# Use $\epsilon$-$\delta$ definition to prove $\lim\limits_{(x,y)\rightarrow (0,0)} \frac{(xy)^4}{ (x^2 + y^4)^3}$ exists.

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

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

To prove this consider $x_n=n^{-2}$, $y_n=n^{-1}$ and recall definition of continuity by Geine.
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 $1/8$.
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.