Question

The limit $\underset{n\to \infty }{\lim }{\Pi }_{r=3}^n\frac{r^3-8}{r^3+8}$ is equal to

• A. $\frac{2}{7}$
• B. $\frac{1}{12}$
• C. $\frac{19}{52}$
• D. None of these

Answer 1

Answer: (A)

$\underset{n\to \infty }{\lim }{\Pi }_{r=3}^n\frac{r^3-8}{r^3+8}=\underset{n\to \infty }{\lim }\left(\frac{3^3-8}{3^3+8}\right)\left(\frac{4^3-8}{4^3+8}\right)........\left(\frac{n^3-8}{n^3+8}\right)$
$=\underset{n\to \infty }{\lim }\left(\frac{3-2}{3+2}.\frac{3^2+4+2\left(3\right)}{3^2+4-2\left(3\right)}\right)$
$\left(\frac{4-2}{4+2}.\frac{4^2+4+2\left(4\right)}{4^2+4-2\left(4\right)}\right)......\left(\frac{n-2}{n+2}.\frac{n^2+4+2n}{n^2+4-2n}\right)$
$=\underset{n\to \infty }{\lim }\left(\frac{3-2}{3+2}.\frac{4-2}{4+2}.\frac{5-2}{5+2}........\frac{n-2}{n+2}\right)\left(\frac{3^2+4+2\left(3\right)}{3^2+4-2\left(3\right)}.\frac{4^2+4+2\left(4\right)}{4^2+4-2\left(4\right)}........\frac{n^2+4+2n}{n^2+4-2n}\right)$
$=\left(\frac{\mathrm{1.2.3.4.5.6.7........}}{\mathrm{5.6.7.8........}}\right)\left(\frac{\mathrm{19.28.39.52.63........}}{\mathrm{7.12.19.28.39.52........}}\right)=\frac{\mathrm{1.2.3.4}}{7.12}=\frac{2}{7}$