# 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.

• Your first method seems wrong as you are considering a case where the expression is already larger than the given expression. Jan 5 at 8:55

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 $$m\geq 0$$.
• Thank you, so if I use the triangle inequality to get $|x|^{\frac{m-3}{3}}|x-y| \leq |x|^{\frac{m-3}{3}}(|x|+|y|)=|x|^{\frac{m}{3}}+|x|^{\frac{m-3}{3}}|y|$ I actually find that the first term is finite when $m \geq 0$ and the second for $x \geq 3$, so both are finite for $m \geq 0$ according to polar coordinates. What do you think? Jan 6 at 1:49