Giải thích các bước giải:
Ta có :
$-1\le \sin\dfrac{n\pi}{5}\le 1$
$\rightarrow -n^2-2n^3\le n^2.\sin\dfrac{n\pi}{5}-2n^3\le n^2-2n^3$
$\rightarrow \lim(-n^2-2n^3)\le\lim ( n^2.\sin\dfrac{n\pi}{5}-2n^3)\le \lim(n^2-2n^3)$
$\rightarrow \lim-n^2(1+2n)\le\lim ( n^2.\sin\dfrac{n\pi}{5}-2n^3)\le \lim n^2(1-2n)$
$\rightarrow -\infty\le\lim ( n^2.\sin\dfrac{n\pi}{5}-2n^3)\le -\infty$
$\rightarrow \lim ( n^2.\sin\dfrac{n\pi}{5}-2n^3)= -\infty$
