a curious limit

[JmsNxn]

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 \( v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\} \)

[nuninho1980]

attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol

[JmsNxn]

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

[bo198214]

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.

[JmsNxn]

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?

[bo198214]

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.

[JmsNxn]

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.

[JmsNxn]

is there any way of letting h approach zero such that:

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

[bo198214]

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 \( 1=e^{vi} \)).

[JmsNxn]

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