# Solving an irrational limit without L'Hopital

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.

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

• The limit doesn't exist. Did you mean $x \to \infty$? Oct 27, 2016 at 12:23
• Factor the argument of the radical and simplify.
– user65203
Oct 27, 2016 at 12:24
• @copper.hat no I do mean the limit to 0. My textbook says the solution in -inf Oct 27, 2016 at 12:40
• @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. Oct 27, 2016 at 13:17

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