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.
2026-04-04 04:22:08.1775276528
Multivariable Delta Epsilon Proof $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$ --- looking for a hint
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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 $$\left|\frac{x^2y^2}{x^4+y^4}\right|\leq\frac{1}{2}\to L=0$$