lim x tiến đến dương vô cùng (x^3+2x^2+1)/(x^2+3x−4) bằng

Ta có: $\underset{\mathrm{x}\to +\infty }{\lim }\frac{{\mathrm{x}}^3+2{\mathrm{x}}^2+1}{{\mathrm{x}}^2+3\mathrm{x}-4}=\underset{\mathrm{x}\to +\infty }{\lim }\frac{{\mathrm{x}}^3(1+\frac{2}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^3})}{{\mathrm{x}}^2(1+\frac{3}{\mathrm{x}}-\frac{4}{{\mathrm{x}}^2})}$$=\underset{\mathrm{x}\to +\infty }{\lim }\mathrm{x}.\underset{\mathrm{x}\to +\infty }{\lim }\frac{1+\frac{2}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^3}}{1+\frac{3}{\mathrm{x}}-\frac{4}{{\mathrm{x}}^2}}$

Vì $\left\{\begin{array}{l}\underset{\mathrm{x}\to +\infty }{\lim }\mathrm{x}=+\infty \\ \underset{\mathrm{x}\to +\infty }{\lim }\frac{1+\frac{2}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^3}}{1+\frac{3}{\mathrm{x}}-\frac{4}{{\mathrm{x}}^2}}=1\end{array}\right.$ nên $\underset{\mathrm{x}\to +\infty }{\lim }\frac{{\mathrm{x}}^3+2{\mathrm{x}}^2+1}{{\mathrm{x}}^2+3\mathrm{x}-4}=+\infty$

Chọn đáp án B.