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Help with formal definition of the limit of a function

  1. Mar 10, 2007 #1
    The problem is not to conduct the proof but how the proof works. (It seems to be a circular argument)



    Please use laymans english to explain, thank you
     
  2. jcsd
  3. Mar 10, 2007 #2

    mjsd

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    what are you asking?
     
  4. Mar 10, 2007 #3
    how the proof works using the formal definition of a limit
     
  5. Mar 10, 2007 #4

    mjsd

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    with epsilon and delta?
     
  6. Mar 10, 2007 #5
    yep thats correct
     
  7. Mar 10, 2007 #6

    Gib Z

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    Formal Definition, so yup. Never really liked epsilon delta proofs myself, but anyway..

    The proof works by showing that in the close neighborhood of the value that the limit is approaching, all the values in the neighborhood as becoming the same.
     
  8. Mar 10, 2007 #7

    mjsd

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    ok, firstly, you ask yourself how may you otherwise define a "limit"? if you think about it you will see that the formal definition of a limit make perfect sense... it is indeed what we meant by a "limit" that is: we want to see what happen to f(x) as x tends to a certain number x_0.
    suppose we want to show
    [tex]\lim_{x\rightarrow x_0} f(x) = L[/tex],
    then we would like to know whether f(x) indeed gets close to L as x gets close to x_0. And that is exactly what the definition is doing.

    now in proofs, you often wanna find a relationship between [tex]\delta[/tex] and [tex]\epsilon[/tex]; and often what that does is to simply ensure that we can ALWAYS find the appropriate values for them so that the the inequalities involving [tex]\delta[/tex] and [tex]\epsilon[/tex] can be satisfied.
     
  9. Mar 10, 2007 #8
    thank you very much
     
  10. Mar 10, 2007 #9

    HallsofIvy

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    When you say "It seems to be a circular argument" you may be thinking of this:

    To prove that [itex]lim_{x\rightarrow 3} 2x+ 1= 7[/itex] a typical argument goes "If [itex]|f(x)-L|= |2x+1- 7|< \epsilon[/itex] then [itex]\|2x- 6|= 2|x-3|< \epsilon[/itex] so it suffices to take [itex]|x-3|< \delta= \epsilon/2[/itex]".

    That seems "circular" because we start from [itex]|f(x)-L|< \epsilon[/itex] which, by the definition of limit, is what we want to show. You are correct that that is not a formal proof- it is more like analyzing the problem to deciding HOW to write a proof- It is deciding how we should choose [itex]\delta[/itex] in order to get the result we want. A "true" proof would go the other way:
    "Given [itex]\epsilon[/itex], take [itex]\delta= \epsilon/2[/itex] (which looks like the professor is picking it out of the air since the student didn't see the analysis above). Then if [itex]0< |x-3|< \delta= \epsilon/2[/itex], we have [itex]0< 2|x-3|= |2x- 6|= |2x+ 1- 7|< \epsilon[/itex]".

    Typically, having done the "analysis" we don't need to write out the "true" proof because it is clear that every step in going from [itex]\epsilon[/itex] to [itex]\delta[/itex] is reversible- it's obvious that we can go from [itex]\delta[/itex] to [itex]\delta[/itex]. A "proof" where you go from the conclusion to the hypothesis by reversible steps, so that it is obvious you could go form hypothesis to conclusion, is sometimes called "synthetic proof". It's used a lot, for example, in proving trig identities.
     
  11. Mar 10, 2007 #10

    mjsd

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    the perception of a "circular argument" appearing only because in those cases (probably cases that you have been exposed to so far) it DOES work and that limit exists and does go to the number we wanna proof. But i think all you need to see is an example of a limit that doesn't exist... and you will see that in this instance, you just can't find a delta and epsilon that work, no matter how hard you try..... hence, this "circular argument" does not appear here. It looked like so before because it worked.
     
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