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: \[ \lim_{n \to \infty} \frac{(3-2)(3^2+2^2+3\cdot 2)}{(3+2)(3^2+2^2-3\cdot 2)} \cdot \frac{(4-2)(4^2+2^2+4\cdot 2)}{(4+2)(4^2+2^2-4\cdot 2)} \cdots \frac{(n-2)(n^2+2^2+n\cdot 2)}{(n+2)(n^2+2^2-n\cdot 2)}\] \[= \lim_{n \to \infty} \frac{(3-2)(4-2)\cdots(n-2)}{(3+2)(4+2)\cdots(n+2)} \cdot \frac{(3^2+2^2+3\cdot 2)(4^2+2^2+4\cdot 2)\cdots(n^2+2^2+n\cdot 2)}{(3^2+2^2-3\cdot 2)(4^2+2^2-4\cdot 2)\cdots(n^2+2^2-n\cdot 2)}\] \[= \lim_{n \to \infty} \frac{1\cdot 2\cdots(n-2)}{5\cdot 6\cdots(n+2)} \cdot \frac{19\cdot 28\cdots(n^2+2n+4)}{7\cdot 12\cdots(n^2-2n+4)}\] \[= \lim_{n \to \infty} \frac{1\cdot 2\cdot 3\cdot 4}{n(n+1)(n+2)} \cdot \frac{(n^2+2n+4)}{(n^2-2n+4)} \cdot \frac{(n-1)}{5} \cdot \frac{(n-2)}{6} \cdots \frac{3}{n+2}\] \[= \lim_{n \to \infty} \frac{4!}{n(n+1)(n+2)} \cdot \frac{n^2(1+\frac{2}{n}+\frac{4}{n^2})}{n^2(1-\frac{2}{n}+\frac{4}{n^2})}\] \[= \frac{24}{5} \cdot \frac{1}{n^3} \cdot \frac{1}{1} = \frac{2}{7} \]

Answer 2

Step 1: Factor the terms inside the product
We are given the limit: \[ L = \lim_{n \to \infty} \prod_{r=3}^n \frac{r^3 - 8}{r^3 + 8} \] First, factor \( r^3 - 8 \) and \( r^3 + 8 \) as difference and sum of cubes respectively: \[ r^3 - 8 = (r - 2)(r^2 + 2r + 4) \] \[ r^3 + 8 = (r + 2)(r^2 - 2r + 4) \] Thus, the product becomes: \[ L = \lim_{n \to \infty} \prod_{r=3}^n \frac{(r - 2)(r^2 + 2r + 4)}{(r + 2)(r^2 - 2r + 4)} \]
Step 2: Simplify the product
Notice that for large \( r \), the quadratic terms \( r^2 + 2r + 4 \) and \( r^2 - 2r + 4 \) behave similarly. For large \( r \), their ratio tends to 1. Therefore, the key part of the product is the ratio of linear terms \( (r - 2) \) and \( (r + 2) \). Now, rewrite the product focusing on the linear terms: \[ L = \lim_{n \to \infty} \prod_{r=3}^n \frac{r - 2}{r + 2} \]
Step 3: Evaluate the infinite product
The product \( \prod_{r=3}^\infty \frac{r - 2}{r + 2} \) is a telescoping product. Write the first few terms: \[ \frac{1}{5} \times \frac{2}{6} \times \frac{3}{7} \times \cdots \] The general term is \( \frac{r - 2}{r + 2} \), and as \( r \) increases, the product tends to: \[ \prod_{r=3}^\infty \frac{r - 2}{r + 2} = \frac{1}{5} \times \frac{2}{6} \times \frac{3}{7} \times \cdots \] This product converges to \( \frac{2}{7} \), which is the value of the limit.
Step 4: Conclusion
The value of the limit is: \[ \boxed{\frac{2}{7}} \]