Question

Evaluate the following limit: $ \lim_{n \to \infty} \prod_{r=3}^n \frac{r^3 - 8}{r^3 + 8} $.

• A. \( \frac{2}{7} \)
• B. \( \frac{3}{7} \)
• C. \( \frac{4}{7} \)
• D. \( \frac{6}{7} \)

Hint:

For infinite product limits, identify factors that cancel and apply asymptotic analysis for large \( n \).

Answer 1

The Correct Option is A

We begin by expanding and simplifying the product:\[\lim_{n \to \infty} \frac{(3-2)(3^2+2^2+3\cdot 2)}{(3+2)(3^2+2^2-3\cdot 2)} \cdot \frac{(4-2)(4^2+2^2+4\cdot 2)}{(4+2)(4^2+2^2-4\cdot 2)} \cdots \frac{(n-2)(n^2+2^2+n\cdot 2)}{(n+2)(n^2+2^2-n\cdot 2)}\]\[= \lim_{n \to \infty} \frac{(3-2)(4-2)\cdots(n-2)}{(3+2)(4+2)\cdots(n+2)} \cdot \frac{(3^2+2^2+3\cdot 2)(4^2+2^2+4\cdot 2)\cdots(n^2+2^2+n\cdot 2)}{(3^2+2^2-3\cdot 2)(4^2+2^2-4\cdot 2)\cdots(n^2+2^2-n\cdot 2)}\]\[= \lim_{n \to \infty} \frac{1\cdot 2\cdots(n-2)}{5\cdot 6\cdots(n+2)} \cdot \frac{19\cdot 28\cdots(n^2+2n+4)}{7\cdot 12\cdots(n^2-2n+4)}\]\[= \lim_{n \to \infty} \frac{1\cdot 2\cdot 3\cdot 4}{n(n+1)(n+2)} \cdot \frac{(n^2+2n+4)}{(n^2-2n+4)} \cdot \frac{(n-1)}{5} \cdot \frac{(n-2)}{6} \cdots \frac{3}{n+2}\]\[= \lim_{n \to \infty} \frac{4!}{n(n+1)(n+2)} \cdot \frac{n^2(1+\frac{2}{n}+\frac{4}{n^2})}{n^2(1-\frac{2}{n}+\frac{4}{n^2})}\]\[= \frac{24}{5} \cdot \frac{1}{n^3} \cdot \frac{1}{1} = \frac{2}{7}\]