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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.


ADDENDUM:

$|\frac{x^4y-xy^4}{x^4+y^4}|\overset{x=\rho\cos\theta\\ y=\rho\sin\theta}{=}\frac{\rho^5 |\cos^4(x)\sin(x)-\cos(x)\sin^4(x)|}{\rho^4 (\cos^4(x)+\sin^4(x))}\leq \rho\ \frac{2}{\min(\cos^4(x)+\sin^4(x))}\overset{\rho\to 0}{\to} 0$ 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.

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  • $\begingroup$ 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
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    $\begingroup$ Yes, it is correct. $\endgroup$
    – Paul Frost
    May 27 at 23:27
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Hint: What is $\lim_{x\to 0} f(x,-x)$?

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  • $\begingroup$ 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
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    $\begingroup$ @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

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