How to find the lim as i->infinity for e^i/(3^i-1) and lim->infinity 2^iπ^i/(3^i+2)?
For the 3^i-1 and 3^1+2 it's 3 to the exponent i-1 and i+2 respectively not just i.
More Questions From This User See All
For the 3^i-1 and 3^1+2 it's 3 to the exponent i-1 and i+2 respectively not just i.
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
(e^i)/(3^(i-1)) = 3(e/3)^i. And since (e/3)<1 , lim(i-->inf.)3(e/3)^i = 0.
2^iπ^i/((3^(i+2)) = (1/9)(2π/3)^i and because, (2π/3) >1 , lim(i-->inf)(1/9)(2π/3)^i = infinity.