The limit of a sequence remains unchanged under the operations of squaring and taking the square root, provided the original limit is 1. Specifically, if a sequence \( a_n \) converges to 1, then both \( \lim_{n \to \infty} a_n^2 = 1 \) and \( \lim_{n \to \infty} \sqrt{a_n} = 1 \). This conclusion is supported by the basic theorem of limits, which states that the limit of a function of a sequence can be derived from the limits of the sequences involved.
PREREQUISITESStudents of calculus, mathematicians interested in sequence behavior, and educators teaching limit concepts in advanced mathematics.
Can you think of a basic theorem of limits that would lead to a one line proof?Karl Porter said:Homework Statement:: I was curious, if I have a sequence that has a limit of 1
Relevant Equations:: Lim an=1 as n tends to inf
Lim of an^2=1 as n tends to inf
Does the square of the sequence also have a limit of 1. Does the square root also equal 1? I've been trying to find some counterexamples but I think the limit doesn't change under these operations?