\(\lim \left( {{n^2}\sin \frac{{n\pi }}{5} - 2{n^3}} \right)\) bằng:

* Hướng dẫn giải

\(\lim \left( {{n^2}\sin \frac{{n\pi }}{5} - 2{n^3}} \right) = \lim {n^3}\left( {\frac{{\sin \frac{{n\pi }}{5}}}{n} - 2} \right) =  - \infty \)

Vì \(\lim {n^3} =  + \infty ;\lim \left( {\frac{{\sin \frac{{n\pi }}{5}}}{n} - 2} \right) =  - 2\)

\(\left| {\frac{{\sin \frac{{n\pi }}{5}}}{n}} \right| \le \frac{1}{n};\lim \frac{1}{n} = 0 \Rightarrow \lim \left( {\frac{{\sin \frac{{n\pi }}{5}}}{n} - 2} \right) =  - 2\)