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Homework Help: Is this an accurate statement of the limit definition? (quantifiers/connectives)

  1. Sep 1, 2011 #1
    [itex]\forall\epsilon\exists\delta[( (\epsilon\wedge\delta) > 0) \wedge ((| x - t| < \delta)\Rightarrow (|f(x) - f(t)|< \epsilon))][/itex]

    The part that seems wrong is the placement of the statement "delta is greater than zero" and "epsilon is greater than zero". It seems like these statements may need to be placed separately after their respective quantifiers for this to be an accurate statement...
  2. jcsd
  3. Sep 1, 2011 #2
    Well, first of all

    [tex](\epsilon\wedge \delta)>0[/tex]

    makes no sense. [itex]\wedge[/itex] is an operator that takes in truth values and returns a truth value. For example "true [itex]\wedge[/itex] true is true". But [itex]\epsilon[/itex] and [itex]\delta[/itex] are not truth values.

    Second point: you never stated what x or t was. I expect another quantifier [itex]\forall t[/itex] in there.
  4. Sep 1, 2011 #3
    Okay. Thanks.

    So if I break ([itex]\epsilon\wedge\delta[/itex] > 0) up into ([itex]\delta > 0) \wedge (\epsilon > 0[/itex]) then THAT part will be okay left inside of the two quantifiers as it is above?

    edit: thanks for pointing that out...it seems so obvious now -.-
  5. Sep 1, 2011 #4
    It will be syntactically ok, but it won't have the meaning you want it to have.
    With your changes, the statement will be always true. Indeed, you can choose [itex]\delta=-1[/itex] each time. Then

    [tex](\epsilon>0)\wedge (\delta >0)\wedge (|x-t|<\delta)[/tex]

    will be false. So the implication will be true. So your statement will always be true!! This is clearly not what you want.
  6. Sep 1, 2011 #5

    thank you
  7. Sep 1, 2011 #6
    If I changed it to [itex]\forall\epsilon[\epsilon > 0]\exists\delta[\delta > 0][/itex]

    Would that be logically correct?

    edit: by logically correct, I really mean to ask "would that logically read: for all epsilon greater than zero, there exists delta greater than zero". It seems to me that it would logically mean that, but I don't want to assume. For the time being I am past trying to get the logical structure of the statement...taking a step back for a minute =0
  8. Sep 1, 2011 #7

    You're on the right track, but it is not syntactically correct what you're stating.

    Try to do something as follows:

    [tex]\forall \epsilon ((\epsilon >0)~\Rightarrow~\exists \delta ((\delta>0)\wedge (......[/tex]

    This is essentially what you mean. But this is syntactically correct.
  9. Sep 1, 2011 #8
    That is very close to what I just did while waiting for you to reply. Combined with what you just said it is:

    [itex]\forall\epsilon[(\epsilon > 0) \Rightarrow \exists\delta((\delta>0) \wedge ((|x-t| < \delta) \Rightarrow (|f(x)-f(t)|<\epsilon))][/itex]

    This feels much better on an intuitive level, but (aside from asking you "IS THIS RIGHT?????") how can I test it to be sure? Is a truth table the only way?

    Also, I am certian that it will change the meaning, but if i combined the "delta greater than zero" and the "|x-t| < delta" into [itex]((0<|x-t|<\delta) \Rightarrow (|f(x)-f(t)|<\epsilon))[/itex], would it still be correct (assuming that I have it right above)?
  10. Sep 1, 2011 #9
    This seems correct. However, you still didn't introduce t (or x) in here. You're missing a quantifier.
    As for checking that it is correct. Well, perhaps a truth table should work...

    No, you need to state separately that [itex]\delta>0[/itex]. Otherwise, you can choose [itex]\delta=-1[/itex] and you'll see that it's always true.
  11. Sep 1, 2011 #10
    Okay. Thanks for your help micro. I should be able to figure the rest out.

    Thanks again

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