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I need to calculate $\lim\limits_{(x,y) \to(0,0)} \frac{x^4y^2}{ (x^4+y)^5}$

I get $[0/0]$. i think it doesn't have a limit but i don't know how to prove it.

Thank you.

roy
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    There are tons of similar questions, have you looked at any of them to get some ideas on how to work on this kind of problem? (see for example this one and this one). The exact expressions don't have to be exactly the same. I suggest to check those questions above for two different kind of attacks in two different situations. – mickep Dec 03 '15 at 20:46
  • i tried to look. didn't see something similar. – roy Dec 03 '15 at 20:46
  • Your limit equals $0$ assuming all variables are real-valued! – Jan Eerland Dec 03 '15 at 20:51
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    You tried. You saw something similar. But you were too lazy to try to do it by yourself... – the_candyman Dec 03 '15 at 20:51
  • The limit cannot exist for the function is not even defined on any punctured neighborhood of $(0,0)$. There are points on the curve $y=-x^4$ arbitrarily close to the origin, and no way to extend the definition of the function on that curve. – Jyrki Lahtonen Dec 03 '15 at 20:57
  • @JanEerland That is clearly wrong. – mickep Dec 03 '15 at 21:10

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Hint: if you don't manage to show the limit exists, try to show it does not. To achieve the latter, the usual trick is to find two "paths" going to zero, along which the function does not have the same limit. Here, two "natural" choices are to set one of the two variables to $0$ (e.g., path $(x,0)$), or to have only one variable by setting for instance $y=x^a$ (path $(x,x^a)$).

  • Along the line $(x,0)$, the quantity $f(x,0)$ is identically zero, hence converges to $0$
  • But along $(x,x^4)$, you have $f(x,x^4) = \frac{x^{12}}{2^5 x^{20}}$, which...
Clement C.
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