Note that
[tex] e^{ix} [/tex]
is bounded:
[tex] e^{ix} = \cos(x) + i\sin(x) [/tex]
so, for large, or arbitrary x, e^{ix}, the imaginary part will never be greater than 1, and the real part will never be greater than 1.
Thus, if you had:
[tex] \lim_{x \rightarrow \infty} e^{(i-1)x} [/tex]
the e^{-x} part forces the whole thing to go to zero.
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling We Value Civility
• Positive and compassionate attitudes
• Patience while debating We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving