I'm supposed to solve this limit without using L'Hopitals rule.

I always find the indeterminate form of $\frac{0}{0}$ but since multiplying by the conjugate is not an option here (atleast I think so) I don't know how to solve this limit.

$\lim \limits_{x \to 0} \frac{\sqrt{-2x-x^2}}{x}$

  • $\begingroup$ The limit doesn't exist. Did you mean $x \to \infty$? $\endgroup$
    – copper.hat
    Commented Oct 27, 2016 at 12:23
  • $\begingroup$ Factor the argument of the radical and simplify. $\endgroup$
    – user65203
    Commented Oct 27, 2016 at 12:24
  • $\begingroup$ @copper.hat no I do mean the limit to 0. My textbook says the solution in -inf $\endgroup$ Commented Oct 27, 2016 at 12:40
  • 1
    $\begingroup$ @TheAlPaca02: The limit only exists if taken from the left (that is, $x \le 0$ in addition to $x \to 0$). If $x>0$ (or $x$ more negative) the expression is complex. Also, it is often (but far from always) the case that having a limit implies that the limit is finite. $\endgroup$
    – copper.hat
    Commented Oct 27, 2016 at 13:17

1 Answer 1


The expression is only definded for $-2 \le x<0\;$ and therefore it is negative. You can write $$\frac{\sqrt{-2x-x^2}}{x}$$ $$=-\frac{\sqrt{-2x-x^2}}{\sqrt{x^2}}$$ $$=-\sqrt{\frac{-2x-x^2}{x^2}}$$ $$=-\sqrt{-\frac{2}{x}-1}$$ From this you can see that the limit $x\rightarrow 0^{-}$ does not exist (or is $-\infty,\;$ if you allow infinite limits).

  • $\begingroup$ How are you allowed to replace $x$ in the denominator by $\sqrt{x^2}$? $\endgroup$ Commented Oct 27, 2016 at 13:33
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    $\begingroup$ @TheAlPaca02: It is $-\sqrt{x^2}$ since the square root is assumend positive and $x$ is negatve, i.e. for $x<0$ you have $x=-\sqrt{x^2}$ $\endgroup$ Commented Oct 27, 2016 at 13:36

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