Question

$\underset{n \rightarrow \infty}{Lim}$$\frac{r^{3} - 8}{r^{3} + 8}$ is equal to

• A. 2/7
• B. 3/7
• C. 7/2
• D. 7/3

Answer 1

Answer: (A)

2/7

Answer 2

$\underset{n \rightarrow \infty}{Lim}$ $\prod _ { r = 3 } ^ { n }$ $\frac{r^{3} - 8}{r^{3} + 8}$ = $\left( \frac{3^{3} - 8}{3^{3} + 8} \right)$ $\left( \frac{4^{3} - 8}{4^{3} + 8} \right)$........$\left( \frac{n^{3} - 8}{n^{3} + 8} \right)$

= $\underset{n \rightarrow \infty}{Lim}$ $\left\lbrack \frac{(3 - 2)(3^{2} + 2^{2} + 3.2)}{(3 + 2)(3^{2} + 2^{2} - 3.2)} \right\rbrack$

$\left\lbrack \frac{(4 - 2)(4^{2} + 2^{2} + 4.2)}{(4 + 2)(4^{2} + 2^{2} - 4.2)} \right\rbrack$.........$\left\lbrack \frac{(n - 2)(n^{2} + 2^{2} + n.2)}{(n + 2)(n^{2} + 2^{2} - n.2)} \right\rbrack$

= $\underset{n \rightarrow \infty}{Lim}$ $\left\lbrack \frac{(3 - 2)(4 - 2).......(n - 2)}{(3 + 2)(4 + 2).......n + 2} \right\rbrack$

$\left\lbrack \frac{(3^{2} + 2^{2} + 3.2)(4^{2} + 2^{2} + 4.2)......(n^{2} + 2^{2} + n.2)}{(3^{2} + 2^{2} - 3.2)(4^{2} + 2^{2} - 4.2).....(n^{2} + 2^{2} - n.2)} \right\rbrack$

= $\left\lbrack \frac{1.2.3.4.5.6........}{5.6.7.8.....} \right\rbrack$ $\left\lbrack \frac{19.28.39.52.63..........}{7.12.19.28.39.52..........} \right\rbrack$

= $\frac{1.2.3.4}{7.12}$= 2/7