<s>[INST] Proving that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$ I need to show that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$. My attempted method was showing that as the distance between $(x,y)$ and $(0,0)$ approaches $0$, so does the function in the limit. I tried to show $\lvert{\frac{x^2y^3}{x^2+y^2}}\rvert \leq \sqrt{x^2+y^2}$ for $(x,y) \neq 0$. I expanded the terms to get $\frac{x^2\lvert y^3\rvert}{x^2+y^2} \leq \frac{(x^2+y^2)\sqrt{x^2+y^2}}{x^2+y^2}$, or $(x^2+y^2)^\frac{3}{2}\geq x^2\lvert y^3\rvert$. I am unsure of how to prove this and whether this method is valid or would work in showing the existence of the above limit at all.
 [/INST] Note :  $x^2+y^2 \ge |2xy|$.
Let $ |x^2+y^2|^{1/2} \lt \delta, $
Choose $\epsilon = \delta^3 /4$.
$(x,y) \ne (0,0)$.
$|\dfrac{x^2y^3}{x^2+y^2}| \le$
$ (1/4)|\dfrac{y(x^2+y^2)^2}{x^2+y^2}| =$
$ (1/4)|y||(x^2+y^2)| \lt $
$ (1/4) (\delta)(\delta)^2 = \epsilon$.
Used: $ |y| \le (x^2+y^2)^{1/2} \lt  \delta $.
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