Proving that $\lim_{(x,y)\rightarrow{}(0,0)}f(x,y)=0$ doesn't hold using the definition.

I realise that a comparable question has been asked in a different thread before but no definitions were used to prove the claim, thus I'd appreciate it if this one stays open.

I've just started a multi-variable calculus course and in one of the exercises we are asked to prove that $$\lim_{(x,y)\rightarrow{}(0,0)}f(x,y)=0$$ doesn't hold for $$f(x,y)=\frac{xy^2}{x^2+y^4}$$. Now I understand that if we were to observe the lines leading to $$(0,0)$$ that the limit is not always equal (for example $$x=0$$ and $$x=y^2$$) so that immediately tells us that the limit doesn't exist. But the question asks for a rigorous proof using the $$\epsilon-\delta$$ definition.

I tried following the negation of the definition to come up with a contradiction and I chose $$\epsilon=1>0$$ such that $$\forall{}\delta>0$$ and $$(x,y)\in\mathbb{R}-{}\{(0,0)\}$$ with $$||(x,y)||<\delta$$ that (and here I got stuck, not knowing how to show that the following inequality holds) $$|f(x,y)-0|=\left|\frac{xy^2}{x^2+y^4}\right|\geq{}1$$.

Edit: Maybe my choice of epsilon is awful..

• @infinity To demonstrate there exists such an $\epsilon$, OP has chosen some $\epsilon$ and is attempting to demonstrate the "for all $\delta$" part. Feb 29, 2020 at 21:31
• @angryavian Yeah I got confused there for a second.. Feb 29, 2020 at 21:32

We need to prove: There exists $$\varepsilon_{0}>0$$ such that for each $$\delta>0$$, there exists $$(x,y)\neq(0,0)$$ satisfying that $$||(x,y)-(0,0)||<\delta$$ and $$|f(x,y)-0|\geq\varepsilon_{0}$$.
Take $$\varepsilon_{0}=\frac{1}{10}$$. Let $$\delta>0$$ be arbitrary. Let $$t=\min(\frac{1}{2},\frac{\delta}{2})>0$$. Let $$(x,y)=(t^{2},t)$$. Clearly, $$(x,y)\neq(0,0)$$. Moreover, $$||(x,y)-(0,0)||=\sqrt{t^{4}+t^{2}}\leq\sqrt{t^{2}+t^{2}}=\sqrt{2}t<\delta$$. Now $$\begin{eqnarray*} f(x,y) & = & \frac{xy^{2}}{x^{2}+y^{4}}\\ & = & \frac{1}{2}. \end{eqnarray*}$$ This shows that $$|f(x,y)-0|\geq\varepsilon_{0}$$.
Fix $$\epsilon = 1/8$$. For any $$\delta > 0$$ you have $$f(0,\delta/2) = 0$$ and $$f(\delta_0^2, \delta_0) = \frac{1}{2}$$, where $$\delta_0$$ is small enough such that $$\|(\delta_0^2, \delta_0)\| < \delta$$ (e.g., choose $$\delta_0 = \min\{1, \delta/2\}$$).
Thus it is impossible for the statement "there exists $$L$$ such that $$|f(x,y) - L| < \epsilon$$ whenever $$\|(x,y)\| < \delta$$" to be true. This is because both $$(0, \delta/2)$$ and $$(\delta_0^2, \delta_0)$$ are within distance $$\delta$$ of $$(0,0)$$, and if the statement were true we would have $$\frac{1}{2}= |f(\delta_0^2, \delta_0) - f(0, \delta/2)| \le |f(\delta_0^2, \delta_0) - L| + |f(0, \delta/2) - L| < 2 \epsilon = \frac{1}{4}$$, a contradiction.