I'm trying to solve the following question. Have tried taking $\log$ on both sides, but didn't get very far. Seems difficult to apply L'Hopital's rule.
$L = \lim_{n \rightarrow \infty} \prod_{r=3}^{n} \frac{r^3 - 8}{r^3 + 8}$
Any help is much appreciated.
$$\prod_{r=3}^{n}\frac{r^3-8}{r^3+8}=\prod_{r=3}^{n}\frac{r-2}{r+2}\prod_{r=3}^{n}\frac{(r+1)^2+3}{(r-1)^2+3}=\frac{\prod_{k=1}^{n-2}k}{\prod_{k=5}^{n+2}k}\cdot\frac{\prod_{h=4}^{n+1}(h^2+3)}{\prod_{h=2}^{n-1}(h^2+3)}$$ for any $n\geq 7$ can be written as $$ \frac{4!}{(n+2)(n+1)n(n-1)}\cdot\frac{((n+1)^2+3)(n^2+3)}{(2^2+3)(3^2+3)} $$ whose limit as $n\to +\infty$ is given by $$ \frac{4!}{(2^2+3)(3^2+3)}=\frac{24}{7\cdot 12}=\color{red}{\frac{2}{7}}.$$