Q44 / 100 · BITSAT 2024 (Memory Based Paper 1) · Mathematics
$\lim _{n \rightarrow \infty} \prod_{r=3}^{\infty} \frac{r^3-8}{r^3+8}$ equals to
- Correct answerA$\frac{2}{7}$
- B$\frac{3}{7}$
- C$\frac{4}{7}$
- D$\frac{6}{7}$
Solution
$\begin{aligned} & \lim _{n \rightarrow \infty} \prod_{r=3}^{\infty} \frac{r^3-8}{r^3+8}=\left(\frac{3^3-8}{3^3+8}\right)\left(\frac{4^3-8}{4^3+8}\right) \\ & \text {......... }\left(\frac{n^3-8}{n^3+8}\right)\end{aligned}$
$\begin{array}{r}=\lim _{n \rightarrow \infty}\left[\frac{(3-2)\left(3^2+2^2+3.2\right)}{(3+2)\left(3^2+2^2-3.2\right)}\right]\left[\frac{(4-2)\left(4^2+2+4.2\right)}{(4+2)\left(4^2+2^2-4.2\right)}\right] \\ \ldots \ldots \ldots . .\left[\frac{(n-2)\left(n^2+2^n+n .2\right)}{(n+2)\left(n^2+2^2-n .2\right)}\right]\end{array}$
$\begin{gathered}\quad=\lim _{n \rightarrow \infty}\left[\frac{(3-2)(4-2) \ldots \ldots \ldots(n-2)}{(3+2)(4+2) \ldots \ldots \ldots(n+2)}\right] \\ {\left[\frac{\left(3^2+2^2+3.2\right)\left(4^2+2^2+4.2\right) \ldots \ldots .\left(n^2+2^2+n .2\right)}{\left(3^2+2^2-3.2\right)\left(4^2+2^2-4.2\right) \ldots \ldots . .\left(n^2+2^2-n .2\right)}\right]}\end{gathered}$
$=\left[\frac{1 \cdot 2 \cdot 3.4 .5 .6 \ldots \ldots \ldots .}{5 \cdot 6.7 .8 \ldots \ldots \ldots . .}\right]\left[\frac{19 \cdot 28 \cdot 39.52 .63 \ldots \ldots \ldots . .}{7 \cdot 12 \cdot 19.28 .39 .52 \ldots \ldots . .}\right]$
$=\frac{2}{7}$
Hence, the correct answer is option: 'A'.