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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2024 · Paper 1 · Mathematics · Definite Integration

_ n arrow ∞ _ r=3 ^ ∞ (r^3-8)/(r^3+8) equals to

lim⁡n→∞∏r=3∞r3−8r3+8\lim _{n \rightarrow \infty} \prod_{r=3}^{\infty} \frac{r^3-8}{r^3+8} equals to

Options

  1. A27\frac{2}{7}
  2. B37\frac{3}{7}
  3. C47\frac{4}{7}
  4. D67\frac{6}{7}

Correct answer

A. 27\frac{2}{7}

Step-by-step solution

lim⁡n→∞∏r=3∞r3−8r3+8=(33−833+8)(43−843+8)......... (n3−8n3+8)\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}=lim⁡n→∞[(3−2)(32+22+3.2)(3+2)(32+22−3.2)][(4−2)(42+2+4.2)(4+2)(42+22−4.2)]………..[(n−2)(n2+2n+n.2)(n+2)(n2+22−n.2)]\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}=lim⁡n→∞[(3−2)(4−2)………(n−2)(3+2)(4+2)………(n+2)][(32+22+3.2)(42+22+4.2)…….(n2+22+n.2)(32+22−3.2)(42+22−4.2)……..(n2+22−n.2)]\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} =[1⋅2⋅3.4.5.6……….5⋅6.7.8………..][19⋅28⋅39.52.63………..7⋅12⋅19.28.39.52……..]=\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] =27=\frac{2}{7}
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