2
$\begingroup$

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 ?

$\endgroup$
  • 5
    $\begingroup$ In Real calculus, $\sqrt x$ is not defined for $x<0$ $\endgroup$ – lab bhattacharjee Jan 6 '16 at 13:58
  • 1
    $\begingroup$ Where is this question from? $\sqrt{x}$ for $x\to-\infty$ seems at best... peculiar, in a question. $\endgroup$ – Clement C. Jan 6 '16 at 13:59
  • $\begingroup$ Rationalize both terms $\endgroup$ – Archis Welankar Jan 6 '16 at 14:00
  • 1
    $\begingroup$ @ArchisWelankar This is not the issue... the question is about taking the square root of a negative number. $\endgroup$ – Clement C. Jan 6 '16 at 14:07
  • $\begingroup$ It will be $2/{\sqrt\infty}{i}=0$ whats the issue Clement C $\endgroup$ – Archis Welankar Jan 6 '16 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.