Minimum integer $m$ such that $\lim_{(x,y)\to(0,0)} \frac{x^{\frac{m}{3}}|x-y|}{\sqrt{x^2+y^2}}\in \mathbb{R}$ Find the minimum $m\in\mathbb{N}$ such that
$$\lim_{(x,y)\to(0,0)} \frac{x^{\frac{m}{3}}|x-y|}{\sqrt{x^2+y^2}}\in \mathbb{R}$$
My attempt: lets consider the modulus of the limit, so it is
$$0 \leq \lim_{(x,y)\to(0,0)} \frac{|x|^{\frac{m}{3}}|x-y|}{\sqrt{x^2+y^2}}\leq\lim_{(x,y)\to(0,0)} \frac{|x|^{\frac{m}{3}}|x-y|}{\sqrt{x^2}}=\lim_{(x,y)\to(0,0)} \frac{|x|^{\frac{m}{3}}|x-y|}{|x|}$$
$$=\lim_{(x,y)\to(0,0)} |x|^{\frac{m}{3}-1}|x-y|=\lim_{(x,y)\to(0,0)} |x|^{\frac{m-3}{3}}|x-y|$$
Since $|x-y| \to 0$ as $(x,y) \to (0,0)$, the only problem can occur when $|x|^{\frac{m-3}{3}}$ goes to the denominator as $m$ varies; so it must be $m-3 \geq 0 \iff m \geq 3$; so if $m \geq 3$ the limit is finite and it is $0$. Is this correct?
Another question: if I try to solve this with polar coordinates I find another value, so one way must be wrong. Let $x=r \cos t$ and $y=r \sin t$, with $r \geq 0$ and $0 \leq t <2\pi$, considering the modulus of the limit it is
$$0 \leq \lim_{r \to 0^+} \frac{|r \cos t|^{\frac{m}{3}}|r\cos t-r \sin t|}{r}0 = \lim_{r \to 0^+} \frac{r^{\frac{m}{3}}|\cos^{\frac{m}{3}} (t)|r |\cos t- \sin t|}{r}=\lim_{r \to 0^+} r^{\frac{m}{3}}|\cos^{\frac{m}{3}}t||\cos t- \sin t|$$
$$\leq \lim_{r \to 0^+} r^{\frac{m}{3}}|\cos t|(|\cos t|+|\sin t|)\leq \lim_{r \to 0^+} r^{\frac{m}{3}}\cdot1\cdot(1+1)=2\lim_{r \to 0^+} r^{\frac{m}{3}}$$
And this limit is finite if $\frac{m}{3} \geq 0 \iff m \geq 0$; where is my mistake? I suspect that it is when I suppose that $|x|^{\frac{m-3}{3}}$ can go to the denominator, because the estimation doesn't give informations for $\frac{m-3}{3} \leq 0$ so I must study what happens for $\frac{x^{\frac{m}{3}}|x-y|}{\sqrt{x^2+y^2}}$ for $m\in\{0,1,2\}$. Thanks.

The problem occurs as even the term $|x-y|$ contains a term of $x$, which will offset the exponent of $|x|^{\frac{m-3}{3}}$ by $1$. If we take this into account, we would get $|x|^{\frac{m}{3}}$, and so $m\geq 0$.
