Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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 $|f(x,y)-f(x_1,y_1)|<\varepsilon$?

share|improve this question
    
What's wrong with the old question? The accepted answer is precisely an epsilon-delta proof... –  user7530 Feb 14 '13 at 7:58
    
@user7530: no it's at point $(0,0)$ not any point $(x_1,x_2)$ –  i_a_n Feb 14 '13 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.

share|improve this answer

$x$ 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...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.