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Thread: Demostration of convergence and limits

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    Demostration of convergence and limits

    Demostrate: If (Sm)--->s then |(Sm)|--->|s|

    It means lím(Sm)=s then lím|(Sm)|=|s|
    .............m-> ........................ m->

    I have some ideas, but nothing concrete. I will explain more in a comment below.
    Last edited by Gabriel0156; Nov 2nd 2015 at 07:54 PM.
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    Re: Demostration of convergence and limits

    Quote Originally Posted by Gabriel0156 View Post
    Demostrate: If (Sm)--->s then |(Sm)|--->|s|

    It means lím(Sm)=s then lím|(Sm)|=|s|
    .............m->∞ ........................ m->∞

    I have some ideas, but nothing concrete. I will explain more in a comment below.
    See if you can prove $(\forall x\& y)[~\left| |x|-|y|\right|\le|x-y|~]$

    If you can then let $x=S_n~\&~y=s$.
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    Re: Demostration of convergence and limits

    Utiliza la definición del limite.

    Tenemos que: para cada \epsilon > 0 existe N > 0 tal que m > N implica |S_m - s| < \epsilon.

    Tambien tenemos que:
    |S_m| = |S_m| - |s| + |s| \ge \big||S_m| - |s|\big| + |s| \implies |S_m| - |s| \ge \big||S_m|-|s|\big|
    |S_m| = |S_m - s + s|  \le |S_m - s| + |s| \implies |S_m| - |s| \le |S_m - s|

    Entonces
    \epsilon > |S_m - s| \ge |S_m| - |s| \ge \big||S_m|-|s|\big|

    Así, para cada \epsilon > 0 existe N > 0 tal que m > N implica \big||S_m|-|s|\big| < \epsilon. Que significa que \lim_{m \to \infty} |S_m| = |s|.
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    Re: Demostration of convergence and limits

    For those that don't speak Español:

    Use the definition of limit.

    We have: For each $\epsilon > 0$ there exists N > 0 such that m > N implies $| S_m - s | < \epsilon$.

    We also have:
    $| S_m | = | S_m | - | s | + | s | \ge \big | | S_m | - | s | \big | + | s | \implies | S_m | - | s | \ge \big | | S_m | - | s | \big |$
    $| S_m | = | S_m - s + s | \le | S_m - s | + | s | \implies | S_m | - | s | \le | S_m - s |$

    So
    $\epsilon> | S_m - s | \ge | S_m | - | s | \ge \big | | S_m | - | s | \big |$

    Thus, for each $\epsilon> 0$ there exists N > 0 such that m > N implies $\big | | S_m | - | s | \big | < \epsilon$. That means that $\lim_{m \to \infty} | S_m | = | s |$.
    @Archie: It think I got that right. Please double check me.

    -Dan
    Last edited by topsquark; Nov 3rd 2015 at 07:40 AM.
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    Re: Demostration of convergence and limits

    "We have:" not "we need:". That's the limit relation we were given.
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    Re: Demostration of convergence and limits

    I assume the point of the exercise is to gain skill with \epsilon-N proofs, so this is a tangent, but...

    Depending on what's already been proven in your text/class, it's possible that the short proof goes like this: "The absolute value function is everywhere continuous. QED." (Ok, I'm joking a little bit there, but only a little bit. Not much more needs to be said.)

    Also, having given an \epsilon-N proof of this, as an immediate corollary you've also proven (assuming you or your class/text has already proven some other basic theorems about continuity) that the absolute value function is everywhere continuous.
    Last edited by johnsomeone; Nov 3rd 2015 at 02:42 AM.
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    Re: Demostration of convergence and limits

    Que bien que algunos hablan español aquí. ¿Es éste el foro indicado para Análisis Real o el de Cálculo es mas apropiado?
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    Re: Demostration of convergence and limits

    Quote Originally Posted by Gabriel0156 View Post
    Que bien que algunos hablan español aquí. ¿Es éste el foro indicado para Análisis Real o el de Cálculo es mas apropiado?
    Translation: It's great that some people speak Spanish here. Is this the best forum for Real Analysis? Or is the Calculus forum more appropriate?
    Pues, no hablo perfectamente el español, pero espero que mejore cada día. Creo que el foro de Cálculo es el mejor para Análisis Real.
    Translation: Well, I don't speak perfect Spanish, but I hope that I improve a little every day. I think that the Calculus forum is best for Real Analysis.
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