Differentiation Question 476
Question: The limit $ \underset{n,\to ,\infty }{\mathop{\lim }}\underset{r=3}{\overset{n}{\mathop{\prod }}},\frac{r^{3}-8}{r^{3}+8} $ is equal to
Options:
A) $ \frac{2}{7} $
B) $ \frac{1}{12} $
C) $ \frac{19}{52} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \underset{n\to \infty }{\mathop{Lim}},\underset{r=3}{\overset{n}{\mathop{II}}},\frac{r^{3}-8}{r^{3}+8} $ $ =\underset{n\to \infty }{\mathop{Lim}},( \frac{3^{3}-8}{3^{3}+8} )( \frac{4^{3}-8}{4^{3}+8} )….( \frac{n^{3}-8}{n^{3}+8} ) $ $ =\underset{n\to \infty }{\mathop{Lim}},( \frac{3-2}{3+2}.\frac{3^{2}+4+2(3)}{3^{2}+4-2(3)} ) $ $ ( \frac{4-2}{4+2}.\frac{4^{2}+4+2(4)}{4^{2}+4-2(4)} )…( \frac{n-2}{n+2}.\frac{n^{2}+4+2n}{n^{2}+4-2n} ) $ $ =\underset{n\to \infty }{\mathop{Lim}},( \frac{3-2}{3+2}.\frac{4-2}{4+2}.\frac{5-2}{5+2}….\frac{n-2}{n+2} ) $ $ ( \frac{3^{3}+4+2(3)}{3^{3}+4-2(3)}.\frac{4^{2}+4+2(4)}{4^{2}+4-2(4)}…\frac{n^{2}+4+2n}{n^{2}+4-2n} ) $ $ =( \frac{1.2.3.4.5.6.7…}{5.6.7.8…..} )( \frac{19.28.39.52.63……}{7.12.19.28.39.52…..} ) $ $ =\frac{1.2.3.4}{7.12}=\frac{2}{7} $
BETA
