- #1

- 241

- 0

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.

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- #1

- 241

- 0

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

- 241

- 0

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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