The discussion centers on proving that if the sequence \( x_n \) approaches 0 as \( n \) approaches infinity, then \( \sqrt{x_n} \) also approaches 0. The proof presented utilizes the epsilon (\( \epsilon \)) definition of limits, establishing that for any \( \epsilon > 0 \), there exists a natural number \( N \) such that for all \( n > N \), \( |x_n| < \epsilon \) implies \( |\sqrt{x_n}| < \epsilon \). A concern is raised regarding the case when \( x_n < 1 \), specifically questioning the validity of the proof under this condition.
PREREQUISITESStudents of calculus, mathematicians focusing on real analysis, and anyone interested in understanding the behavior of sequences and limits.
What if ##x_n < 1\,?## E.g. ##\sqrt{\frac{1}{100}} = \frac{1}{10} > \frac{1}{100}##.Mathematicsresear said:Homework Statement
Suppose sequence x_n tends to 0 as n approaches infinity, prove that sqrt(x_n) also tends to 0
x_n is a sequence of non negative real numbers
Homework Equations
The Attempt at a Solution
Proof. Let e>0. There exists an N in the naturals such that for n>N Ix_nI < e So if I choose an N such that x_n < e whenever n>N then Isqrt(x_n)I<x_n<e
e is epsilon.
Is this proof correct?