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Show that $$\lim_{(x,y)\to(0,0)}\frac{x^4y^4}{(x^2+y^4)^3}$$ D.N.E

In order to do this I should consider two paths and show that they do not reach the same limit.

Consider when $x=0$.

Then we have $\lim_{y\to 0}\dfrac{y^4}{y^{12}}=\dfrac{1}{y^8}=\infty$

Similarly if we choose $y=0$, then we have

Then we have $\lim_{x\to 0}\dfrac{x^4}{x^{6}}=\dfrac{1}{x^2}=\infty$

Since both of these limits approach an asympote, can I even use them to show that the limit does not exist? Or must they approach a finite number?

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    $\begingroup$ The limit is zero as $x=0$. $\endgroup$ – Nosrati Sep 24 '17 at 16:05
  • $\begingroup$ If you choose $x = 0$, then what happened to the $x^4$ in the numerator. $\endgroup$ – A---B Sep 24 '17 at 16:06
  • $\begingroup$ I see my mistake. THe entire numerator becomes $0$ in that case resuling in the limit becoming $0$ $\endgroup$ – K Split X Sep 24 '17 at 16:18
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Take the path $(x=0, y\rightarrow0)$, limit is $0$.

Take the path $(x^2=y^4;x,y\rightarrow 0)$, limit $=\frac1{8}$ and you are DONE!

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Write the function as $$\frac{x^4 y^4}{(x^2 + y^4)^3} = \frac{x^4 y^4 / x^6}{(1 + (y^4/x^2))^3} = \frac{y^4/x^2}{(1+(y^4/x^2))^3}.$$ Now we can see that if $(x,y) \to (0,0)$ in such a way that the ratio $y^4/x^2$ does not tend to $0$, the resulting limit will not be zero. This furnishes the motivation behind the choice of path $x = y^2$. In fact, we can see that for any real constant $c$, if we choose the parametrization $(x,y) = (t^2, ct)$, then $y^4/x^2 = c^4$ is independent of the parameter $t$, and $$\lim_{(x,y) \to (0,0)} \frac{x^4 y^4}{(x^2 + y^4)^3} = \lim_{t \to 0} \frac{c^4}{(1+c^4)^3} = \frac{c^4}{(1+c^4)^3}.$$

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