Tính giới hạn limx(x+ căn x^2+1)

* Hướng dẫn giải

Ta có: 

$$\begin{gathered} \underset{x\to -\infty }{\lim }x\left(x+\sqrt{x^2+1}\right) \\ =\underset{x\to -\infty }{\lim }x\left(x+\left|x\right|+\sqrt{1+\frac{1}{x^2}}\right) \\ =\underset{x\to -\infty }{\lim }x\left(x-x\sqrt{1+\frac{1}{x^2}}\right) \\ =\underset{x\to -\infty }{\lim }{x}^2\left(1-\sqrt{1+\frac{1}{x^2}}\right) \\ =\underset{x\to -\infty }{\lim }{x}^2\frac{1-\left(1+{\displaystyle \frac{1}{x^2}}\right)}{1+\sqrt{1+{\displaystyle \frac{1}{x^2}}}} \\ =-\frac{1}{2} \end{gathered}$$

Đáp án B