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Homework Help: I cant understand this prove explanation on limit series

  1. Jan 21, 2009 #1
    there is a continues function f(x) and bounded on (x_0,+infinity)
    proove that for every T there is a sequence
    X_n=+infinity
    so
    lim [f(x_n +T) - f(x_n)]=0
    n->+infinity

    i was told:
    uppose that [itex]\lim_{x\to\infty}f(x)=a[/itex]. So we know that given any [itex]\varepsilon>0[/itex] there exists a [itex]\eta>0[/itex] such that [itex]\eta< x\implies |f(x)-a|<\varepsilon~(1)[/itex], and since [itex]x_n\to \infty[/itex] we may find a [itex]\delta>0[/itex] such that [itex]\delta<n\implies \eta<x_n~(2)[/itex]. Now suppose that [itex]T>0[/itex] (the proof for the other cases is analgous), then choose [itex]\delta[/itex] such that [itex](2)\implies (1)[/itex] then [itex]\left[f\left(x+T\right)-f(x)\right|\leqslant \left|f\left(x+T\right)-a\right|+|f(x)-a|<2\varepsilon[/itex], this implies the result.


    but whrn i read this article on limit proves
    http://www.mathhelpforum.com/math-h...e-never-learnt-well-epsilon-delta-proofs.html

    i learned from the delta proofes article that when you define the delta

    [itex]
    \delta>0
    [/itex]
    it needs to come with
    [itex]
    |x-x_3|<\delta
    [/itex]
    in our case x_3 goes to infinity
    so the inqueality that i presented not logical

    but on the other hand
    it how its done on the article limit proove

    ??
     
  2. jcsd
  3. Jan 22, 2009 #2

    Mark44

    Staff: Mentor

    Limits in which x approaches infinity are proved differently than those for which x approaches some finite number a.

    [tex]\lim_{x \rightarrow \infty} f(x) = L[/tex]

    means that for any [itex]\epsilon > 0[/itex] there exists a number M > 0, such that for any x > M, then |f(x) - L| < [itex]\epsilon[/itex]

    If I want to prove the limit above to you, you tell me how close f(x) has to be to L (you give me [itex]\epsilon[/itex]), and I tell you a number M.

    If you're satisfied, we shake hands and go about our business.

    If you're not satisfied, you tell me another [itex]\epsilon[/itex] that's even smaller, and I have to find another M (even larger).

    And so on, until you're convinced that I can force f(x) as close to L as you like, by specifiying how big x has to be.

    Got it?
     
  4. Jan 22, 2009 #3
    in your explanation M is delta>0
    ??
     
  5. Jan 22, 2009 #4

    Mark44

    Staff: Mentor

    No. M is generally a pretty large number, while [itex]\delta[/itex] is usually very small. A big difference is the definition I showed doesn't try to get x within [itex]\delta[/itex] of infinity.
     
  6. Jan 22, 2009 #5
    but he does use delta

    for what purpose
    ??
     
  7. Jan 22, 2009 #6

    Mark44

    Staff: Mentor

    If the limit is as x approaches a finite number a, you use [itex]\delta[/itex], since you want to make x very close to a. I.e., you want to make |x - a| < [itex]\delta[/itex].
    If the limit is as x approaches infinity, that's a different matter, since as I explained earlier, you can't get x within [itex]\delta[/itex] of infinity. You can, however, make x larger than some (presumably large) number M.

    I don't think I can make it any clearer than that.
     
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