I have the limit $$\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4},$$ and would like to show with an $\epsilon-\delta$ proof that it is zero. I know with a situation like $$\left|\frac{x^4y}{x^4+y^4}\right|\leq y$$ or something similar, but I can't find a way to do the same thing here, as no single term in the numerator is of sufficient degree, although I think I could get this with a small hint.

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    $\begingroup$ You can use $2ab \leqslant a^2 + b^2$. $\endgroup$ – Daniel Fischer Oct 1 '13 at 21:48
  • $\begingroup$ Or this is one place a polar coordinates substitution can work really nicely, as $|\cos^3\theta\sin^2\theta|\le 1$ (we could get a better upper bound, but why bother?). $\endgroup$ – Ted Shifrin Oct 1 '13 at 21:51
  • $\begingroup$ Thanks guys, the first hint especially was quite helpful. Feeling rather silly now... $\endgroup$ – Stefan Dawydiak Oct 1 '13 at 22:01
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    $\begingroup$ @Imbilio I suggest you answer your own question so this doesn't come up as unanswered. $\endgroup$ – Git Gud Oct 1 '13 at 22:30
  • $\begingroup$ To see a proof based on Daniel Fischer's hint, see this answer to a duplicate question. $\endgroup$ – Martin Sleziak Jun 21 '16 at 20:57

My hint:

$$L=\lim_{(x,\: y)\to (0,\: 0)}\frac{x^3y^2}{x^4+y^4}=\lim_{(x,\: y)\to (0,\: 0)}\frac{x^2y^2}{x^4+y^4}\: .x$$

Because $$\left|\frac{x^2y^2}{x^4+y^4}\right|\leq\frac{1}{2}\to L=0$$


Let $z=\max (|x|,|y|).$ Then $$|x^3 y^2|\leq z^5.$$ And for $(x,y)\ne (0,0)$ we have $x^4+y^4\geq z^4>0$, implying $$0<1/(x^4+y^4)\leq 1/z^4 .$$ So for $(x,y)\ne (0,0)$ we have $$|(x^3 y^2)/(x^4+y^4|\leq z^5(1/z^4)=z.$$


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