What is the value of the following limit? $$\lim_{x\to 0} x^i$$ Wolfram Alpha gives an insane result, so does Mathematica.
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
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. |
|||||||||||||
|
|
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. |
|||||||
|
|
|
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. |
|||||||
|
