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If f is a function defined throughout a disk centred at $(x_0, y_0)$, and if $f_x(x_0, y_0)$ and $f_y(x_0, y_0)$ both exist, then f is continuous at $(x_0,y_0)$.

I know to be a function to be continuous at a point its limit approaching that point and its value at that point are equal, but the function isn't defined so it feels like data is insufficient.

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This is not true. Consider the function $f$ defined on a disk of radius $1$ centred at $(0,0)$, by the formula $f(x,y)=\frac{xy}{x^2+y^2}$ for $(x,y) \neq 0$ and $f(0,0)=0$.

It is clear that all the partial derivatives exist (with value $0$) at $(0,0)$. But $f$ is discontinuous at $(0,0)$, as can be easily checked by calculating $\lim f(x,y)$ where $(x,y)$ moves through different line segments approaching $(0,0)$.

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  • $\begingroup$ I agree with your answer, but how do I prove it in general (without taking an example and keeping it x and y)? $\endgroup$
    – user562600
    Aug 25, 2019 at 14:47
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    $\begingroup$ I don't think you can prove a false statement. $\endgroup$
    – J. Doe
    Aug 25, 2019 at 14:48
  • $\begingroup$ Not prove it, I meant to prove a counterexample. $\endgroup$
    – user562600
    Aug 25, 2019 at 14:48
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    $\begingroup$ @redx: The point of counterexamples is that you just need one, in order to falsify whatever it is that you want to falsify. What do you even mean by “proving a counterexample in general”? $\endgroup$ Aug 25, 2019 at 17:52

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