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Question

For what values of $s$ (where s is real and positive) is:
$f(x,y)=\dfrac{x^4+y^4-6x^2y^2}{(x^2+y^2)^s}$ when $(x,y) \neq (0,0)$
and
$f(x,y) = 0$ when $(x,y)=(0,0)$ ,

(a) Continuous at $(0,0)$
(b) Differentiable at $(0,0)$ .

Attempt at answer

Looking at the line $y=x$, I get:

$$f(t,t)=\frac{-4t^4}{(2t^4)^s} = \frac{-4}{(2)^s(t^{4(s-1)})}$$

However for any positive $s$, $f(t,t)$ does not tend to zero as $t$ tends to zero. Therefore $f(x,y)$ is not continuous or differentiable at $(0,0)$ for any positive real $s$.

My question is, does this prove the function is not differentiable at $(0,0)$ for any $s$?

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Replace $x=rcos(\theta)$ and $y=rsin(\theta)$ in polar coordinates. Then $f(x,y)={{\cos 2\theta}\over {r^{2s}}}$. It has a limit in 0 if an only if the expression is independent from $\theta$ and bounded when $r\to 0$ which is possible only if $s<0$. It would have 1st order derivations in origin if $s\le -1$.

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  • $\begingroup$ The question asked for positive s and I found there are none. We are agreeing in answer however my method is different but is it valid? $\endgroup$ – Matthew Jan 4 '18 at 14:33
  • $\begingroup$ That method is true and if it responses to you is also sufficient but in general case there are more conditions needed to be held, roughly say, what if the method doesn't imply enough for us to argue? But is it replies to you it's correct because at this case the limit must exist from any direction (including y=x) and be equal but about another function may this case hold but for example y=x^2 doesn't hold. Anyway you need to check all directions in which function argument can tend to some certain point... $\endgroup$ – Mostafa Ayaz Jan 4 '18 at 14:48
  • $\begingroup$ That makes sense thank you! $\endgroup$ – Matthew Jan 4 '18 at 14:55
  • $\begingroup$ How did you arrive at the expression $\cos(2\theta)/r^{2s}\>$? $\endgroup$ – Christian Blatter Jan 4 '18 at 15:07
  • $\begingroup$ It's a simple substitution and using $cos^4(\theta)+sin^4(\theta)+2cos^2(\theta)sin^2(\theta)=1$ $\endgroup$ – Mostafa Ayaz Jan 4 '18 at 15:10

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