04/14/2011, 10:14 PM
(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.
