# Finding a limit with negative infinity (Square root)

I'm given this question $$\lim_{x\rightarrow -\infty }\left(\sqrt{x}-\frac{2+x}{\sqrt{x}}\right)$$

My attempt,

$\lim_{x\rightarrow -\infty }(\sqrt{x}-\frac{2+x}{\sqrt{x}})=\lim_{x\rightarrow -\infty }(-\frac{2}{\sqrt{x}})$

How to I substitute negative infinity to square root of $x$? Wouldn't it be an imaginary number ?

• In Real calculus, $\sqrt x$ is not defined for $x<0$ – lab bhattacharjee Jan 6 '16 at 13:58
• Where is this question from? $\sqrt{x}$ for $x\to-\infty$ seems at best... peculiar, in a question. – Clement C. Jan 6 '16 at 13:59
• Rationalize both terms – Archis Welankar Jan 6 '16 at 14:00
• @ArchisWelankar This is not the issue... the question is about taking the square root of a negative number. – Clement C. Jan 6 '16 at 14:07
• It will be $2/{\sqrt\infty}{i}=0$ whats the issue Clement C – Archis Welankar Jan 6 '16 at 14:17