a curious limit

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