Convergence Proof: Showing (\sqrt{x_{n}})\rightarrow0

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Homework Help Overview

The original poster attempts to prove that if a sequence \( (x_n) \) converges to 0, then the sequence \( (\sqrt{x_n}) \) also converges to 0. The problem is situated within the context of real analysis and convergence of sequences.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of convergence and its implications. There are attempts to relate properties of sequences, such as inequalities and the behavior of functions like the square root. Some participants question the assumptions made about the sequences and their values.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have offered guidance on starting points, while others are clarifying definitions and questioning the validity of certain statements made. There is no explicit consensus yet on the best approach to the proof.

Contextual Notes

There are mentions of different limit theorems and their applicability, with some participants noting that the theorem in question may not be standard across all analysis curricula. Additionally, the constraints of the problem include the non-negativity of the sequence \( x_n \).

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Homework Statement


Let [tex]x_{n\geq}[/tex]0 for all n in the natural numbers.
If ([tex]x_{n}[/tex])[tex]\rightarrow[/tex]0, show that ([tex]\sqrt{x_{n}}[/tex])[tex]\rightarrow[/tex]0.



Homework Equations





The Attempt at a Solution


So far, I have started with [tex]\left|\sqrt{x_{n}}-0\right|[/tex]. Not sure if that's the right way to start.
 
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You should probably start with the definition of convergence
 
A sequence converges to a real number a if for every positive [tex]\epsilon[/tex], there exists an N element of the natural numbers such that whenever n[tex]\geq[/tex]N, it follows that [tex]\left|a_{n}-a\right|[/tex]<[tex]\epsilon[/tex].
 
[tex]a^{2} \leq b^{2}[/tex] iff [tex]a \leq b[/tex]

[tex]a,b \geq 0[/tex]

Can you use this ?
 
Then the sequence is less than 0 and thus converges to 0
 
How did you arrive at such a conclusion? Btw what you said not correct.

How is the sequence less than zero ? In your definition [tex]x_{n} \geq 0[/tex].
 
so the sequence is greater than 0 because x is greater than 0.
 
All I wanted you to do wanted you to do was take the square root of both sides of the inequality...
[tex]x_{n} < \epsilon[/tex].
 
If g(x) --> A when x --> a then p(g(x)) --> p(A) when x --> a.
 
  • #10
Hmm... what if p was the square root function and A was negative. ?
 
  • #11
Well A needs to be in the domain of p for it to make sense. Guess I should have written that..
 
  • #12
It's fine. Btw this theorem is not one of the 4 limit theorems given in most analysis books so I doubt the OP can use it. OP would need to prove it to use it.
 
  • #13
We had 5 limit theorems when I did analysis in first year at uni. The proof is like 3 lines and not harder than the rest so I think it is strange it isn't standard at other places
 
  • #14
I think you are referring to limits of functions not limit theorems. The limit theorems are for sequences and they are later generalized to functions.
I am taking analysis right now and the thoerem you mentioned is in the limit of functions section
 

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