04/15/2011, 05:09 PM
(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.
