<s>[INST] Show limit of function $\frac{xy(x+y)}{x^2-xy+y^2}$ for $(x,y)\to(0,0)$ Show $$\lim_{(x,y)\to (0,0)} xy \frac{(x+y)}{x^2-xy+y^2}=0$$
If I approach from $y=\pm x$, I get $0$. Is that sufficient?
 [/INST] From $\bigl(|x|-|y|\bigr)^2\geq0$ we get $2|xy|\leq x^2+y^2$ and then
$$|x|+|y|\leq\sqrt{2(x^2+y^2)},\qquad x^2-xy+y^2\geq{1\over2}(x^2+y^2)\ .$$
This implies
$$\bigl|f(x,y)\bigr|=\left|{xy(x+y)\over x^2-xy+y^2}\right|\leq{{1\over2}(x^2+y^2)\sqrt{2(x^2+y^2)}\over{1\over2}(x^2+y^2)}=\sqrt{2(x^2+y^2)}\ .$$
Therefore, given an $\epsilon>0$, we have $\bigl|f(x,y)\bigr|<\epsilon$ as soon as $\sqrt{x^2+y^2}<\delta:={\displaystyle{\epsilon\over\sqrt{2}}}$.
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