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Prove a limit doesn't exist

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data does not exist.

    Prove that the limit as x approaches 0 of 1/(x2+x3)
    2. Relevant equations



    3. The attempt at a solution
    I know that I have to prove that the absolute value of 1/(x2+x3) - L is greater than or equal to epsilon for some delta. What do I do next?
     
  2. jcsd
  3. Oct 2, 2009 #2

    CompuChip

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    The definition of limit says:
    [tex]\exists L, \forall \epsilon > 0, \exists \delta > 0, \forall x \text{ s.t.} |x| < \delta: |f(x) - L| < \epsilon[/tex]
    where L is the supposed limit.

    What is the negation of this?
     
  4. Oct 2, 2009 #3
    The negation would be that the absolute value of f(x)-L is greater than or equal to epsilon. But how do I prove that there exists a delta for which that is true?
     
  5. Oct 2, 2009 #4
    Why don't you just show that for all positive real numbers M, |1/(x^2 + x^3)| > M if x is taken sufficiently small?
     
  6. Oct 2, 2009 #5

    CompuChip

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    Yes, but you should take care with the quantifiers: the negation of
    [tex]
    \exists L, \forall \epsilon > 0, \exists \delta > 0
    [/tex]
    is
    [tex]
    \forall L, \exists \epsilon > 0, \forall \delta > 0
    [/tex]

    I was stressing this because I think it is important that you do not fall into such logical traps.

    Of course, if you just want to solve the question, follow JG's advice :)
     
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