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Limit of sequence equal to limit of function

  1. Apr 5, 2006 #1
    Suppose

    lim (when n -> oo) of sup {f(x) / x belongs to (0, 1/n) } = L_1, and


    lim (when e -> 0+) of sup {f(x) / x belongs to (0, e) } = L_2.


    (e is simply epsilon).

    It seems pretty obvious, but it is truth that L_1 = L_2 ?

    Thanks for your help.
     
  2. jcsd
  3. Apr 5, 2006 #2
    Maybe this theorem works.

    If:
    1. Lim g(e) = L ( when e -> a+ )
    2. The sequence {a_n} -> "a"+ (but there is not a_n = a),

    Then the sequence { g(a_n) } -> L (when n -> oo).

    So in the case I stated previously I have these facts:

    1. Let "g" be the function / g(e) = sup { f(x) / x belongs to (0,e) }
    2. Lim g(e) = L ( when e -> 0+ ).
    3. The sequence {1/n} -> 0+

    So by the theorem I got { g(1/n) } -> L (when n -> oo), or in other words
    sup { f(x) / x belongs to (0,1/n) } -> L (when n -> oo).

    Now I would thank if someone can bless this.
     
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