A) 0
B) 2
C) 4
D) \[\infty \]
Correct Answer: C
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{n\,{{(2n+1)}^{2}}}{(n+2)\,\,({{n}^{2}}+3n-1)}=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{4{{n}^{3}}+4{{n}^{2}}+n}{{{n}^{3}}+5{{n}^{2}}+5n-2}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{n}^{3}}\,\left( 4+\frac{4}{n}+\frac{1}{{{n}^{2}}} \right)}{{{n}^{3}}\left( 1+\frac{5}{n}+\frac{5}{{{n}^{2}}}-\frac{2}{{{n}^{3}}} \right)}=4\]
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