<s>[INST] How to show this limit exists I know the limit is 0, but how would I prove the limit? $$\lim_{(x,y)\to(0,0)} {\frac{x^3y^2}{x^4+y^6}}$$
 [/INST] As 
\begin{align*}
x^4+y^6 = \frac{1}{2} x^4 + \frac{1}{2} x^4 + y^6 \geq \frac{3}{\sqrt[3]{4}}x^{8/3}y^2,
\end{align*}
then
\begin{align*}
\Big|\frac{x^3y^2}{x^4+y^6}\Big| \leq \frac{\sqrt[3]{4}}{3}|x|^{1/3} \rightarrow 0.
\end{align*}
</s>