The limit of x^i as x->0+

  • #1
Hi,

I'm having to teach myself math and I guess I'm skipping ahead here but am really interested. For f(x)=x^i, the limit from the positive side I see from WolframAlpha is e^(2 i 0 to pi). Can someone please explain what the "2 i 0 to pi" means? Is it 2 times i times the set 0 to pi? Or something like that? And why is it 0 to pi? As it spins around going to zero, wouldn't it 'touch' the y-imaginary axis at one particular point?

Just thinking aloud...any help would be much appreciated :)
 

Answers and Replies

  • #2
Mute
Homework Helper
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You can rewrite xi as

$$x^i = e^{i\ln x},$$
assuming x is real. Do you know Euler's identity?

$$e^{iy} = \cos y + i \sin y.$$

With these two pieces of information, can you see what happens as x tends to zero from above?
 
  • #3
D H
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Can someone please explain what the "2 i 0 to pi" means?
The limit doesn't exist. However, for all positive real x, xi is constrained to lie on the unit circle. That is all you can say about the limit, and that is all that Mathematica is saying. e2i (0 to pi) might be better written as [itex]e^{i\theta}, 0\le\theta<2\pi[/itex].
 
  • #4
798
34
Hi,

I'm having to teach myself math and I guess I'm skipping ahead here but am really interested. For f(x)=x^i, the limit from the positive side I see from WolframAlpha is e^(2 i 0 to pi). Can someone please explain what the "2 i 0 to pi" means? Is it 2 times i times the set 0 to pi? Or something like that? And why is it 0 to pi? As it spins around going to zero, wouldn't it 'touch' the y-imaginary axis at one particular point?

Just thinking aloud...any help would be much appreciated :)
"2 i (0 to pi)" means "oscillating on the range 0 to 2*i*pi".
The real part of x^i is oscillating on the range -1 to 1
The imaginary part of x^i is oscillating on the range -i to i
 
  • #5
You can rewrite xi as

$$x^i = e^{i\ln x},$$
assuming x is real. Do you know Euler's identity?

$$e^{iy} = \cos y + i \sin y.$$

With these two pieces of information, can you see what happens as x tends to zero from above?
Yes! I had actually tried to change x^i into Euler's identity, but not only was I unable to do it at the time, I see now that it likely wouldn't have helped without the x^i = e^i*ln(x) relationship.

Basically ln(x) goes to -∞ and e^i(-∞)=cos(-∞) + i*sin(-∞), and Euler's identity only gives specific y-imaginary values for particular values of x. But since ∞ isn't a particular value of x, but an arbitrarily large value, you can't say anything specific about the limit except that it lies somewhere on the unit circle. Is that the right way to think about it?
 
  • #6
Mute
Homework Helper
1,388
10
Yes! I had actually tried to change x^i into Euler's identity, but not only was I unable to do it at the time, I see now that it likely wouldn't have helped without the x^i = e^i*ln(x) relationship.

Basically ln(x) goes to -∞ and e^i(-∞)=cos(-∞) + i*sin(-∞), and Euler's identity only gives specific y-imaginary values for particular values of x. But since ∞ isn't a particular value of x, but an arbitrarily large value, you can't say anything specific about the limit except that it lies somewhere on the unit circle. Is that the right way to think about it?
You can say something specific about the limit: it doesn't exist. The functions ##\sin x## and ##\cos x## have no limit as x approaches infinity, so this tells you that ##x^i## has no limiting value, its value just twirls around the unit circle in the complex plane faster and faster as x grows larger.
 

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