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Homework Help: Question of proof of analysis

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that f(x)>=0 in some deleted neighborhood of c, and that lim x->a f(x)=R. Prove that lim x->a sqrt{f(x)}=sqrt{R} under the assumption that R>0.

    2. Relevant equations

    if 0<|x-c|<delta, then |f(x)-L|<epsilon.

    3. The attempt at a solution

    I don't know how to start with this.

    I tried to work on lsqrt{f(x)}-sqrt(L)l=lsqrt{f(x)}-sqrt(L)llsqrt{f(x)}+sqrt(L)l/lsqrt{f(x)}+sqrt(L)l

    But I don't know how to go from here, I'm not sure if it's a correct start as well.

    Any help would be appreciated. Thanks!
  2. jcsd
  3. Sep 30, 2011 #2
    No, you calculated wrong. What is

    [tex]\frac{|\sqrt{f(x)}-\sqrt{L}||\sqrt{f(x)}+\sqrt{L}|}{ |\sqrt{f(x)}+\sqrt{L}|}[/tex]
  4. Sep 30, 2011 #3
    I actually don't know what it is...I just don't know how to proceed. Could you help me how I get to start with this question?
  5. Sep 30, 2011 #4


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    You have a deleted neighborhood of c, so isn't this lim x ➙ c , not lim x ➙ a ?
  6. Sep 30, 2011 #5
    Yes, you're right.
  7. Sep 30, 2011 #6
    Since f(x) > 0, R>0, Hence 0 < |√(f(x)) - √(R)| < |√(f(x)) + √(R)|.

    for ε^2 given there exists δ>0 such that when |x-c| <δ, |f(x) - R| < ε^2.


    |√(f(x)) - √(R)|^2 < |√(f(x)) - √(R)||√(f(x)) + √(R)| = |f(x) - R| < ε^2,

    taking square roots give the desired result.
  8. Sep 30, 2011 #7
    Thank you for your help. But, sorry, how can I conclude this? Why can I say that lim x->a sqrt{f(x)}=sqrt{R}?
  9. Sep 30, 2011 #8


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    (a - b)(a + b) = a2 - b2 .

    What is [itex]|\sqrt{f(x)}-\sqrt{L}||\sqrt{f(x)}+\sqrt{L}|\,?[/itex]

    Now can you answer micromass's question?
  10. Oct 1, 2011 #9

    Sorry, which part you're asking? Is it for what Trevor Vadas said?

    I'm now just wondering how I can conclude this problem from |√(f(x)) - √(R)|^2 < |√(f(x)) - √(R)||√(f(x)) + √(R)| = |f(x) - R| < ε^2.
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