I want evaluate this $\lim_{(x,y)\rightarrow (0,0)}\dfrac{xy^3}{x^2+y^4}$. I tried polar coordinates and I got this $$\lim_{r\rightarrow 0^+}\dfrac{r^2\cos\theta \sin^3\theta}{\cos^2\theta+r^2\sin^4\theta}$$
I would like to show that $\dfrac{\cos\theta \sin^3\theta}{\cos^2\theta+r^2\sin^4\theta}$ is bounded, but I was not able to do that. Any hint?
We have $(|x|-y^2)^2\geq 0$, hence $2|x|y^2\leq (x^2+y^4)$. So if $f(x,y)$ is your function we have for $(x,y)\not =(0,0)$ the inequality: $$2|f(x,y)|=|y|\frac{2|x|y^2}{x^2+y^4}\leq |y|\leq \sqrt{x^2+y^2}$$ and it is easy to finish.
Let's fix $\theta$ and take $r\rightarrow 0$.
\begin{eqnarray} lim_{r\rightarrow 0}\frac{c_\theta s^3_\theta}{c_\theta^2 +r^2s^4_\theta}&=&\frac{c_\theta s^3_\theta}{c_\theta^2} \\ &=& \frac{s^3_\theta}{c_\theta} \end{eqnarray}
We see that this depends on $\theta$ and is not bounded, for example when $\theta=\frac{\pi}{2}$. However, this does not fully answer the original question since for any nonzero $r$ the quantity considered 0.
Let us consider $\theta=\frac{\pi}{2}+\epsilon$, and we will take $r\rightarrow 0$.
\begin{eqnarray} \frac{s^3_\theta}{c_\theta} &=& \frac{1-\epsilon^2}{-\epsilon} \end{eqnarray} This is clearly unbounded as $\epsilon\rightarrow 0$.
