Limit at (0,0) of $\frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$ $$\text{Let}\ f(x,y) = \frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$$
WolframAlpha tells me that $\lim_{(x,y) \to (0,0)} f(x,y)$ does not exist. To prove the non-existence of the limit, I tried three different paths ($y=kx, y=kx^2, y=kx^3)$ and they all equal zero. From the graph of the function it looks like the limit is indeed zero. Also I thought I was able to prove the limit does indeed exist at $(0,0)$ with the epsilon-delta definition:
$$y^2 \leq x^2+y^2 \leq x^2 + y^4 \iff \frac{y^2}{x^2+y^4} \leq 1 \overset{|x|<1}{\iff} \frac{|x|y^2}{x^2+y^4} \leq 1 \iff \frac{|x|y^2\sqrt{x^2+y^2}}{x^2+y^4} \leq \sqrt{x^2+y^2} \iff \left|\frac{xy^2\sqrt{x^2+y^2}}{x^2+y^4} - 0\right|\leq \left|\sqrt{(x-0)^2+(y-0)^2}\right| \lt δ := ε $$
So for $δ = ε$ we have that $\|x -(0,0)\| < δ \implies |f(x,y) - 0| < ε$
I'm confused

Hint: $\dfrac{xy^2}{x^2+y^4} \le \dfrac{1}{2}$. This is true by the AM-GM inequality. Note that the inequality you claimed is wrong because the line $y^2 \le y^4$ is not true !
