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) $|f(x,y)-0|=\left|\frac{xy^2}{x^2+y^4}\right|\geq{}1$.

How should I go about proving this? Thank you in advance,

Edit: Maybe my choice of epsilon is awful..

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

2 Answers 2


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .