Is $\lim\limits_{x,y\to-\infty}\frac{\sqrt x\sqrt y}{\sqrt{xy}}=1$?

Question

WolframAlpha is suggesting (judging by the plot given) that the limit is actually $-1$.

I would think the following manipulations would be okay to conclude that is the opposite. $$\lim_{x,y\to-\infty}\frac{\sqrt x\sqrt y}{\sqrt{xy}}=\lim_{x,y\to-\infty}\frac{\sqrt x\sqrt y}{\sqrt x\sqrt y}=\lim_{x,y\to-\infty}1=1$$ I usually trust WA's conclusions about things like this, so I'm fairly certain I'm missing something, obvious or otherwise.

Answer 1

If you insist that $\sqrt 1=1$ and $\sqrt {-1}=i$ then $\sqrt{xy}=\sqrt x\sqrt y$ does not always hold, for example when $x=y=-1$.

Answer 2

It is common for a CAS like WA to definitively take the square root of a negative real to be the imaginary square root with positive imaginary component. So WA is probably seeing $$\frac{\sqrt{x}\sqrt{y}}{\sqrt{xy}}=\frac{(i\sqrt{|x|})(i\sqrt{|y|})}{\sqrt{xy}}=-\frac{\sqrt{|x||y|}}{\sqrt{xy}}=-1$$ where throughout, keep reminding yourself both $x$ and $y$ are negative.

Answer 3

To add on to Hagen's answer. We can use the principle branch of the square root and say that $u=-x$, giving us $\lim_{x\to-\infty} \sqrt x = \lim_{u\to\infty} \sqrt{-1}\sqrt u = \lim_{u\to\infty} i\sqrt u$

and $v = -y$ giving us $\lim_{v\to\infty} i\sqrt v$

But we also have $lim_{x,y\to-\infty}\sqrt{xy} = lim_{u,v\to\infty}\sqrt{uv} $

This gives us $$\lim_{x,y\to-\infty}\frac{\sqrt x\sqrt y}{\sqrt{xy}} = i^2 \lim_{u,v\to\infty}\frac{\sqrt u\sqrt v}{\sqrt{uv}}=-\lim_{u,v\to\infty}\frac{\sqrt u\sqrt v}{\sqrt{uv}}= -1$$