Question

$\underset{n\to \infty }{lim}\frac{n(2n+1{)}^2}{(n+2)(n^2+3n-1)}$ is equal to

• A. $0$
• B. $2$
• C. $4$
• D. $\infty$

Answer 1

The correct option is C $4$

$\underset{n\to \infty }{lim}\frac{n(2n+1{)}^2}{(n+2)(n^2+3n-1)}$
$=\underset{n\to \infty }{lim}\frac{{(2+\frac{1}{n})}^2}{(1+\frac{2}{n})(1+\frac{3}{n}-\frac{1}{n^2})}$
$=\frac{(2+0{)}^2}{(1+0)(1+0+0)}=4$