Functions Question 681
Question: $ \underset{n\to \infty }{\mathop{\lim }},\frac{n{{(2n+1)}^{2}}}{(n+2)(n^{2}+3n-1)}= $
Options:
A) 0
B) 2
C) 4
D) $ \infty $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{n\to \infty }{\mathop{\lim }}\frac{n,{{(2n+1)}^{2}}}{(n+2)(n^{2}+3n-1)}=\underset{n\to \infty }{\mathop{\lim }},\frac{4n^{3}+4n^{2}+n}{n^{3}+5n^{2}+5n-2} $ $ =\underset{n\to \infty }{\mathop{\lim }},\frac{n^{3},( 4+\frac{4}{n}+\frac{1}{n^{2}} )}{n^{3}( 1+\frac{5}{n}+\frac{5}{n^{2}}-\frac{2}{n^{3}} )}=4 $
BETA
