$\lim _\limits{n\to \infty }\left(4n^2\left(\frac{n^3+3n^2+3}{2n^3+n-1}\right)^n\right)$ How can I calculate this limit?
$\lim_\limits{n\to \infty }\left(4n^2\left(\frac{n^3+3n^2+3}{2n^3+n-1}\right)^n\right)\:$
Can I apply the rule $a_n+1/a_n$ to proove its convergence?

Hint. Note that
$$a_n:=4n^2\left(\frac{n^3+3n^2+3}{2n^3+n-1}\right)^n=4n^2\left(\frac{1+3/n+3/n^3}{2+1/n^2-1/n^3}\right)^n.$$
Can we say that $a_n\sim\frac{4n^2}{2^n}$?
