<s>[INST] Show that $\lim_{(x,y)\to (0,0)}\frac{x^4y^3}{x^2+y^4} = 0$. Show that
\begin{equation*}
\lim_{(x,y)\to (0,0)}\frac{x^4y^3}{x^2+y^4} = 0.
\end{equation*}
Let's show this formally using an $\epsilon-\delta$ proof. For $(x,y)\neq (0,0)$, let $\epsilon > 0$. Then if $(x,y)\in \mathbb{R}^2$ and $|(x,y)| < \frac{\epsilon}{3}$, then
\begin{equation*}
|y|^3\leq x^2+y^4 < \epsilon^3,
\end{equation*}
so $|y| < \sqrt[3]{\epsilon}$. Thus,
\begin{equation*}
\begin{split}
\left|\frac{x^4y^3}{x^2+y^4}-0\right| &= \left|\frac{x^4y^3}{x^2+y^4}\right| \\
&= \frac{x^4|y|^3}{x^2+y^4} \\
&\leq \frac{x^4y^3}{x^4} \\
&= |y|^3 \\
&< \epsilon
\end{split}
\end{equation*}
and we are done.
Where have I gone wrong? Thanks.
 [/INST] Let $|x|<1$, $|y| <1$, $(x,y) \not = (0,0)$;
$x^2+y^4 \ge x^4+y^4\ge 2x^2y^2$ (AM-GM);
$|\dfrac {x^4y^3}{x^2+y^4}|\le |\dfrac {x^4y^3}{2x^2y^2}|=$
$(1/2)x^2|y| \le (x^2+y^2)\cdot 1$;
Choose $ \delta= \epsilon^{1/2}$.
</s>