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$$f(x,y)={xy^2\over x^2+y^2}, \text{ with } f(0,0)=0$$ Show that $f$ is not differentiable at $(0,0)$.

I know that by definition, I need to show that $$\lim_{h->0} \frac{||f(x+h,y+h)-f(0,0)||}{||h||}=0$$ (both partial derivatives are 0).

However when I plugged in $x+h$ and $y+h$, I was stuck. Would really appreciate the help.

mathreadler
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Annie Fannie
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  • Hi and welcome to the site. In the long run you are encouraged to learn typesetting with MathJax and LaTeX as it will increase chances of getting good responses. I have helped you with this first one. Also if questions look like they could be homework questions you are encouraged to show any own work on solving the problem. – mathreadler Mar 13 '17 at 12:26
  • No, it is not enough to show that the limit exists when $x$ and $y$ both increase by the same $h$ -- the definition should be something like $$ \lim_{(h_1,h_2)\to(0,0)}\frac{f(x+h_1,y+h_2)-f(x,y)}{|(h_1,h_2)|} $$ (with no norm signs in the numerator too) -- where $h_1$ and $h_2$ go towards $(0,0)$ together but are not necessarily equal while they do so. – hmakholm left over Monica Mar 13 '17 at 12:32

2 Answers2

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The point is $(x,y) = (0,0)$, so your limit is really $\lim_{(u,v) \to (0,0)} \frac{| f(u,v) |}{\sqrt{u^2+v^2}}$. To show that the limit is not 0, use polar coordinates $u=r\cos{t}$, $v=r\sin{t}$ and change the limit into one involving $\lim_{r\to 0}$.

Catalin Zara
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You have to investigate $\lim_{(u,v) \to (0,0)} \frac{\| f(u,v) \|}{\sqrt{u^2+v^2}}$

For $v=u>0$ we have

$\frac{\| f(u,u) \|}{\sqrt{u^2+u^2}}=\frac{1}{2 \sqrt{2}}$

Your turn !

Fred
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  • thanks! this is very straight forward. I really don't want to use the polar coordinate method since I'm always confused. – Annie Fannie Mar 13 '17 at 12:34