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.

I have $f(x,y) =\frac{2xy^2 }{x^2+y^4}$ if $(x,y) \neq (0,0)$ and $0$ if $(x,y) = (0,0)$. I figured I could just find that the limit as $\frac{2xy^2 }{x^2+y^4}$ does not equal to $0$, which should show discontinuity, but it does and I don't know how else one can show discontinuity. I showed it was not differentiable with the definition of the derivative if that helps at all, but I can't reuse that. Thanks for any help.

share|improve this question
    
"but it does": But what does what? If the limit of $f(x,y)$ as $(x,y)\to (0,0)$ were to exist and be $0$, then the function would be continuous. But the limit does not exist. –  Jonas Meyer Sep 26 '12 at 4:59
add comment

1 Answer

By taking the path $\mathbf{r}_a(t)=(t,\sqrt{\frac{t}{a}})$ reaching to $(0,0)$, you will find out that the limit is not exist at the origin. In fact, it depends on the value of $a$.

share|improve this answer
add comment

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.