Use the metric $(x,y)$ = $\rho(x,y)=|x-y|$ for the reals and use the metric $\rho((x_1,y_1),(x_2,y_2)) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$ for the plane.

Define $f:R\times R \to R$ as $f(x,y)=\frac{xy^2}{(x^2 + y^2)}$ for $f(x,y)$ $\neq$ $(0,0)$ and set $f(0,0) = 0$. Determine whether $f$ is continuous using $\epsilon$-$\delta$ proof.


I've tried using a direct approach by using the definition of a continuous function:

$f:R\times R \to R$ is continuous at $(x_1,y_1) \in R \times R$, $\forall \epsilon>0, \exists \delta >0$ such that $\forall (x_2,y_2) \in R \times R$ if $\sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2 } <\delta$, then $|f(x_1,y_1)-f(x_2,y_2)|<\epsilon$

However, with the $f(x,y)$ as defined above, it seems impossible with the messy algebra since I want to find a $\delta$ that satisfies $|\frac{x_1y_1^2}{(x_1^2 + y_1^2)}-\frac{x_2y_2^2}{(x_2^2 + y_2^2)}| < \epsilon$ using the metrics defined above.

Is there a trick I am supposed to be seeing? Would really appreciate if anyone could show me the proof for this question. Thank you!


marked as duplicate by Jack, Davide Giraudo real-analysis Jan 8 '17 at 22:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See that $\rho((x_1, y_1), (x_2, y_2)) = \Vert \vec x_1 - \vec x_2 \Vert_2 \leq 2 \max (|x_1 - x_2|, |y_1 - y_2|) = 2 \Vert \vec x_1 - \vec x_2 \Vert_\infty$ $\endgroup$ – AlexR Oct 31 '13 at 9:21

(I realize this question was asked a long time ago, but I recently had it assigned as homework and thus maybe answering it now could still be useful to somebody.)

Check out what happens when you approach (0, 0) from the direction $y = \sqrt{x^3 - x^2}. f(x, \sqrt{x^3 - x^2}) = x -1 $, so $f(0, \sqrt{x^3 - x^2}) = -1$.

  • $\begingroup$ You seem to be suggesting that the function is not continuous. This isn't true, as it is actually continuous. Note that you can't approach (0,0) (at least, you can't get particularly close) from the direction $y = \sqrt{x^3-x^2}$, as $x^2 > x^3$ for small $x$. $\endgroup$ – user2258552 Oct 30 '14 at 22:04
  • $\begingroup$ If x = 0, $y = \sqrt{x^3−x^2} = \sqrt{0^3−0^2} = 0.$ Therefore, if f is continuous, f(0, 0) should equal f(x, $\sqrt{x^3−x^2}$) when x = 0. $\endgroup$ – rakhil11 Nov 1 '14 at 18:40

Here is how you advance

$$ \frac{xy^2}{x^2+y^2} \leq \sqrt{x^2+y^2} < \epsilon=\delta.$$

Note: we used the inequality

$$ |a| \leq \sqrt{a^2+b^2}. $$


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