Demostrate: If (S_{m})--->s then |(S_{m})|--->|s|
It means lím(S_{m})=s then lím|(S_{m})|=|s|
............._{m->∞ ........................ }_{m->∞}_{ }I have some ideas, but nothing concrete. I will explain more in a comment below.
Demostrate: If (S_{m})--->s then |(S_{m})|--->|s|
It means lím(S_{m})=s then lím|(S_{m})|=|s|
............._{m->∞ ........................ }_{m->∞}_{ }I have some ideas, but nothing concrete. I will explain more in a comment below.
Utiliza la definición del limite.
Tenemos que: para cada $\displaystyle \epsilon > 0$ existe $\displaystyle N > 0$ tal que $\displaystyle m > N$ implica $\displaystyle |S_m - s| < \epsilon$.
Tambien tenemos que:
$\displaystyle |S_m| = |S_m| - |s| + |s| \ge \big||S_m| - |s|\big| + |s| \implies |S_m| - |s| \ge \big||S_m|-|s|\big|$
$\displaystyle |S_m| = |S_m - s + s| \le |S_m - s| + |s| \implies |S_m| - |s| \le |S_m - s|$
Entonces
$\displaystyle \epsilon > |S_m - s| \ge |S_m| - |s| \ge \big||S_m|-|s|\big|$
Así, para cada $\displaystyle \epsilon > 0$ existe $\displaystyle N > 0$ tal que $\displaystyle m > N$ implica $\displaystyle \big||S_m|-|s|\big| < \epsilon$. Que significa que $\displaystyle \lim_{m \to \infty} |S_m| = |s|$.
For those that don't speak Español:
@Archie: It think I got that right. Please double check me.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 $\lim_{m \to \infty} | S_m | = | s |$.
-Dan
I assume the point of the exercise is to gain skill with $\displaystyle \epsilon-N$ 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 $\displaystyle \epsilon-N$ 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.
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.