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Math Help - possible theorem in limits??

  1. #1
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    possible theorem in limits??

    Is the following a possible theorem in limits??


    If a is not an accumulation point of the domain of .......  f:A\subseteq R\rightarrow R then f has a limit over a
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by archidi View Post
    Is the following a possible theorem in limits??


    If a is not an accumulation point of the domain of .......  f:A\subseteq R\rightarrow R then f has a limit over a
    Note: You may want to wait for a more senior member to validate this answer.


    My interpretation of your question

    "If f:A\longmapsto \mathbb{R} with A\subseteq\mathbb{R}, and if a is not a limit point of A then does \lim_{x\to a}f(x) make sense?"

    Answer to that question: No. For \lim_{x\to a}f(x) to exist there must be a point q\in\mathbb{R} such that for every \varepsilon>0 there exists a \delta>0 such that d_{A}(x,a)<\delta\implies d_{\mathbb{R}}(f(x),q)=|f(x)-q|<\varepsilon\quad x\in A. But now suppose that a is not a limit point of A then we can pick an open neighborhood of a lets call it N_{\delta_1}(a) such that there does not exist an element of A in that neighborhood. Now choose an value of \varepsilon that requires \delta<\delta_1 and we run into a contradiction.


    EDIT: I am sorry, I completely for some reason disregarded the A\subseteq\mathbb{R} part. This can be made slightly simpler

    By the defintion of limits in \mathbb{R}\mapsto\mathbb{R} for \lim_{x\to a}f(x) to exist there must exist a point q\in\mathbb{R} such that for every \varepsilon>0 there exsits a \delta>0 such that |x-a|<\delta\implies|f(x)-q|<\varepsilon~~x\in A. But now suppose that a is not a limit point of A then there exists a neighborhood (x-\delta_1,x+\delta_1) such that there does not exist an element of A within that neighborhood. Now if we choose an \varepsilon that requires \delta<\delta_1 we run into a contradiction.
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Note: You may want to wait for a more senior member to validate this answer.


    My interpretation of your question

    "If f:A\longmapsto \mathbb{R} with A\subseteq\mathbb{R}, and if a is not a limit point of A then does \lim_{x\to a}f(x) make sense?"

    Answer to that question: No. For \lim_{x\to a}f(x) to exist there must be a point q\in\mathbb{R} such that for every \varepsilon>0 there exists a \delta>0 such that d_{A}(x,a)<\delta\implies d_{\mathbb{R}}(f(x),q)=|f(x)-q|<\varepsilon\quad x\in A. But now suppose that a is not a limit point of A then we can pick an open neighborhood of a lets call it N_{\delta_1}(a) such that there does not exist an element of A in that neighborhood. Now choose an value of \varepsilon that requires \delta<\delta_1 and we run into a contradiction.


    EDIT: I am sorry, I completely for some reason disregarded the A\subseteq\mathbb{R} part. This can be made slightly simpler

    By the defintion of limits in \mathbb{R}\mapsto\mathbb{R} for \lim_{x\to a}f(x) to exist there must exist a point q\in\mathbb{R} such that for every \varepsilon>0 there exsits a \delta>0 such that |x-a|<\delta\implies|f(x)-q|<\varepsilon~~x\in A. But now suppose that a is not a limit point of A then there exists a neighborhood (x-\delta_1,x+\delta_1) such that there does not exist an element of A within that neighborhood. Now if we choose an \varepsilon that requires \delta<\delta_1 we run into a contradiction.
    NO NO the theorem is:

    if a is not accumulation of the domain of a function:  f:A\subseteq R------->R then ,THERE exists m such that

    ....................... \forall\epsilon[\epsilon>0\rightarrow\exists r(r>0\wedge\forall x( x\epsilon A\wedge 0<|x-a|<r\rightarrow |f(x)-m|<\epsilon))] .

    .......................or in words : given ε>0 there exists r>0 such that......


    .....................if xεA and 0<|x-a|<r then |f(x)-m|<ε...........................
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