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.  
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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?

If the limit exists, then it must be equal to the limit along any line, for example, $y=x/4$.
In that case,
$$ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} =\lim_{x\downarrow 0} \frac{8}{15}\frac{x}{x^2}=\frac{8}{15}\lim_{x\downarrow 0}\frac{1}{x}=\infty.$$
