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I Epsilon-Delta definition property.

  1. Mar 22, 2016 #1
    he d2d5e6d4a64dd4c644db1fc0cae964e7.png definition of the limit of a function is as follows:[5]

    Let 0be7555a4a4967cb39a53624642583a4.png be a function defined on a subset [PLAIN]https://upload.wikimedia.org/math/a/1/b/a1b67abab803e714098f3e69a33900da.png, [Broken] let 4a8a08f09d37b73795649038408b5f33.png be a limit point of [PLAIN]https://upload.wikimedia.org/math/f/6/2/f623e75af30e62bbd73d6df5b50bb7b5.png, [Broken] and let d20caec3b48a1eef164cb4ca81ba2587.png be a real number. Then

    the function 8fa14cdd754f91cc6554c9e71929cce7.png has a limit d20caec3b48a1eef164cb4ca81ba2587.png at 4a8a08f09d37b73795649038408b5f33.png
    is defined to mean

    for all [PLAIN]https://upload.wikimedia.org/math/b/0/f/b0f19c5714fe9f9891ed26ff783cf639.png, [Broken] there exists a https://upload.wikimedia.org/math/1/c/b/1cb24dafc8035d2c720256620066ae73.png such that for all 9dd4e461268c8034f5c8564e155c67a6.png in f623e75af30e62bbd73d6df5b50bb7b5.png that satisfy [PLAIN]https://upload.wikimedia.org/math/6/e/e/6ee6dc9ee03f04da6ff784b6eca416e5.png, [Broken] the inequality https://upload.wikimedia.org/math/2/c/4/2c412a6f49514db4dbd890537413bcc6.png holds.
    .

    Taken from https://en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit

    The problem what Iam facing is the statement ," for all 9dd4e461268c8034f5c8564e155c67a6.png in f623e75af30e62bbd73d6df5b50bb7b5.png that satisfy [PLAIN]https://upload.wikimedia.org/math/6/e/e/6ee6dc9ee03f04da6ff784b6eca416e5.png". [Broken] Though I have understood this statement, this hinders my proving this property of limit

    lim f(c+h)=L =lim f(x).
    h->0 x->c

    I am trying to prove the LHS from the RHS.

    We know,

    "the function f has a limit L at 4a8a08f09d37b73795649038408b5f33.png
    is defined to mean
    for every ε>0,there exists a δ>0, sucht that, for all x in D satisfying, o<|x-a|<δ, implies |f(x)-L|<ε.


    The LHS can be proven from the trivial fact taking
    x-c=h.
    A question arises to me. How do we know the existance of such an h? Also, for what all h does this inequality o<|h|<δ implies |f(c+h)-L|<ε hold?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 22, 2016 #2

    HallsofIvy

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    You are taking h= x- c so "h goes to 0" is the same as "x goes to c" or [itex]|h|< \delta[/itex] is the same as [itex]|c+ h|< \delta[/itex]. In more detail, if [itex]|h|< \delta[/itex] then [itex]-\delta< h< \delta[/itex] so that [itex]-\delta< x- c< \delta[/itex] and vice-versa.
     
  4. Mar 22, 2016 #3
    How is this possible?
     
  5. Mar 22, 2016 #4

    Ssnow

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    The fact is that you change variable but the limit remains the same, by assumptions you have ##\lim_{x\rightarrow c} f(x)=L## if you change variable, you can set as example ##x=c+h##, so when ##x\rightarrow c## you have ##h\rightarrow 0## and the limit become:
    ##\lim_{h\rightarrow 0} f(c+h)=L## ... , such ##h## exists (because the difference ##x-c## always exists and you call it with the letter ##h##), therefore ##0<|h|<\delta ## and from ##\left|f(x)-L\right|<\varepsilon## you have ##\left|f(c+h)-L\right|<\varepsilon## ...

    It is only a way to rewrite the limit from the variable ##x## to the new variable ##h##, in fact ##c## is a number always fixed and I suppose finite ...
     
  6. Mar 22, 2016 #5
    @Ssnow
    You are right. Such an h exists like h=x-c.
    But what is the condition 'h' should satisfy? How do convert the statement "for all x satisfying 0<|x-c|<δ" to "for all h satisfying 0<|h|<δ"
    Other statements in the epsilon-delta definition make sense except this statement. Can we convert from "all x" to "all h"?
     
  7. Mar 22, 2016 #6

    Ssnow

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    Yes, now you point is ##0## so for all ##h## satisfying ##0<|h-0|<\delta## ...
     
  8. Mar 22, 2016 #7

    HallsofIvy

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    Oops! I meant to write [itex]|h|< \delta[/itex] is the same as [itex]|x- c|< \delta[/itex].
     
  9. Mar 25, 2016 #8
    Yes, logically it is right. But, i need conversion of statement. I thought of this proof. Please say whether it is right.
    All the numbers are real in this discussion.
    Because of exsistance property of h for any real numbers x and c, we can write:
    For all real x and all real c, there exists an unique(one and only one) h such that h=x-c.
    Let us take an x which satisfies 0<|x-c|< δ. Let it be x1. By definition, we have "for all x satisfying the property P". So, we can take an arbitray x =x1 satisying that property.
    For this x1 and c, we have an unique h=h1 such that h1=x1-c.
    By induction,we can prove that for a xn(satisfying P) and c, we have an unique hn such that hn=xn-c.
    x can be x1,x2....xn(which satisfying P)
    h can be h1,h2,....hn(which are unique for a given x).
    So, by induction we can say, "for every x, we can find, every h."
     
  10. Mar 25, 2016 #9

    Ssnow

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    yes, but I think the existance it is clear because ##x,c## exists so also the difference exists in ##\mathbb{R}##. To prove the unicity of ##h## you can procede by contradiction argument. Assume that there are two ##h_{1},h_{2}## such that ##h_{1}=x-c## and ##h_{2}=x-c##, then it is clear that ##h_{1}=h_{2}##, so is unique ...
     
  11. Mar 25, 2016 #10
    Good proof of uniqueness theorem....
     
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