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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.

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

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    $\begingroup$ Did you look at the path $(x,\sqrt x)$? $\endgroup$ – zhw. Oct 4 '15 at 19:59
  • $\begingroup$ Funny you mention, just thought this right now for other question, thanks. Edit: worked. $\endgroup$ – João Pedro Oct 4 '15 at 20:00
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    $\begingroup$ 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. $\endgroup$ – João Pedro Oct 4 '15 at 20:03
  • $\begingroup$ 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? $\endgroup$ – the_candyman Oct 4 '15 at 20:39
  • $\begingroup$ @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 $\endgroup$ – user190080 Oct 4 '15 at 20:46
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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.

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  • $\begingroup$ I think you are right. Anyway, why my approach yields to different results? $\endgroup$ – the_candyman Oct 4 '15 at 20:18

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