lim(n^2 sin $\frac{n\ pi }{5}$ – 2n^3) bằng?

Giải thích các bước giải:

Ta có :

$-1\le \sin\dfrac{n\pi}{5}\le 1$

$\rightarrow -n^2-2n^3\le n^2.\sin\dfrac{n\pi}{5}-2n^3\le n^2-2n^3$ 

$\rightarrow \lim(-n^2-2n^3)\le\lim ( n^2.\sin\dfrac{n\pi}{5}-2n^3)\le \lim(n^2-2n^3)$ 

$\rightarrow \lim-n^2(1+2n)\le\lim ( n^2.\sin\dfrac{n\pi}{5}-2n^3)\le \lim n^2(1-2n)$ 

$\rightarrow -\infty\le\lim ( n^2.\sin\dfrac{n\pi}{5}-2n^3)\le -\infty$ 

$\rightarrow \lim ( n^2.\sin\dfrac{n\pi}{5}-2n^3)= -\infty$