Tính 

- Hướng dẫn giải

$$\begin{gathered} \underset{x\to {\left(-1\right)}^{+}}{\lim }\frac{x^2+4x+3}{\sqrt{x^3+x^2}}=\underset{x\to {\left(-1\right)}^{+}}{\lim }\frac{\left(x+1\right)\left(x+3\right)}{\sqrt{x^2\left(x+1\right)}} \\ =\underset{x\to {\left(-1\right)}^{+}}{\lim }\frac{\sqrt{x+1}\left(x+3\right)}{\sqrt{x^2}}=\frac{0}{1}=0. \end{gathered}$$

Chọn đáp án D