Results 1 to 3 of 3

Thread: possible theorem in limits??

  1. #1
    Banned
    Joined
    Oct 2008
    Posts
    71

    possible theorem in limits??

    Is the following a possible theorem in limits??


    If a is not an accumulation point of the domain of .......$\displaystyle f:A\subseteq R\rightarrow R$ then f has a limit over a
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    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 .......$\displaystyle f:A\subseteq R\rightarrow R$ 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 $\displaystyle f:A\longmapsto \mathbb{R}$ with $\displaystyle A\subseteq\mathbb{R}$, and if $\displaystyle a$ is not a limit point of $\displaystyle A$ then does $\displaystyle \lim_{x\to a}f(x)$ make sense?"

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


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

    By the defintion of limits in $\displaystyle \mathbb{R}\mapsto\mathbb{R}$ for $\displaystyle \lim_{x\to a}f(x)$ to exist there must exist a point $\displaystyle q\in\mathbb{R}$ such that for every $\displaystyle \varepsilon>0$ there exsits a $\displaystyle \delta>0$ such that $\displaystyle |x-a|<\delta\implies|f(x)-q|<\varepsilon~~x\in A$. But now suppose that $\displaystyle a$ is not a limit point of $\displaystyle A$ then there exists a neighborhood $\displaystyle (x-\delta_1,x+\delta_1)$ such that there does not exist an element of $\displaystyle A$ within that neighborhood. Now if we choose an $\displaystyle \varepsilon$ that requires $\displaystyle \delta<\delta_1$ we run into a contradiction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2008
    Posts
    71
    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 $\displaystyle f:A\longmapsto \mathbb{R}$ with $\displaystyle A\subseteq\mathbb{R}$, and if $\displaystyle a$ is not a limit point of $\displaystyle A$ then does $\displaystyle \lim_{x\to a}f(x)$ make sense?"

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


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

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

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

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

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


    .....................if xεA and 0<|x-a|<r then |f(x)-m|<ε...........................
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Greens theorem limits
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 8th 2011, 12:26 PM
  2. Divergence Theorem limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: May 8th 2011, 06:34 AM
  3. Theorem on the composition of limits?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Oct 10th 2010, 12:19 PM
  4. Theorem in limits
    Posted in the Math Challenge Problems Forum
    Replies: 0
    Last Post: Jun 28th 2010, 04:14 PM
  5. Bayes Theorem Limits
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Sep 15th 2009, 10:17 AM

Search Tags


/mathhelpforum @mathhelpforum