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What is the value of the following limit? $$\lim_{x\to 0} x^i$$ Wolfram Alpha gives an insane result, so does Mathematica.

Anixx
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  • @DonAntonio I usually ask difficult questions that little people can answer. I prefer asnwering easy questions myself. – Anixx Nov 04 '12 at 17:29
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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you... Oops! Seems you are not new at all, so what is preventing you to say in what context you encountered the problem, and what your thoughts on it are? – Did Nov 04 '12 at 17:30
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    Do you consider this as a difficult question? – Did Nov 04 '12 at 17:31
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    @Annix, knock yourself out then. Many people here though could think that you don't like the answers you've received so far and thus they won't also bother to answer the present one. – DonAntonio Nov 04 '12 at 17:32
  • @did I do not know, but many of the questions I ask indeed difficult. – Anixx Nov 04 '12 at 17:34
  • All is clear now, @Anixx: you don't like the answers you get here. Fine, that's of course your right. – DonAntonio Nov 04 '12 at 17:37
  • @amWhy it is insane because it is neither an algebraic expression, nor indeterminacy. – Anixx Nov 04 '12 at 17:38
  • Anixx: if indeed you do not like the answers you get here, why do you keep coming back to ask questions? – amWhy Nov 04 '12 at 17:39
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    Anixx: Q: In what context did you encounter the problem? A: (None.) Q: What are your thoughts on it? A: (None.) Please explain. – Did Nov 04 '12 at 17:44
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    @did People are always asking questions here in the way he did, there's no reason to single him out for objections. – Zarrax Nov 04 '12 at 18:01
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    @Zarrax Strange reasoning. First, always is wrong. Second, yes, similar objections should be, and indeed are, raised at other questions similarly flouting the rules of the site. – Did Nov 04 '12 at 18:07
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    @did You don't make the rules for this site. If you dislike the way students normally ask questions, maybe you should work on getting over your addiction to this site, rather than harassing the questioners. – Zarrax Nov 04 '12 at 18:12
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    @Zarrax Where did I refer to rules of my own? Another strange assertion. (But I note you downplayed always to normally.) – Did Nov 04 '12 at 18:19
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    @DonAntonio Please see my prior comment. – Bill Dubuque Nov 04 '12 at 20:32
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    @BillDubuque , I see your prior comment, yet I fail to see your point in recommending me to see it. – DonAntonio Nov 05 '12 at 02:36

3 Answers3

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Take $x = e^{-2n \pi}$, where $n \in \mathbb{N}$, then $n \to \infty \implies x \to 0$. Hence, $$\lim_{n \to \infty} e^{-2n \pi i} = \lim_{n \to \infty} (\cos(2n \pi) - i \sin(2 n \pi)) = 1$$

Take $x = e^{-2n \pi - \pi/2}$, where $n \in \mathbb{N}$, then $n \to \infty \implies x \to 0$. Hence, $$\lim_{n \to \infty} e^{-2n \pi i - i \pi/2} = \lim_{n \to \infty} (\cos(2n \pi + \pi/2) - i \sin(2 n \pi + \pi/2)) = -i$$

Hence, limit doesn't exist.

Note that $\vert x ^i \vert = 1$. I think WA is just telling you that the value keeps going around in the unit circle.

5

By definition, for $x > 0$ you have $$x^i = e^{i \ln x} = \cos(\ln x) + i \sin(\ln x)$$ As $x$ goes to zero, $\ln x$ goes to $-\infty$. So while $|x^i| = 1$ for all $x > 0$, the argument of $x^i$ decreases to $-\infty$ as $x$ goes to $0$ from above.

Geometrically, this means the vector $(\cos(\ln x),\sin(\ln x))$ rotates clockwise around the unit circle over and over again as $x$ goes to zero from above.

Zarrax
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Remember that when $z$ is complex, the expression $x^z$ (with positive real $x$, which is the only case where it it unambiguously defined) is just an abbreviation for $\exp(z\ln x)$. So you're looking at $\lim_{t\to-\infty}\exp(ti)$, and you can see for yourself that the limit does not exist. The value spins aroung the unit circle as $t\to-\infty$, and I think that is what the CAS is trying to tell you.

Marc van Leeuwen
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