# How does one evaluate $\lim\limits _{n\to \infty }\left(\prod_{x=2}^{n}\frac{x^3-1}{x^3+1}\right)$?

I tried this form:
$$\lim_{n\to+\infty}\left(\prod_{x=2}^{n}\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)$$ but it doesn’t ring any bell.

#### Solutions Collecting From Web of "How does one evaluate $\lim\limits _{n\to \infty }\left(\prod_{x=2}^{n}\frac{x^3-1}{x^3+1}\right)$?"

Write out the fraction

$\frac {(x-1)(x^2+x+1)}{(x+1)(x^2-x+1)}$

for $x\in \{2,3,…,10\}$. Multiply these together and note that most factors cancel; you have a telescoping product. Once you figure out how the product telescopes, you can extrapolate to $n\rightarrow \infty$ and extract the terms at the “front end” of the telescope to get the limit.