lim x->∞ 3n^2 +4n-1/3n^2 +2n +7)^2n +5

Решите задачу:

$\lim_{n \to \infty} \left(\cfrac{3n^2+4n-1}{3n^2+2n+7}\right)^{2n+5}= \lim_{n \to \infty} \left(1+\cfrac{2n-8}{3n^2+2n+7}\right)^{2n+5}=\\= \lim_{n \to \infty} \left(1+\cfrac{1}{\cfrac{2n-8}{3n^2+2n+7}}\right)^{\tfrac{3n^2+2n+7}{2n-8}\cdot(2n+5)\cdot\tfrac{2n-8}{3n^2+2n+7}}=\\= \lim_{n \to \infty} e^{(2n+5)\cdot\tfrac{2n-8}{3n^2+2n+7}}=e^{\tfrac{4}{3}}=\sqrt[3]{e^4}$