What is the value of the following limit? $$\lim_{x\to 0} x^i$$ Wolfram Alpha gives an insane result, so does Mathematica.
What is the value of $\lim_{x\to 0} x^i$?
-
0@DonAntonio I usually ask difficult questions that little people can answer. I prefer asnwering easy questions myself. – 2012-11-04
-
6*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*? – 2012-11-04
-
3Do you consider this as a *difficult question*? – 2012-11-04
-
1@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. – 2012-11-04
-
0@did I do not know, but many of the questions I ask indeed difficult. – 2012-11-04
-
0All is clear now, @Anixx: you don't like the answers you get here. Fine, that's of course your right. – 2012-11-04
-
0@amWhy it is insane because it is neither an algebraic expression, nor indeterminacy. – 2012-11-04
-
0Anixx: if indeed you do not like the answers you get here, why do you keep coming back to ask questions? – 2012-11-04
-
5Anixx: Q: In what context did you encounter the problem? A: (None.) Q: What are your thoughts on it? A: (None.) Please explain. – 2012-11-04
-
1@did People are always asking questions here in the way he did, there's no reason to single him out for objections. – 2012-11-04
-
3@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. – 2012-11-04
-
4@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. – 2012-11-04
-
2@Zarrax Where did I refer to rules of my own? Another strange assertion. (But I note you downplayed *always* to *normally*.) – 2012-11-04
-
1@DonAntonio Please see my prior comment. – 2012-11-04
-
1@BillDubuque , I see your prior comment, yet I fail to see your point in recommending me to see it. – 2012-11-05
3 Answers
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.
-
0So the limit of the absolute value is 1 and the limit of the argument in infinity, right? – 2012-11-04
-
0the argument doesn't have a limit – 2012-11-04
-
0@Zarrax: Indeed the argument diverges, and takes on all possible values infinitely many times. – 2012-11-04
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.
-
0Thanks, but what does mean the result which Wolfram Alpha returns? – 2012-11-04
-
0What does WA return? Can you paste the result here? – 2012-11-04
-
0there is a link in the question http://tinyurl.com/ckwmayc – 2012-11-04
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.
-
2The argument is only defined up to multiples of $2\pi$, saying it goes to minus infinity has no meaning. – 2012-11-04
-
2View the argument here as the number inside the cosine or sine function. – 2012-11-04
-
0Seeing the argument as you wish, @Zarrax, Marc's comment's still true: it is defined *only* up to multiples of $\,2\pi\,$ – 2012-11-05
-
0@DonAntonio But the second part "has no meaning" is not true. – 2012-11-05