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I'm trying to evaluate $$ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} $$ over the domain $x>0$, $y>0$.

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My attempt:
$f(x,x^2)\to +\infty$; so if the limit exists it must be $+\infty$.

I tried to evaluate the limits "near" $(x,x)$ where, I thought, there's may be some problems:
$f(x, x-x^2)\to +\infty$.

Then I convinced myself the limit could be $+\infty$:
since $f(x,y)>0$ over the domain, I had to find such $g(x,y)$ that:
1. $f(x,y) \ge g(x,y)$
2. $ \lim_{(x,y)\to (0,0)} g(x,y)\to +\infty $

$$ f(x,y)=\frac{x-\sqrt{xy}}{x^2-y^2}=\frac{x-\sqrt{xy}+y-y}{x^2-y^2}=\frac{x-\sqrt{xy}+y}{x^2-y^2}-\frac{y}{x^2-y^2}=\frac{\sqrt{\left(x-\sqrt{xy}+y\right)^2}}{x^2-y^2}-\frac{y}{x^2-y^2} $$ Where the last step follows by $(x-\sqrt{xy}+y) \ge 0$ with $x>0$, $y>0$. $$ \frac{\sqrt{\left(x-\sqrt{xy}+y\right)^2}}{x^2-y^2}-\frac{y}{x^2-y^2} = \frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2}. $$ From $\left[3\left(x-\sqrt{xy}+y\right)^2\right] \ge \left[x+xy+y^2\right]$, for every $(x,y)$ with $(x>y)$: $$ \frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2} \ge \frac{\sqrt{x^2+xy+y^2}}{\sqrt{3}(x^2-y^2)}+\frac{y}{y^2-x^2}. $$ From here I observated that $ \left[\lim_{(x,y)\to (0,0)} g(x,y)\to +\infty \right]$ and eventually $ \left[\lim_{(x,y)\to (0,0)} f(x,y)\to +\infty \right]$ for $(x>y)$.

I thought that for $(y>x)$, the inequality was formally equivalent when I replace $(x)$ with $(y)$ and viceversa: $$ \frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2} \ge \frac{\sqrt{x^2+xy+y^2}}{\sqrt{3}(y^2-x^2)}+\frac{x}{x^2-y^2}. $$ However I could see, through an online grapher, that it is false!!
So I remained without any chance to conclude the limit.

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Is there anybody who knows why the last inequality isn't correct?
And also, has anybody some hints to evaluate the limit?

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  • $\begingroup$ Hint : change to polar coordinates. $\endgroup$
    – Mick
    Apr 3, 2020 at 17:49
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    $\begingroup$ Try $u=\sqrt{x}$ and $v=\sqrt{y}$, expression is then $\frac{u}{u^3+u^2v+uv^2+v^3}$ It looks easier to work with.. $\endgroup$ Apr 3, 2020 at 17:59

2 Answers 2

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In order to disprove your conjecture that $\lim_{(x, y) \to (0, 0)} f(x, y) = +\infty$, take the limit along the curve $(x, y) = (t^5, t)$ as $t \to 0^+$. Then we have: $$f(t^5, t) = \frac{t^5 - t^3}{t^{10} - t^2} = \frac{t(1-t^2)}{1-t^8}$$ and from the last expression, we see that $f(t^5, t) \to 0$ as $t \to 0^+$.

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If the limit exists, then it must be equal to the limit along any line, for example, $y=x/4$.

In that case,

$$ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} =\lim_{x\downarrow 0} \frac{8}{15}\frac{x}{x^2}=\frac{8}{15}\lim_{x\downarrow 0}\frac{1}{x}=\infty.$$

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    $\begingroup$ This does nothing to disprove the OP's conjecture that $\lim_{(x, y) \to (0, 0), x > 0, y > 0} f(x, y) = +\infty$. $\endgroup$ Apr 3, 2020 at 18:05
  • $\begingroup$ 1. You are right. I didn't look at the whole thing. 2. I see your example listed below the other answer, which shows that along the curve $(t^5,t)$, the limit is zero, hence, with the other examples shows the limit does not exist. Thanks! $\endgroup$
    – mjw
    Apr 3, 2020 at 18:07
  • $\begingroup$ @DanielSchepler, why not post your answer? $\endgroup$
    – mjw
    Apr 3, 2020 at 18:08

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