Functions Question 313
Question: $ \underset{n\to \infty }{\mathop{\lim }},[ \frac{1}{n^{3}+1}+\frac{4}{n^{3}+1}+\frac{9}{n^{3}+1}+……..+\frac{n^{2}}{n^{3}+1} ]= $
Options:
A) $ 1 $
B) 2/3
C) 1/3
D) $ 0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Given limit $ =\underset{n\to \infty }{\mathop{\lim }},\frac{1^{2}+2^{2}+3^{2}+…..+n^{2}}{1+n^{3}}=\underset{n\to \infty }{\mathop{\lim }},\frac{\Sigma n^{2}}{1+n^{3}} $ $ =\underset{n\to \infty }{\mathop{\lim }},\frac{1}{6}\frac{n,(n+1),(2n+1)}{1+n^{3}} $ $ =\underset{n\to \infty }{\mathop{\lim }}\frac{1}{6}\frac{( 1+\frac{1}{n} ),( 2+\frac{1}{n} )}{( \frac{1}{n^{3}}+1 )} $ $ =\frac{1}{6},.,1,.,\frac{2}{(1)}=( \frac{1}{3} ). $
BETA
