Consider the function:

$$f:\mathbb{R^2}\rightarrow\mathbb{R}$$ $$f(x,y)=\frac{x^2y^2}{x^4+y^2}\forall (x,y)\neq(0,0)$$ $$f(0,0)=0$$

It's clearly differentiable for all $(x,y)\neq(0,0)$. I have shown that both partial derivatives at the origin are $0$, however, I'm having trouble showing whether is is differentiable or not. I haven't even been able to prove whether its continuous at the origin or not.

By guess on the continuity is that it is indeed continuous, based on the graph itself and on the fact that both iterated limits are $0$, and so are the limits when choosing $y=\pm x$ and $y=\pm x^2$. However, I can't bound the expression by any arbitrary $\epsilon>0$:


which would be a sufficiente condition for the continuity of $f$ in $(0,0)$

On the other hand, since all partial derivatives are $0$, if the function were differentiable, the differential $L$ would have to be $0$, and so, we would have that:

$$f((0,0)+(h,k))-f((0,0))=L(h,k) + ||(h,k)||\cdot\rho(h,k)$$

Where $L(h)$ is the linear differential $0$ and $\lim_{(h,k)\to(0,0)}\rho(h,k)=0$ That is:


So we have to prove that the function $\rho(h,k)$ defined by:

$$\rho(h,k)=\frac{h^2k^2}{||(h,k)||\cdot h^4+k^2}=\frac{h^2k^2}{\sqrt{h^2+k^2}\cdot (h^4+k^2)}$$

Does or does not converge to $(0,0)$ when $(h,k)$ tends towards $(0,0)$. How to show both of these concerns? (Continuity and differentiability).


1 Answer 1


Since the function is equal to $0$ on both axes, we need only consider values with $x,y\not=0$. In that case

$${x^2y^2\over x^4+y^2}={1\over{x^2\over y^2}+{1\over x^2}}$$

Note that both terms in the denominator on the right hand side are positive, and the $1/x^2$ term clearly goes to infinity as $x$ goes to $0$. So the denominator goes to infinity, hence the expression goes to $0$. So at the very least the function is continuous.

Added later: A simpler way to get the same thing, still relying on nonnegativity of all the terms, is to note

$$0\le {x^2y^2\over x^4+y^2}\le{x^2y^2\over y^2}=x^2\to0$$

This works for $\rho(h,k)$ as well:


so the function is differentiable as well.

  • $\begingroup$ Thanks Barry, that was very clear ;) I'll leave the question open to see if anyone has an idea regarding the differentiability. $\endgroup$
    – F.Webber
    Nov 30, 2014 at 22:10
  • $\begingroup$ @LMartin, actually can't you just make the exact same argument with your expression for $\rho(h,k)$, since $\sqrt{h^2+k^2}$ is also positive? $\endgroup$ Nov 30, 2014 at 22:17
  • $\begingroup$ If I do that, I end up with $$\frac{1}{\sqrt{h^2+k^2}(\frac{h^2}{k^2}+\frac{1}{h^2})}$$, in which, the second term in the denominator does indeed go to infinity as before, but it is multiplied by the square root which goes to $0$, so I'm not sure how to proceed here. $\endgroup$
    – F.Webber
    Nov 30, 2014 at 22:25
  • $\begingroup$ @LMartin, oh you're right. My bad. $\endgroup$ Nov 30, 2014 at 22:32
  • $\begingroup$ @LMartin, it works after all (but not quite the way I thought when I put up my first comment). I've edited the answer. $\endgroup$ Nov 30, 2014 at 22:45

You must log in to answer this question.

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