Analysis Proof: prove that sqrt(x_n) also tends to 0

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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?
 
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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?
What if ##x_n < 1\,?## E.g. ##\sqrt{\frac{1}{100}} = \frac{1}{10} > \frac{1}{100}##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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