Tính lim từ x đến âm vô cùng (căn bậc hai (x^2+1)+x-1) bằng? A.-1 B.0 C.1/2 D.1

* Hướng dẫn giải

$\underset{x\to -\infty }{\lim }\left(\sqrt{x^2+1}+x-1\right)$

$=\underset{x\to -\infty }{\lim }\frac{(\sqrt{x^2+1}+x-1)(\sqrt{x^2+1}-x+1)}{\sqrt{x^2+1}-x+1}$

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