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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.

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    $\begingroup$ 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? $\endgroup$
    – Did
    Nov 4, 2012 at 17:30
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    $\begingroup$ Do you consider this as a difficult question? $\endgroup$
    – Did
    Nov 4, 2012 at 17:31
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    $\begingroup$ Anixx: Q: In what context did you encounter the problem? A: (None.) Q: What are your thoughts on it? A: (None.) Please explain. $\endgroup$
    – Did
    Nov 4, 2012 at 17:44
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    $\begingroup$ @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. $\endgroup$
    – Did
    Nov 4, 2012 at 18:07
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    $\begingroup$ @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. $\endgroup$
    – Zarrax
    Nov 4, 2012 at 18:12

3 Answers 3

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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.

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  • $\begingroup$ Thanks, but what does mean the result which Wolfram Alpha returns? $\endgroup$
    – Anixx
    Nov 4, 2012 at 17:49
  • $\begingroup$ What does WA return? Can you paste the result here? $\endgroup$
    – user17762
    Nov 4, 2012 at 17:51
  • $\begingroup$ there is a link in the question tinyurl.com/ckwmayc $\endgroup$
    – Anixx
    Nov 4, 2012 at 17:53
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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.

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    $\begingroup$ The argument is only defined up to multiples of $2\pi$, saying it goes to minus infinity has no meaning. $\endgroup$ Nov 4, 2012 at 19:32
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    $\begingroup$ View the argument here as the number inside the cosine or sine function. $\endgroup$
    – Zarrax
    Nov 4, 2012 at 20:07
  • $\begingroup$ Seeing the argument as you wish, @Zarrax, Marc's comment's still true: it is defined only up to multiples of $\,2\pi\,$ $\endgroup$
    – DonAntonio
    Nov 5, 2012 at 2:37
  • $\begingroup$ @DonAntonio But the second part "has no meaning" is not true. $\endgroup$
    – Phira
    Nov 5, 2012 at 17:49
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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.

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  • $\begingroup$ So the limit of the absolute value is 1 and the limit of the argument in infinity, right? $\endgroup$
    – Anixx
    Nov 4, 2012 at 18:21
  • $\begingroup$ the argument doesn't have a limit $\endgroup$
    – Zarrax
    Nov 4, 2012 at 18:26
  • $\begingroup$ @Zarrax: Indeed the argument diverges, and takes on all possible values infinitely many times. $\endgroup$ Nov 4, 2012 at 19:34

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