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  4. The limit displaystyle limn→∞ Πnr = 3 (r3-8/r3+8) is equal to

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Q. The limit $\displaystyle\lim_{n\to\infty} \Pi^{n}_{r = 3} \frac{r^{3}-8}{r^{3}+8}$ is equal to

Limits and Derivatives

Solution:

$\displaystyle\lim_{n\to\infty} \Pi^{n}_{r = 3} \frac{r^{3}-8}{r^{3}+8} = \displaystyle\lim _{n\to \infty } \left(\frac{3^{3}-8}{3^{3}+8}\right)\left(\frac{4^{3}-8}{4^{3}+8}\right)........\left(\frac{n^{3}-8}{n^{3}+8}\right)$
$= \displaystyle\lim _{n\to \infty }\left(\frac{3-2}{3+2}. \frac{3^{2}+4+2\left(3\right)}{3^{2}+4-2\left(3\right)}\right)$
$\left(\frac{4-2}{4+2}. \frac{4^{2}+4+2\left(4\right)}{4^{2}+4-2\left(4\right)}\right)\,......\left(\frac{n-2}{n+2}. \frac{n^{2}+4+2n}{n^{2}+4-2n}\right)$
$= \displaystyle\lim _{n\to \infty }\left(\frac{3-2}{3+2}. \frac{4-2}{4+2}. \frac{5-2}{5+2}........ \frac{n-2}{n+2}\right)\left(\frac{3^{2}+4+2\left(3\right)}{3^{2}+4-2\left(3\right)}. \frac{4^{2}+4+2\left(4\right)}{4^{2}+4-2\left(4\right)}........\frac{n^{2}+4+2n}{n^{2}+4-2n}\right)$
$= \left(\frac{1.2.3.4.5.6.7........}{5.6.7.8........}\right)\left(\frac{19.28.39.52.63........}{7.12.19.28.39.52........}\right) = \frac{1.2.3.4}{7.12} = \frac{2}{7}$