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Find the limit

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

This limit does not exists since when convert it into polar we get $\frac {1 }{ r^2 (1-2 \sin^2 \theta \cos^2 \theta)}$ which is one over zero right?

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    Well, it is *not* "one over zero" all the time, but for $$1-2\sin\theta\cos\theta=0\iff\sin2\theta=1\iff \theta=\frac\pi4$$ it would be, and thus indeed the limit doesn't exist. – 2017-01-11
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    But r approach zero we do not care about what in () – 2017-01-11
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    @DonAntonio: But one should perhaps note that the expression in terms of polar coordinates is clearly incorrect to begin with, since the original denominator $x^4+y^4$ is zero only at the origin... – 2017-01-11
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    @HansLundmark Well, yes: the denominator is zero but *because of* $\;r\;$, which is zero at the origin. The trigonometric expression is correct, though: $$\cos^4\theta+\sin^4\theta=(\cos^2\theta+\sin^2\theta)^2-2\cos \theta\sin\theta=1-\sin2\theta$$ – 2017-01-11
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    @DonAntonio: No, $\cos^4 \theta+\sin^4 \theta$ can obviously never be zero, because when one term equals one, the other equals zero. You forgot some squares in $$\cos^4 \theta+\sin^4 \theta = (\cos^2\theta+\sin^2\theta)^2-2\cos^2 \theta\sin^2 \theta=1-\frac12 \sin^2 2\theta \ge \frac12.$$ – 2017-01-11
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    And to avoid future confusion it should perhaps also be said that the question has been edited so that the expression there is correct now. – 2017-01-11

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You are right indeed. One faster way (or at least what I would do) would be to consider one of the following:

  • $x = 0$;

  • $y = 0$;

  • $x = y$;

either of those would turn the limit into something of the form

$$\lim_{t \to 0} \frac1t$$

which does not exist.

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    What if we take the first one we get infinity is not that enough to conclude that the limit DNE – 2017-01-11
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    @Ameryr if you take the first one, $x = 0$, you get $\lim_{y \to 0} \frac{y^2}{y^4} = \lim_{y \to 0} \frac1{y^2} = \infty$ and therefore the limit cannot exist. – 2017-01-11
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    That's enough so we do not need to find the other directions? – 2017-01-11
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    @Ameryr precisely. If one specific direction gives $\infty$ then the limit does not exist. – 2017-01-11
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plugging $$x=\frac{1}{\sqrt{n}},y=\frac{1}{n^2}$$ in the term above we get $${\frac { \left( n+1 \right) \left( {n}^{2}-n+1 \right) {n}^{4}}{ \left( {n}^{2}+1 \right) \left( {n}^{4}-{n}^{2}+1 \right) }} $$ and the limit is $$+\infty$$

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    If we choose one direction with limit result equal to infinity then we conclude that the limit DNE? – 2017-01-11
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    May I ask why did you choose such complicated expressions for $x, y$? Does it have any particular purpose, or where they just random expressions? – 2017-01-11
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    The problem in these choices they dose not tend to zero as n goes to zero, so we need to let n goes to infinity – 2017-01-11