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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2You can use $2ab \leqslant a^2 + b^2$. – Daniel Fischer Oct 01 '13 at 21:48
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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?). – Ted Shifrin Oct 01 '13 at 21:51
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Thanks guys, the first hint especially was quite helpful. Feeling rather silly now... – Stefan Dawydiak Oct 01 '13 at 22:01
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1@Imbilio I suggest you answer your own question so this doesn't come up as unanswered. – Git Gud Oct 01 '13 at 22:30
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To see a proof based on Daniel Fischer's hint, see this answer to a duplicate question. – Martin Sleziak Jun 21 '16 at 20:57
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Two more closely related questions: http://math.stackexchange.com/questions/856098/proving-fracx3y2x4y4-is-continuous and http://math.stackexchange.com/questions/1822811/continuity-of-fracx3y2x4y4-at-0-0 – Martin Sleziak Jun 22 '16 at 12:10
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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 $$\left|\frac{x^2y^2}{x^4+y^4}\right|\leq\frac{1}{2}\to L=0$$
Iloveyou
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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 $$|(x^3 y^2)/(x^4+y^4|\leq z^5(1/z^4)=z.$$
DanielWainfleet
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