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Proving a Limit.

  1. Feb 17, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]f(x)=\frac{\sqrt{x+2}-\sqrt{2}}{\sqrt{x}}[/tex]


    2. Relevant equations

    prove that: [tex]\lim_{x\rightarrow0}f(x)=0[/tex]


    3. The attempt at a solution

    Alright i let ε>0 and i have to find α>0 (which depends on ε) so that if:

    0<l x-a l< α, then l f(x)-L l<ε

    Any help guys thank you before hand.
     
  2. jcsd
  3. Feb 17, 2012 #2
    differentiate both numerator and denominator
     
    Last edited: Feb 17, 2012
  4. Feb 17, 2012 #3
    I haven't learned differentials yet, is there any other way to do it?
     
  5. Feb 17, 2012 #4
    If you let [itex]u=\sqrt{x+2}-\sqrt{2}[/itex], you'll find
    [itex]f=\frac{1}{\sqrt{1+2\sqrt{2}/u}}[/itex]
    the rest is obvious
     
  6. Feb 19, 2012 #5
    Yes but i have to prove that its true using the limit rule:

    For every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − p | < δ implies | f(x) − L | < ε.
     
  7. Feb 19, 2012 #6
    I multiplied the numerator and the denominator by the conjugate of the numerator and at the end i got:

    [tex]f=\frac{\sqrt{x}}{\sqrt{x-2}+\sqrt{2}}[/tex]

    and i substitute with the zero and i get that the limit as x approaches 0 of f(x) is 0 is that correct?
     
  8. Feb 19, 2012 #7

    vela

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    Your idea is correct, but the numerator should be just x, not the square root of x.
     
  9. Feb 19, 2012 #8
    alright is that all that needs to be done?
     
  10. Feb 19, 2012 #9

    Mark44

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    Your denominator is incorrect. If you take the limit as x approaches 0, the first radical in the denominator is undefined. Check your work.
     
  11. Feb 20, 2012 #10
    sorry i wanted to put √(x+2)
     
  12. Feb 20, 2012 #11

    SammyS

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    With the correction given by Mark44, the result is:
    [itex]\displaystyle f(x)=\frac{\sqrt{x}}{\sqrt{x+2}+\sqrt{2}}\ .[/itex]​
    And indeed, the numerator is √(x) as you have.

    I gather from your Original Post, that you need to come up with a δ, ε -proof.

    So, given an arbitrary ε > 0, you need to show that there is a δ ( which usually depends upon ε ) such that for any x, in the domain of f, that satisfies 0 < |x - 0| < δ, it's true that |f(x) - L|< ε. In this case L = 0.
     
  13. Feb 20, 2012 #12

    vela

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    Oops, forgot about the ##\sqrt{x}## on the bottom. Just ignore me. :redface:
     
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