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Homework Help: Convergence proof

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data
    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.



    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Sep 28, 2010 #2

    Office_Shredder

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    You should probably start with the definition of convergence
     
  4. Sep 28, 2010 #3
    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].
     
  5. Sep 28, 2010 #4
    [tex] a^{2} \leq b^{2}[/tex] iff [tex] a \leq b[/tex]

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

    Can you use this ?
     
  6. Sep 28, 2010 #5
    Then the sequence is less than 0 and thus converges to 0
     
  7. Sep 28, 2010 #6
    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].
     
  8. Sep 28, 2010 #7
    so the sequence is greater than 0 because x is greater than 0.
     
  9. Sep 28, 2010 #8
    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].
     
  10. Sep 28, 2010 #9
    If g(x) --> A when x --> a then p(g(x)) --> p(A) when x --> a.
     
  11. Sep 28, 2010 #10
    Hmm... what if p was the square root function and A was negative. ?
     
  12. Sep 28, 2010 #11
    Well A needs to be in the domain of p for it to make sense. Guess I should have written that..
     
  13. Sep 28, 2010 #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.
     
  14. Sep 28, 2010 #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
     
  15. Sep 28, 2010 #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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