# 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?

• Hint : change to polar coordinates.
– Mick
Apr 3, 2020 at 17:49
• 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.. Apr 3, 2020 at 17:59

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 $$t \to 0^+$$.

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

In that case,

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

• This does nothing to disprove the OP's conjecture that $\lim_{(x, y) \to (0, 0), x > 0, y > 0} f(x, y) = +\infty$. Apr 3, 2020 at 18:05
• 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!
– mjw
Apr 3, 2020 at 18:07