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.

As the topic, Use $\epsilon - \delta$ definition to prove $\lim\limits_{(x,y)\rightarrow (0,0)} \frac{(xy)^4}{ (x^2 + y^4)^3}$ exists. I tried to use the inequalities $|x+y|>|xy|$ and $x^2+y^4>(xy^2)$ but I am not not sure how to set up the inequality only with $|x+y|^n<\delta ^n< \epsilon$

share|improve this question
1  
Ok, let us $\varepsilon >0$... –  nikita2 Oct 21 '12 at 14:46

3 Answers 3

It doesn't exist.

To prove this consider $x_n=n^{-2}$, $y_n=n^{-1}$ and recall definition of continuity by Geine.

share|improve this answer

Besides to Norbert's answer; you can take two different paths approaching the origin: $$y=x,\\\ y=\sqrt{x}$$ First one gives the limit, zero and another path gives us $1/8$.

share|improve this answer

Let $(x,y)\to(0,0)$ along the path $y^2=mx$. Then the given limit reduces to: $\lim \limits_{(x,y)\to(0,0)} \frac{m^2x^6}{(x^2+m^2x^2)^3}$ = $\lim \limits_{(x,y)\to(0,0)} \frac{m^2}{(1+m^2)^3}$ which is clearly dependent on m i.e. the path of approach.

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.