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a curious limit
#1
I'm wondering if the following limit is non-zero; v E R



and if so, what is it equal to? Thanks

I know it doesn't converge for
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#2
attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol
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#3
(04/14/2011, 08:10 PM)nuninho1980 Wrote: attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol

no no, I put 0, but latex just designs it to look like o
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#4
(04/14/2011, 08:01 PM)JmsNxn Wrote: I'm wondering if the following limit is non-zero; v E R



and if so, what is it equal to? Thanks

I know it doesn't converge for

The powers with non-integer exponents are not uniquely defined in the complex plane.
In your case you would need to put:



But then the standard logarithm has a cut on , which is quite arbitrary: one could put a cut however one likes. For example could spiral around 0, while moving towards 0 and would increase/decrease its imaginary part by in each round.
I guess it really depends on how approaches 0.
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#5
let's take the limit from positive (keep it simple first)

so:


Is there any way of re-expressing this limit?
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#6
(04/14/2011, 10:55 PM)JmsNxn Wrote: let's take the limit from positive (keep it simple first)

so:


Is there any way of re-expressing this limit?

But then its not difficult, since on the reals, the whole limit goes to (complex) except for , for which the whole limit is 0.
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#7
(04/15/2011, 07:14 AM)bo198214 Wrote: But then its not difficult, since on the reals, the whole limit goes to (complex) except for , for which the whole limit is 0.

Alright, how about



where is taken to mean approaching along the axis.

I think it's the equivalent of:


which I guess converges to negative infinity again, except for 1-e^{vi}=0

hmm, seems this is less interesting than I thought.
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#8
is there any way of letting h approach zero such that:

?
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#9
(04/16/2011, 07:22 PM)JmsNxn Wrote: is there any way of letting h approach zero such that:

?

The logarithm of is .
So regardless how you approach 0, i.e. , you will allways have that .
So the answer is no (except ).

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#10
(04/16/2011, 07:41 PM)bo198214 Wrote: The logarithm of is .
So regardless how you approach 0, i.e. , you will allways have that .
So the answer is no (except ).

that's what I thought
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