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To show that it is continuous I used polar coordinates to show that the limit at the origin is indeed 0 and so it must be continuous everywhere since the limit obviously equals the function everywhere else

To determine if it is differentiable, I found the partial derivative with respect to x and said that it does not exist at (0,0) therefore the function is not differentiable at the origin.

Thanks

mmmmo
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    What did you do with polar coordinates? Existence of the limit for all lines (constant $\theta$) does not imply existence of the limit for all curves through $(0,0)$. Also, the partial derivative is $$f_x(0,0)=\lim_{\Delta x\to0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=\lim_{\Delta x\to0}\frac{0-0}{\Delta x}=0$$ Same with $f_y(0,0)$. – mr_e_man Sep 09 '18 at 16:49
  • so in this case because the function is defined as 0 at (0,0) I have to use the limit definition as opposed to just finding the partial derivative normally? – mmmmo Sep 09 '18 at 21:04
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    You have to use the definition of partial derivative because $f(x,y)$ is a piecewise function, more over it's easier to evaluate the values of partial derivatives in $(0,0)$ with the limit. :) – Ixion Sep 10 '18 at 01:18

1 Answers1

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It's easy to prove that $\lim_{(x,y)\to (0,0)}f(x,y)=0=f(0,0)$ (see this) hence $f(x,y)$ is continous at $(0,0)$. To determine if $f(x,y)$ is differentiable, we need the values of the derivatives $f_{x}(0,0)$ and $f_{y}(0,0)$. By definition of partial derivatives:

$\\ f_{x}(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}0=0 \\ \\ f_{y}(0,0)=\lim_{k\to 0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to 0}0=0$

Now $f(x,y)$ is differentiable in $(0,0)$ if and only if $$\lim_{(h, k)\to (0,0)}\frac{f(h,k)-f(0,0)-f_{x}(0,0)k-f_{y}(0,0)k}{\sqrt{h^2+k^2}}=0$$

which becomes

$$\lim_{(h, k)\to (0,0)}\frac{h k^3}{(h^2+k^4)\sqrt{h^2+k^2}}=0$$

but if you take $h=k^2$, the restriction is $$\lim_{k\to 0}\frac{k}{2\sqrt{k^2+k^4}}=\lim_{k\to 0}\frac{k}{2|k|\sqrt{1+k^2}}=\begin{cases}\frac{1}{2}&\mbox{if} \ k\to 0^{+}\\ -\frac{1}{2}&\mbox{if}\ k\to 0^{-}\end{cases}$$

On the path $(k^2,k)$ the limit $$\lim_{(h, k)\to (0,0)}\frac{h k^3}{(h^2+k^4)\sqrt{h^2+k^2}}\ne 0$$ hence $f(x,y)$ is not differentiable in $(0,0)$.

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Ixion
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  • this might be a silly question but doesnt the top part have to be an absolute value? – mmmmo Sep 12 '18 at 01:22
  • Well, in general for $(x,y)\to(x_0,y_0)$ it's easy to show that $$g(x,y)→0\iff |g(x,y)|→0$$ so you can avoid the absolute value if you want. – Ixion Sep 12 '18 at 17:01