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

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}$.

Answer 2

Yes your way is correct, more simply as $|x|<1$

$$\left|\frac{x^4y^3}{x^2+y^4}\right|\le\left|\frac{x^4y^3}{x^4+y^4}\right|=r^3\left|f(\theta)\right|\to 0$$

since $f(\theta)$ is bounded.

Answer 3

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.