Delta-Epsilon Limit Test for Sqrt(X) > 0

[olds442]

Homework Statement

Ok, so I was reading my Real Analysis Text and there was a proof that Lim sqrt(Xn)=Lim sqrt (X). They had two cases, one where Xn=0 where they use the Delta-Epsilon limit test and it makes perfect sense to me. However, they also show the example where x>0. obviously since x>0, Sqrt(x) is also >0. They prove it without using the E-D method and I am trying to prove it using it. Ill show my first few calculations ill use S(Xn) to denote sq. rt. and e to denote epsilon:
e>0 is given
|S(Xn)-S(x)|<e after multiplying by (S(Xn)+S(x))/(S(Xn)+S(x)) i get: Xn-X/(S(Xn)+S(x))<e now it would also be true that Xn-X/(S(Xn)<e from here, i do not know how to continue the proof in order to solve for it. If i squared both sides, i would have a nasty polynomial on top and that wouldn't offer much help.. any advice?
thanks

 

[CompuChip]

Welcome to PF olds442.
Your notation is not very clear, so let us first find out what you meant

I suppose you meant we have a sequence (X_n)_{n \in \mathbb{N}} of real numbers whose limit for n to infinity is X, and you want to prove that
\lim_{n \to \infty} \sqrt{X_n} = \lim_{n \to \infty} \sqrt{X},
which is \sqrt{X}.
What do you mean by "Xn = 0", that each element in the sequence is zero (for all n?). Then you talk about x > 0, what is x? Is it X ?

 

[olds442]

I am sorry for the terrible notation.. Your latex example is what i meant. Here is how it is written: Let X= (Xn) be the sequence of real numbers that converges to x and suppose that Xn is greater or equal to 0. Then the sequence Sqrt(Xn) of positive roots converges an lim(sqrt(Xn)=sqrt(x)

It proves the case where x=0 using d-e, and i get that no problem.. i am having problems simplifying and showing the limit for the case where x is strictly greater than 0 using d-e. does that help? sorry for tripping you up with poor notation.. i am a rookie at trying to type it on computers!