Tính 

- Hướng dẫn giải

$\begin{array}{l}\underset{x\to -\infty }{\lim }\left(\sqrt{x^2+1}+x-1\right)\\ =\underset{x\to -\infty }{\lim }\frac{\left(\sqrt{x^2+1}+x-1\right)\left(\sqrt{x^2+1}-x+1\right)}{\left(\sqrt{x^2+1}-x+1\right)}\\ =\underset{x\to -\infty }{\lim }\frac{x^2+1-{(x-1)}^2}{\sqrt{x^2+1}-x+1}\\ =\underset{x\to -\infty }{\lim }\frac{x^2+1-x^2+2x-1}{\sqrt{x^2+1}-x+1}\\ =\underset{x\to -\infty }{\lim }\frac{2x}{\left|x\right|\sqrt{1+\frac{1}{x^2}}-x+1}=\underset{x\to -\infty }{\lim }\frac{2x}{-x\sqrt{1+\frac{1}{x^2}}-x+1}\\ =\underset{x\to -\infty }{\lim }\frac{2}{-\sqrt{1+\frac{1}{x^2}}-1+\frac{1}{x}}=\frac{2}{-1-1+0}=-1\end{array}$

Đáp án cần chọn là: A