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Calculus, Definition of limit, Concept

  1. Jul 12, 2012 #1
    Hi comrades.

    According to spivak, the defition of limit goes as follows:

    " For every ε > 0, there is some δ > 0, such that, for every x, if 0 < |x-a| < δ,
    then |f(x) - l |< ε. "

    After some exercices, I came across with a doubt.

    Say that I could prove that | f(x) - l |< 5ε, for some δ[itex]_{1}[/itex] such that 0 < |x-a| < δ[itex]_{1}[/itex].
    Since ε > 0, and thus 5ε > 0, could I say that lim[itex]_{x→a}[/itex]f(x) = l based on this proof?

    Regards,
     
  2. jcsd
  3. Jul 12, 2012 #2

    HallsofIvy

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    Yes, if [itex]|f(x)- L|< 5\epsilon[/itex], for any [itex]\epsilon> 0[/itex] then, taking [itex]\epsilon_1= \epsilon/5[/itex] where [itex]\epsilon[/itex] is any given number, we have [itex]|f(x)- L|< 5\epsilon_1= 5(\epsilon/5)= \epsilon[/itex]
     
  4. Jul 12, 2012 #3
    Thanks!
     
  5. Jul 13, 2012 #4
    Well, two days later and I have another question.

    I tougth in creating another post, but since the question is similar and fairly simple, I will jsut continue on this post.

    When introducing Integrals, Spivak reaches the following during a proof:

    inf{U(f, P`)} - sup{L(f, P´)} < ε , for any ε > 0.

    Since this is true for all ε > 0, it follows that sup{L(f, P´)}= inf{U(f, P`)}∴

    First Comment : Althoug Spivak doesn't say it, we shoud keep in mind that inf{U(f, P`)}[itex]\geq[/itex] sup{L(f, P´)}, otherwise one would have to use the absolute value, right?

    Second comment, during the study of limits, we used |f(x) - L | < ε, for any ε > 0. So according with this part, for a limit to exist that should be a δ > 0, such that | f(x) - L | = 0. But this doesn't seem right.

    Could you shed some light in this?

    regards,
     
  6. Jul 13, 2012 #5

    HallsofIvy

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    Yes, that's the important case but Spivak doesn't have to say it because if it is negative, it is trivially less than any positive number.

    No, that doesn't follow. The set of all positive numbers does not contain 0- no number in it is equal to 0 but the inf (greatest lower bound) is 0.

     
  7. Jul 13, 2012 #6
    I am not sure If I follow.

    The set of all positive numbers doesn't coutain zero, however
    | f(x) - L| = 0, would respect | f(x) - L| < ε for any ε > 0. right?

    So, altough I am alredy a bit confused, I still don't understand why he can say that if

    inf{U(f, P`)} - sup{L(f, P´)} < ε , for any ε > 0, then it follows that sup{L(f, P´)}= inf{U(f, P`)}∴

    But we cannot say that if | f(x) - L| < ε for any ε > 0, then |f(x) - L| = 0.

    thanks,
     
  8. Jul 14, 2012 #7
    Let me rearrange my line of thought, so I can make my doubt clearer in order for you to help me out.

    So, I actually understand that :we cannot say that if | f(x) - L| < ε for any ε > 0, then |f(x) - L| = 0.

    I guess this may be explained to the fact that for any ε > 0, the is a natural number n with [itex]\frac{1}{n}[/itex] < ε. Is this a suitable explanation?

    Anyway, to my main question:

    So, from | f(x) - L| < ε for any ε > 0, it doesn't follow that |f(x) - L| = 0, why can Spivak assure that if inf{U(f, P`)} - sup{L(f, P´)} < ε , for any ε > 0, then it follows that sup{L(f, P´)}= inf{U(f, P`)}∴

    I hope you have understood,

    Regards,
     
  9. Jul 14, 2012 #8

    HallsofIvy

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    If it were true that [itex]|f(x)- L|< \epsilon[/itex] for every [itex]\epsilon> 0[/itex] then, yes, we would have to have f(x)= L. But that is NOT true. We are only saying that, given some [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] so that if [itex]|x- a|< \delta[/itex], then [itex]|f(x)- L|<\epsilon[/itex] for that particular [itex]\epsilon[/itex], not for all [itex]epsilon[/itex].

    While the general statement "inf{U(f, P)}-sup{L(f,P)}<[itex]\epsilon[/itex]" is true for all [itex]\epsilon> 0[/itex]", the particular P depends upon the particular [itex]\epsilon[/itex].
     
    Last edited by a moderator: Jul 14, 2012
  10. Jul 14, 2012 #9

    arildno

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    In my student days, I made the revolutionary, and highly applicable definition of the hypercontinuous function:
    A function f(x) is hypercontinuous, if, for EVERY epsilon>0 and EVERY delta>0, there exists an L so that |f(x)-L| is less than epsilon.
     
    Last edited by a moderator: Jul 14, 2012
  11. Jul 14, 2012 #10
    Limits, sure are a dificult concept to grasp completly.

    Althoug, you seem quite confident about explanation, I don't undersant it. I have tought about it most of my day.

    The defition of lim:

    "for every ε > 0, there is some δ > 0 , such that, for all x, if 0 < | x - a |< δ, then | f(x) - L | < ε".

    So, there should be and δ[itex]_{1}[/itex] > 0, for wich | f(x) - L | < 10[itex]^{-20}[/itex], given that |x -a| < δ[itex]_{1}[/itex], and there should be another δ[itex]_{2}[/itex], for wich | f(x) - L | < 10[itex]^{-1000}[/itex], and in the end, since it is valid for any ε > 0, shouln't it exist an δ[itex]_{3}[/itex] for wich | f(x) - L | = 0, given that |x -a| < δ[itex]_{3}[/itex] ?
    Because, this is the line of though we use to explain:

    → if inf{U(f, P`)} - sup{L(f, P´)} < ε , for any ε > 0, then it follows that sup{L(f, P´)}=inf{U(f, P`)}∴?, or is it not this tipe of reasoning?


    Thank you for your most valued explanations.
     
  12. Jul 15, 2012 #11
    I have actually tried to prove this by myself. Not even knowing if such thing of aplicable to the case.
    Here it goes.

    Let A = { ε : ε > 0}; Then 0 would be the greatest lower bound for the set A.
    But, inf{U(f, P`)} - sup{L(f, P´)} < ε, for any ε. Thus meaning inf{U(f, P`)} - sup{L(f, P´)} is a lower bound for the set A. Consequenly, inf{U(f, P`)} - sup{L(f, P´)}≤ 0, since 0 is the greatest lower bound. Hence sup{L(f, P´)} = inf{U(f, P`)}, because on the other hand inf{U(f, P`)} - sup{L(f, P´)} ≥ 0. ∴

    By how much did I miss the target?

    Regards,
     
  13. Jul 16, 2012 #12
    That actually seems right to me.

    This is actually the line of reasoning used to explain that sup{L(f, P´)}=inf{U(f, P`)}

    Since for any partition P, U(f,P) is greater than or equal to L(f,P') for any partition P', it follows that U(f,P) is an upper bound on {L(f,P)} From this, it follows that inf{U(f,P)} is greater than or equal to sup{L(f,P)} (Otherwise, we would be able to produce some P such that U(f,P) were not an upperbound for {L(f,P)}) Thus, inf{U(f,P)} - sup{L(f,P)} ≥ 0.

    Now to the part I think you're confused about. inf{U(f,P)} - sup{L(f,P)} < ε for all ε > 0 because of the following.

    Given any ε > 0, there exist two partitions P and P' such that U(f,P) - L(f,P') < ε. Since U(f,P) ≥ inf{U(f,P)}, and L(f,P) ≤ sup{L(f,P)}, we have that

    ε > U(f,P) - L(f,P') ≥ inf{U(f,P)} - L(f,P') ≥ inf{U(f,P)} - sup{L(f,P)} ≥ 0

    Since we can do this for all ε > 0, we have that

    ε > inf{U(f,P)} - sup{L(f,P)} ≥ 0 for all ε > 0, so that inf{U(f,P)} = sup{L(f,P)} holds.

    The difference between saying that Lim x-> a of f(x) is L and |a - b| < ε for all ε > 0 is that, in the first case, we are only guaranteed the existence of an interval centered around a so that |f(x) - L| < ε holds for all values in this interval, and it is crucial to note that the interval may or may not depend on ε. While in saying that |a - b| < ε for all ε > 0, we are saying that two constants are arbitrarily close to each other.

    I apologize for my lack of LaTex, but I honestly have no clue how to use it. I hope I helped!
     
    Last edited: Jul 16, 2012
  14. Jul 22, 2012 #13
    Yes, you have helped me. It is much clearer know.

    Thanks,
     
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