<s>[INST] Suggestions for $ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} $? 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.  
============
Is there anybody who knows why the last inequality isn't correct?
And also, has anybody some hints to evaluate the limit?
 [/INST] 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 $t \to 0^+$.
</s>