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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.

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  • $\begingroup$ Your first method seems wrong as you are considering a case where the expression is already larger than the given expression. $\endgroup$
    – N.S.JOHN
    Jan 5 at 8:55
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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 $m\geq 0$.

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  • $\begingroup$ 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? $\endgroup$
    – ZaWarudo
    Jan 6 at 1:49

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