0 votes 0 votes The limit of the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ is $1$ $2$ $2\sqrt{2}$ $\infty$ Calculus isi2017-dcg calculus limits + – gatecse asked Sep 18, 2019 • recategorized Nov 15, 2019 by Lakshman Bhaiya gatecse 352 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply `JEET commented Nov 15, 2019 reply Follow Share $2$ is the answer. 0 votes 0 votes `JEET commented Nov 15, 2019 reply Follow Share http://zvihrosen.com/S2012-1B/quiz1b-4s.pdf 0 votes 0 votes Please log in or register to add a comment.