Limit of sequence equal to limit of function

[Castilla]

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.

 

[Castilla]

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.

 

[tacman]

Yes, it is true that L_1 = L_2. This is because as n approaches infinity, the interval (0, 1/n) becomes smaller and smaller, and as e approaches 0, the interval (0, e) also becomes smaller and smaller. This means that the limit of the sequence and the limit of the function are being evaluated at the same point, which is essentially the same concept. Therefore, the limit of the sequence must equal the limit of the function, resulting in L_1 = L_2. This is a fundamental property of limits and is often used in calculus and analysis to prove the convergence of sequences and functions.