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Homework Help: Proof of lim(1/x) x->0 by negating epsilon delta definition of limit

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data
    I want to show that [itex]\lim_{x \rightarrow 0}\frac{1}{x}[/itex] does not exist by negating epsilon-delta definition of limit.

    2. Relevant equations



    3. The attempt at a solution
    We say limit exists when:
    [itex]\forall \epsilon > 0, \exists \delta > 0 : \forall x(0< \left| x\right| < \delta \Rightarrow \left| \frac{1}{x} - L \right| < \epsilon) [/itex]
    And limit does not exist corresponds to:
    [itex]\exists \epsilon > 0, \forall \delta > 0 : \exists x(0< \left| x\right| < \delta \wedge \left| \frac{1}{x} - L \right| > \epsilon) [/itex] (for any L ?)

    We look for an \epsilon such that

    [itex]\left| \frac{1}{x} - L \right| > \epsilon [/itex]

    then using triangle inequality,

    [itex]\left| \frac{1}{x}\right| + \left| L \right| ≥ \left| \frac{1}{x} - L \right| > \epsilon [/itex]

    therefore

    [itex]\left| \frac{1}{x}\right| + \left| L \right| > \epsilon [/itex]

    and

    [itex]| \frac{1}{x}| > \epsilon - \left| L \right| [/itex]

    for [itex] \epsilon > |L|[/itex], ( the other possibility gives no info since |x| > 0 already)

    [itex]|x| < \frac{1}{\epsilon - |L|}[/itex]

    we also had

    [itex]0<|x|<\delta[/itex]

    hence,

    [itex]|x| < min( \delta , \frac{1}{\epsilon - |L|} )[/itex]

    we found such x satisfies the above condition, for some epsilon and L.



    Is that approach correct?? Especially, I am not really sure about the negation of the definition...
     
    Last edited: Oct 6, 2012
  2. jcsd
  3. Oct 6, 2012 #2

    Zondrina

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    Homework Helper

    Ah now you have to make some alterations to the definition for it to work properly. You have a limit involving an infinite quantity which requires a change in your definition to be :

    [itex]\forall ε>0, \exists δ>0 | 0<|x-a|<δ \Rightarrow f(x)>ε[/itex]

    So for f(x) = 1/x, you desire a δ which ensures that f will be bigger than epsilon no matter what. You can start finding this delta by massaging the expression f(x) > ε to suit your needs.
     
  4. Oct 6, 2012 #3
    alright,

    for limits not dealing with infinity, is my negation correct??
     
  5. Oct 6, 2012 #4

    Zondrina

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    Homework Helper

    I'm not sure why you would want to negate the statement because |f(x)-L| which translates into |f(x)-∞| is for lack of better term, a garbage statement.

    That's why I suggested using the modified form of the definition so it at least makes sense.

    To answer your question though, I don't see a problem with your negation.
     
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