Determine for which $a$ it holds that $\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=0$ Let $a>0$.
Determine for which values of $a$ it holds that:
$$\lim_{(x,y)\to(0,0)}\frac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=0$$
The solution indicated is $a>3$.
Passing in polar coordinates, I obtain:
$$\frac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=\frac{\rho^{a-3}|\cos\theta-\sin\theta|^{a-1}}{(\cos\theta+\sin\theta)}$$
If $a > 3$, when $\rho \to 0$, the fraction goes to $0$, and this does not depend on $\theta$.
However, this is not enough for the limit to exist and it must exist a $g(\rho)$ such that:
$$\frac{\rho^{a-3}|\cos\theta-\sin\theta|^{a-1}}{\cos\theta+\sin\theta}\leqslant g(\rho)\to 0 $$
But
$$\frac{|\cos\theta-\sin\theta|^{a-1}}{\cos\theta+\sin\theta}$$
is not limited, because, for example, when $\theta \to \pm \frac{3}{4}\pi$, it goes to $\mp \infty$.
Where is the mistake?
Thanks in advance.
 A: 
Determine for which values of $\;a\;$ it holds that:
$$\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=0\;.$$

By letting $\;X=x-y\;$ and $\;Y=x+y\;,\;$ it follows that
$\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}\;.$
First case : $\;a\le1\;.$
$\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2}{Y|X|^{1-a}\sqrt{X^2+Y^2}}\;,$
but the last limit does not exist because we get different results when we calculate the limits of the restrictions of the function to two different subsets of its domain, indeed
$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y>0\end{align}}\dfrac{\sqrt2}{Y|X|^{1-a}\sqrt{X^2+Y^2}}=$
$=\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y>0\end{align}}\dfrac1{Y|X|^{1-a}}$$\cdot\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y>0\end{align}}\dfrac{\sqrt2}{\sqrt{X^2+Y^2}}=$
$=+\infty\cdot(+\infty)=+\infty\;,$
$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y<0\end{align}}\dfrac{\sqrt2}{Y|X|^{1-a}\sqrt{X^2+Y^2}}=$
$=\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y<0\end{align}}\dfrac1{Y|X|^{1-a}}$$\cdot\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y<0\end{align}}\dfrac{\sqrt2}{\sqrt{X^2+Y^2}}=$
$=-\infty\cdot(+\infty)=-\infty\;.$
Second case : $\;a>1\;.$
The limit $\;\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}\;$ does not exist because we get different results when we calculate the limits of the restrictions of the function to two different subsets of its domain, indeed
$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&X=0\end{align}}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}=$$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&X=0\end{align}}\dfrac0{Y\big|Y\big|}=0\;,$
$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y=|X|^{a-1}\end{align}}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}=$
$=\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y=|X|^{a-1}\end{align}}\dfrac{\sqrt2}{\sqrt{X^2+X^{2(a-1)}}}=+\infty\;.$
Conclusion :
In any case, for any $\;a\in\mathbb{R}\;,$ the limit
$\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}$
does not exist.
