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I tried several paths but all leads to the same value. Since the question wants me to prove, it need to exist, but I can't find two paths.

$$\lim_{(x,y)\to (0,0)}\frac{x^4 \cdot y^4}{(x^2+y^4)^3}$$

Aleksandar
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João Pedro
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    Did you look at the path $(x,\sqrt x)$? – zhw. Oct 04 '15 at 19:59
  • Funny you mention, just thought this right now for other question, thanks. Edit: worked. – João Pedro Oct 04 '15 at 20:00
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    I think in questions like this we need to think a path that makes the exponent in numerator be equal to the exponent in denominator. Thats why paths like $y=x$ and $y=x^2$ leads to nowhere. – João Pedro Oct 04 '15 at 20:03
  • using polar coordinates, you get: $$\lim_{\rho \to 0}\frac{\rho^8 \cos^4(\theta)\sin^4(\theta)}{(\rho^2\cos^2(\theta)+\rho^4\sin^4(\theta))^3} = \lim_{\rho \to 0}\frac{\rho^8 \cos^4(\theta)\sin^4(\theta)}{\rho^6(\cos^2(\theta)+\rho^2\sin^4(\theta))^3} = \ = \lim_{\rho \to 0}\frac{\rho^2 \cos^4(\theta)\sin^4(\theta)}{(\cos^2(\theta)+\rho^2\sin^4(\theta))^3} = 0 ~\forall \theta$$. Where is the error? – the_candyman Oct 04 '15 at 20:39
  • @the_candyman I still wonder myself...I don't know who downvoted but you were faster with deleting than me with upvoting your answer. maybe one can ask this one as a question on its own. calculations are correct, double checked, so is the conclusion, so there must be something wrong with the argument itself – user190080 Oct 04 '15 at 20:46
  • @user190080 thanks man, this is a good place only thanks to people like you – the_candyman Oct 04 '15 at 20:47
  • @the_candyman thanks, I am totally with you on this! you could check out this answer, the very last, it looks like a very similar approach except that in this one, it really fails – user190080 Oct 04 '15 at 20:54
  • @the_candyman Can you construct a function discontinuous at $0$ with $\lim_{\rho \to 0} =0 \forall \theta$ ? – Keith McClary Oct 05 '15 at 03:15
  • @the_candyman: the problem is really that you fixed the $\theta$, so while it is true that $\forall\theta\lim_{\rho \to 0}f =0$ this is not sufficient - graphical this means that you are always taking a straight line to the origin. The problem arises when we have a look to the denominator, which can become arbitrarily small with $\rho\to0$ and a proper sequence of $\theta_n$ which finally leads to something like $\frac00$. The polar coordinates is a valid argument, if one can control the growth, like for example if the resulting term is independent of $\theta$. cheers – user190080 Oct 05 '15 at 20:17
  • @KeithMcClary the given function has exactly this property...if you like please have a look to my comment above which should explain the situation and why the polar coordinate access here is not such an approriate tool – user190080 Oct 05 '15 at 20:18
  • @the_candyman We only know this (I think?) because zhw and HasanSaad pulled $x=y^2$ out of their hats. I'm not sure if I have learned anything to help me approach the next limits problem (except to be careful with polar coordinates, which I already knew from another question ). – Keith McClary Oct 05 '15 at 21:50

1 Answers1

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Take $x=y$, then

$\lim=\lim_{x\to0}\frac{x^8}{(x^4+x^2)^3}=0$

Take $x=y^2$, then

$\lim=\lim_{y\to0}\frac{y^{12}}{8y^{12}}=\frac{1}{8}$

Thus, the limit does not exist. QED.

Marconius
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Hasan Saad
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