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$f(x,y)=y\frac{x^2-y^2}{x^2+y^2}$

How do you determine if $f$ is continuous everywhere except (0,0)?

Do you look at each component individually?

e.g $g(x,y)=y,h(x,y)=x^2-y^2,k(x,y)=\frac{1}{x^2+y^2}$ are all continuous (except (0,0)) so the original function is continuous?

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    In practice, that is essentially correct. Compositions of continuous functions is continuous, and the same applies with products, divisions, etc. You just have to be cautious around zeros when using division. – 2017-02-28
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    Please don't use the same notation $f(x,y)$ for different functions in the same question. – 2017-02-28
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    What do you mean by "each component"? – 2017-02-28
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    What I believe to be components I have expressed as individual functions – 2017-02-28

2 Answers 2

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If $g(x,y)$ and $h(x,y)$ are continuous functions, and $h(x,y)$ is bounded away from zero in a neighbourhood of $(x_0,y_0)$, then $r(x,y)=\frac{g(x,y)}{h(x,y)}$ is continuous at $(x_0,y_0)$. Just pick your favourite proof of this well-known lemma.

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    Indeed, it is sufficient that $h(x_0, y_0) \ne 0$. – 2017-02-28
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Use the polar coordinate and write $x=rcos(u), y=rsin(u)$, $f(r,u)=rsin(u){{r^2(cos^2(u)-sin^2(u))}\over r^2}=rsin(u)cos(2u)$ and the limit exists when $r\rightarrow 0$ and is zero.

On $R^2-\{(0,0)\}$, $f$ is the quotient of two polynomial functions so it is continuous, so you can extend $f$ to a continuous function on the plane. I believe, this is the next question of the exercice:

  1. Show that the function is continuous on $R^2-\{(0,0)\}$

  2. Show that $lim_{(x,y)\rightarrow (0,0)}f(x,y)$ exists

  3. Deduce that you can extend $f$ to a continuous function on $R^2$.

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    Doesn't this only show (if $f(0,0)=0$ but it isn't) that $f$ is continuous at $(0,0)$? – 2017-02-28