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

## 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

## 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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## 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

## 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. $\sqrt{\frac{1}{100}} = \frac{1}{10} > \frac{1}{100}$.