My book uses epsilon to solve this and the answer should be zero but I'd like another easier way
$$\lim_{(x,y) \to (0,0)} \frac{4xy^2}{x^2+y^2}$$
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1How about polar coordinates? – Oct 04 '15 at 01:15
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@Jack Exactly what I was thinking. – Aleksandar Oct 04 '15 at 01:18
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2See also this post and other questions linked there. – Martin Sleziak Dec 07 '15 at 06:23
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Substitute $x=r \sin \theta$ and $y=r \cos \theta$.
$$\lim_{r \to 0} 4 r^3 \cos \theta \sin^2 \theta=0 $$
Aleksandar
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Does this method work always whenever i come across similar examples or just under certain conditions? – Square-root Oct 04 '15 at 01:28
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1@Square-root It always works however sometimes it's more useful than others. – Aleksandar Oct 04 '15 at 01:32
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Recall that $x^2+y^2\geqslant 2xy$ for any choice of $x,y$, so that $$\left|\frac{4xy^2}{x^2+y^2}\right|\leqslant \left|\frac{4xy^2}{2xy}\right| = |2y|$$
Pedro
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