\frac{2}{3}\left(\frac{n^2\left(1+1/n+1/n^2\right)}{n^2\left(1+1/n\right)}\right)$

$\Rightarrow \underset{x\to \infty }{lim}\frac{2}{3}\left(\frac{1+\frac{1}{n}+\frac{1}{n^2}}{1+\frac{1}{n}}\right)$

$\Rightarrow \frac{2}{3}\left(\frac{1+\frac{1}{\infty }+\frac{1}{\infty }}{1+\frac{1}{\infty }}\right)$

$\Rightarrow \frac{2}{3}\left(\frac{1+0+0}{1+0}\right)$

$\Rightarrow \frac{2}{3}\times 1=\frac{2}{3}$