Show that $\lim_{(x,y)\to (0,0)}\frac{x^4y^3}{x^2+y^4} = 0$.

Question

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.

Answer 1

By AM-GM we get $$x^2+y^4\geq 2|x|y^2$$ so we get $$\frac{x^4y^3}{x^2+y^4}\le \frac{x^4|y|^3}{2|x|y^2}=\frac{|x|^3|y|}{2}$$ this tends to zero if $x,y$ tends to zero.