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
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?