Functions Question 66
Question: $ \underset{n\to \infty }{\mathop{\lim }},
[ \frac{\Sigma n^{2}}{n^{3}} ]= $ [AMU 1999; RPET 1999, 2002]
Options:
A) $ -\frac{1}{6} $
B) $ \frac{1}{6} $
C) $ \frac{1}{3} $
D) $ -\frac{1}{3} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{n\to \infty }{\mathop{\lim }}[ \frac{n,(n+1),(2n+1)}{6n^{3}} ]=\underset{n\to \infty }{\mathop{\lim }}\frac{( 1+\frac{1}{n} ),( 2+\frac{1}{n} )}{6}=\frac{1}{3}. $ Note : Students should remember that $ \underset{n\to \infty }{\mathop{\lim }}\frac{\sum n}{n^{2}}=\frac{1}{2},,\underset{n\to \infty }{\mathop{\lim }}\frac{\sum n^{2}}{n^{3}}=\frac{1}{3} $ and $ \underset{n\to \infty }{\mathop{\lim }}\frac{\sum n^{3}}{n^{4}}=\frac{1}{4}. $
BETA
