$$ f(x,y) = \begin{cases} \frac{x^3-xy^2}{x^2+y^2}, & \text{if }(x,y)\not= (0,0) \\ 0, & \text{if } (x,y)=(0,0) \end{cases} $$

how to prove $f(x,y)$ is continuous on $\mathbb{R}^2$? I know it must be continuous on $\mathbb{R}^2$, but according to the definition of continuity, how to use the $\delta- \varepsilon $way to prove?

I know how to prove the continuity at $(0,0)$ but want to know how to prove in the general case other than at $(0,0)$. I can't find a $\delta$ to fit the $\varepsilon$. I mean, let $|f(x,y)-f(x_1,y_1)|<\varepsilon$

$(x_1,y_1)$)is any point but not $(0,0)$.

and how to transform it to relate to a $\delta$ in order to have
$|(x,y)-(x_1,y_1)|<\delta$ then $|f(x,y)-f(x_1,y_1)|<\varepsilon$?

  • $\begingroup$ What's wrong with the old question? The accepted answer is precisely an epsilon-delta proof... $\endgroup$
    – user7530
    Feb 14, 2013 at 7:58
  • $\begingroup$ @user7530: no it's at point $(0,0)$ not any point $(x_1,x_2)$ $\endgroup$
    – i_a_n
    Feb 14, 2013 at 7:59

2 Answers 2


You can argue by using the following inequality at $(0,0)$:

$$\left|\frac{x^3-xy^2}{x^2+y^2} \right|\leq |x|\frac{x^2}{x^2+y^2}+|x|\frac{y^2}{x^2+y^2}=|x|$$

Therefore, as $(x,y) \to (0,0)$ you have that $x \to 0$ and by the previous inequality $f(x,y) \to 0=f(0,0)$.

In other points than $(0,0)$ the continuity is guaranteed because all the operations made are elementary and well defined.


$x$ and $y$ are continuous functions. Moreover, the sum, product, and quotient (at points where the denominator is non-zero) of continuous functions are continuous.

If you really want a self-contained $\epsilon-\delta$ proof, you can use the proofs of the above lemmas and pump $\epsilon$ through the compositions. But that sounds like an awful lot of work for little gain...


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