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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 $|x+y|^n<\delta ^n< \epsilon$

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    $\begingroup$ Ok, let us $\varepsilon >0$... $\endgroup$ Commented Oct 21, 2012 at 14:46

3 Answers 3

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It doesn't exist.

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

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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 $1/8$.

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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.

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