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Prove A Left Hand Limit

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data

    Proove the statement using the [tex]\epsilon[/tex] and [tex]\delta[/tex] definition of a limit.

    The limit of x as it approaches 9 from the left side of the x axis is [tex]\sqrt[4]{9-x}[/tex]=0

    2. Relevant equations

    The limit of x as it approaches a from the left side of the x axis is f(x) = L

    if for every number [tex]\epsilon[/tex]>0 there is a number [tex]\delta[/tex]>0 such that if

    a-[tex]\delta[/tex]<x<a then |f(x)-L|<[tex]\epsilon[/tex]

    3. The attempt at a solution

    9-[tex]\delta[/tex]<x<9 and |[tex]\sqrt[4]{9-x}[/tex]-0|<[tex]\epsilon[/tex]

    Using algebra, I've worked my way to |9-x|<[tex]\epsilon^{4}[/tex]. I'm not sure where to go with it now. Do I subtract the 9 over and then change the sign of x, ending up with |x|>-[tex]\epsilon[/tex][tex]^{4}[/tex]+9 ? That doesn't seem right to me.
  2. jcsd
  3. Aug 31, 2010 #2
    You can pick delta=episilon^4 and you are done.
  4. Aug 31, 2010 #3
    That's great! However, I don't know why it's OK to stop there. Would anyone be kind enough to explain?

    More specifically, why can I stop at |9-x| and how do I work my way back to verify that [tex]\epsilon[/tex][tex]^{4}[/tex] can be substituted for [tex]\delta[/tex]?
  5. Aug 31, 2010 #4
    You can stop there because this is the precise definitionof a limit. We say limit as x approaches a f(x)=L if
    When|f(x)-L|< episilon then |x-a|< delta.

    If your case a=7 and delta = episilon^4.

    Too lazy to use latex of my phone, sorry!
  6. Aug 31, 2010 #5
    I believe it's because if we assume that d = E^4, then we have 9 - E^4 < x < 9. This means that 9 - E^4 - x < 0 < 9 - x. Then 9 - x < E^4 < 9 - x + E^4.

    Then you have the algebra: since |(9 - x)^(1/4)| < E, then |9 - x| < E^4 in order for the limit to be true. We've found a delta such that this is true, so our proof is complete.
  7. Aug 31, 2010 #6
    Thank you both so much for explaining it for me! I understand why it works now. There were no good examples in the textbook, so I can use this as a guide for similar problems.
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