I am studying the following function: $$f(x,y)=\begin{cases} \frac{x^4y-xy^4}{x^4+y^4}, & (x,y)\neq (0,0)\\ 0, & (x,y)=(0,0) \end{cases}$$ to see if it is continuous, if the partial derivatives exist, compute them, see if they are continuous and if $f$ is differentiable.

What I have done:

The domain of $f$ is $\mathbb{R^2}$.

$f$ is clearly continuous in $\mathbb{R^2}\setminus\{(0,0)\}$.

The partial $x$ derivative $\frac{\partial f}{\partial x}(x,y)=\frac{3x^4y^4+4x^3y^5-y^8}{(x^4+y^4)^2}$ is continuous in $\mathbb{R^2}\setminus \{(0,0)\}, \frac{\partial f}{\partial x} (0,0)=\lim_{t\to 0}\frac{f(0+t,0)-f(0,0)}{t}=0$ but $\frac{\partial f}{\partial x}(x,0)=0$ and $\frac{\partial f}{\partial x}(x,x)=\frac{3}{2}$ so it is not continuous in $(0,0)$.

The partial $y$ derivative $\frac{\partial f}{\partial y}(x,y)=\frac{x^8-3x^4y^4-4x^5y^3}{(x^4+y^4)^2}$ is continuous in $\mathbb{R^2}\setminus \{(0,0)\}, \frac{\partial f}{\partial y} (0,0)=\lim_{t\to 0}\frac{f(0+t,0)-f(0,0)}{t}=0$ but $\frac{\partial f}{\partial y}(x,0)=1$ and $\frac{\partial f}{\partial x}(x,x)=-\frac{3}{2}$ so it is not continuous in $(0,0)$.

Now, to see if $f$ is continuous in $(0,0)$ I have tried to compute the $\lim_{(x,y)\to (0,0)} f(x,y)$ in various ways (like using polar coordinates) but to no avail, so I would like an hint about how to do this, thanks.


$|\frac{x^4y-xy^4}{x^4+y^4}|\overset{x=\rho\cos\theta\\ y=\rho\sin\theta}{=}\frac{\rho^5 |\cos^4(x)\sin(x)-\cos(x)\sin^4(x)|}{\rho^4 (\cos^4(x)+\sin^4(x))}\leq \rho\ \frac{2}{\min(\cos^4(x)+\sin^4(x))}\overset{\rho\to 0}{\to} 0$ so, since $\lim_{(x,y)\to (0,0)} |f(x,y)|=0$ we have that $\lim_{(x,y)\to (0,0)}f(x,y)=0$ and $f$ is thus continuous at the origin too.

$\lim_{(h,k)\to (0,0)}\frac{f(0+h,0+k)-f(0,0)-\vec{\nabla}f(0,0)\cdot (h,k)}{\sqrt{h^2+k^2}}=\lim_{(h,k)\to (0,0)}\frac{h^4k-hk^4}{h^4+k^4}\cdot\sqrt{h^2+k^2}=\lim_{\rho\to 0}\rho^2 \frac{\cos^4(\theta)\sin(\theta )-\cos(\theta)\sin^4(\theta)}{\cos^4(\theta)+\sin^4(\theta)}=0$ so $f$ is differentiable at the origin.

  • $\begingroup$ Ah, you changed the definition of $f$ ;-) $\endgroup$
    – Paul Frost
    May 27 at 23:14
  • $\begingroup$ @Paul Frost Yes, the denominator shouldn't have been squared so I corrected $f$ and solved the problem (correctly, I hope). $\endgroup$
    – lorenzo
    May 27 at 23:27
  • 1
    $\begingroup$ Yes, it is correct. $\endgroup$
    – Paul Frost
    May 27 at 23:27

Hint: What is $\lim_{x\to 0} f(x,-x)$?

  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$
    – amWhy
    May 20 at 17:36
  • 3
    $\begingroup$ @amWhy The OP wanted a hint and this is what I did. I think he will be able to compute the limit. $\endgroup$
    – Paul Frost
    May 20 at 17:42
  • $\begingroup$ @PaulFrost thank you very much for your hint: I have amended my question accordingly. $\endgroup$
    – lorenzo
    May 20 at 18:34
  • $\begingroup$ @amWhy thank you for your interest in my question. The answer by Paul Frost was just what I was looking for. $\endgroup$
    – lorenzo
    May 20 at 18:35
  • $\begingroup$ Dear Paul Frost, and lorenzo: the comment was auto-generated From Review. Please read comments fully. $\endgroup$
    – amWhy
    May 20 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.