Tính 

A

- Hướng dẫn giải

Đáp án A

Ta có:

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

$=\underset{x\to 1}{\lim }\frac{4x+3}{\left(2x+\sqrt{x+3}\right)\left(x+1\right)}=\frac{7}{8}.$