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

$\lim_{h\to\0}h^{1-e^{vi}}$

and if so, what is it equal to? Thanks

I know it doesn't converge for $v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\}$
attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol
(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
(04/14/2011, 08:01 PM)JmsNxn Wrote: [ -> ]I'm wondering if the following limit is non-zero; v E R

$\lim_{h\to\0}h^{1-e^{vi}}$

and if so, what is it equal to? Thanks

I know it doesn't converge for $v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\}$

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

$\lim_{h\to\0} e^{(1-e^{vi})\log(h)}$

But then the standard logarithm has a cut on $(-\infty,0]$, which is quite arbitrary: one could put a cut however one likes. For example $h$ could spiral around 0, while moving towards 0 and would increase/decrease its imaginary part by $2\pi i$ in each round.
I guess it really depends on how $h$ approaches 0.
let's take the limit from positive (keep it simple first)

so:
$\lim_{h\to\0^+}\, (1-e^{vi})ln(h)\, =\, f(v)$

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

so:
$\lim_{h\to\0^+}\, (1-e^{vi})ln(h)\, =\, f(v)$

Is there any way of re-expressing this limit?

But then its not difficult, since $\ln(h)\to -\infty$ on the reals, the whole limit goes to (complex) $\infty$ except for $1-e^{vi}=0$, for which the whole limit is 0.
(04/15/2011, 07:14 AM)bo198214 Wrote: [ -> ]But then its not difficult, since $\ln(h)\to -\infty$ on the reals, the whole limit goes to (complex) $\infty$ except for $1-e^{vi}=0$, for which the whole limit is 0.

Alright, how about

$\lim_{h\to\0^{v+}} (1-e^{vi})ln(h)$

where $\lim_{h\to\0^{v+}}$ is taken to mean approaching along the $e^{vi}$ axis.

I think it's the equivalent of:
$= \lim_{h\to\0^{+}} (1-e^{vi})ln(he^{vi})$
$= \lim_{h\to\0^{+}} (1-e^{vi})(ln(h) + vi)$
which I guess converges to negative infinity again, except for 1-e^{vi}=0

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

$\lim_{h\to\0} (1-e^{vi})ln(h) = 0$?
(04/16/2011, 07:22 PM)JmsNxn Wrote: [ -> ]is there any way of letting h approach zero such that:

$\lim_{h\to\0} (1-e^{vi})ln(h) = 0$?

The logarithm of $z=r e^{i\phi}$ is $\log( r)+i\phi$.
So regardless how you approach 0, i.e. $r\to 0$, you will allways have that $|\log(z)|=\sqrt{\log( r)^2+\phi^2}\to \infty$.
So the answer is no (except $1=e^{vi}$).

(04/16/2011, 07:41 PM)bo198214 Wrote: [ -> ]The logarithm of $z=r e^{i\phi}$ is $\log( r)+i\phi$.
So regardless how you approach 0, i.e. $r\to 0$, you will allways have that $|\log(z)|=\sqrt{\log( r)^2+\phi^2}\to \infty$.
So the answer is no (except $1=e^{vi}$).

that's what I thought
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