lim x tiến đến dương vô cùng (x^3+2x^2−1) bằng

Ta có: $\underset{\mathrm{x}\to +\infty }{\lim }\left({\mathrm{x}}^3+2{\mathrm{x}}^2-1\right)$$=\underset{\mathrm{x}\to +\infty }{\lim }{\mathrm{x}}^3\left(1+\frac{2}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^3}\right)$

Vì $\underset{\mathrm{x}\to +\infty }{\lim }{\mathrm{x}}^3=+\infty$  và $\underset{\mathrm{x}\to +\infty }{\lim }\left(1+\frac{2}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^3}\right)=1>0$

Suy ra $\underset{\mathrm{x}\to +\infty }{\lim }{\mathrm{x}}^3\left(1+\frac{2}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^3}\right)=+\infty$

Vậy $\underset{\mathrm{x}\to +\infty }{\lim }\left({\mathrm{x}}^3+2{\mathrm{x}}^2-1\right)=+\infty$

Chọn đáp án C.