- Hướng dẫn giải

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

Vì: $\underset{x\to -1}{\lim }{x}^2=1>0;\underset{x\to -1}{\lim }{\left(x+1\right)}^2=0$ và ${\left(x+1\right)}^2>0,\forall x\ne -1$

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

Chọn đáp án D