Proving The Continuous Theorem for Sequences

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Homework Help Overview

The discussion revolves around proving the continuous theorem for sequences, specifically that if a sequence \( a_n \) converges to \( L \) and \( f \) is continuous at \( L \), then \( b_n = f(a_n) \) converges to \( f(L) \). Participants are exploring the nuances of this theorem and the challenges in constructing a formal proof.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definitions of limits and continuity, with some expressing confusion about transitioning from \( a_n \) to \( f(a_n) \) and from \( f(x_0) \) to \( f(L) \). There is mention of using epsilon-delta definitions to clarify the proof structure.

Discussion Status

Several participants are actively engaging with the problem, questioning the relationships between the sequences and the function values. Some guidance has been offered regarding the epsilon-delta approach, but there is no clear consensus on the next steps or resolution of confusion.

Contextual Notes

Participants note the potential confusion arising from using different symbols for limits and the abstract nature of the problem. There is an acknowledgment of the need for clarity in definitions and relationships between the variables involved.

Lucretius
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Homework Statement


4.8 Show the following continuous theorem for sequences: if [tex]a_n \rightarrow L[/tex] and f is a real valued function continuous at L, then [tex]bn = f(a_n) \rightarrow f(L)[/tex].

Homework Equations



No real relevant equations here. Just good old proof I'm thinking.

The Attempt at a Solution



Well, I stared at this for an hour today. I was able to complete the rest of the assignment but this one has me stumped. I realize that [tex]\displaystyle\lim_{n\rightarrow\infty}a_n=L[/tex] and that for a real-valued function to be continuous at L that [tex]\displaystyle\lim_{x\rightarrow x_0}f(x)=f(x_0)=L.[/tex] I don't know what to do from here though. How do I get f(L) from f(x0)=L, and then get f(a_n) from just plain old a_n. This thing makes intuitive sense to me; it's blatantly obvious it's right - proving it has ... well.. proven to be really hard!
 
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When something seems like it should be obvious, you might want to write out some definitions. Epsilons, deltas, and all that.
 
Well that's the thing; I thought I could just approach it nicely, but the concepts seem to be two different things; it's as if I want to prove a finite limit at infinity. There is a proof I thought was similar to it, in the notes he gave us. However, I fleshed out a few epsilon delta definitions; drew a nice picture so I could see what was going on. My main problem is I have no clue how to get to f(a_n) from a_n, and from f(x0)=L to f(L).
 
It looks like you might be getting confused by using two different [itex]L[/itex]'s. Typically, the [itex]a_n[/itex] and [itex]b_n[/itex] will not converge to the same value.

Really, this is straightforward stuff... given some [itex]\epsilon[/itex] greater than [itex]0[/itex] can you show that there is some [itex]N[/itex] so that [itex]n>N \Rightarrow |f(a_n)-f(L)| < \epsilon[/itex]?
 
Hmm, I must not be seeing it...

I get that it's going to follow the basic form of a proof. For all epsilon > 0, there exists an N such that, for all n>N, then |f(a_n)-f(L)|<epsilon. The N is more than likely going to be some integer part of some number plus one. I get how all that works. The problem is... I'm used to an |f(x)-L|, where i know what the L and f(x) are. I have no clue as to what f(L) or f(a_n) are, and I don't see how I can understand what they are from the given information. Sorry; I hate missing obvious stuff =(
 
It's a little abstract. Do you know if there's a [itex]\delta[/itex] so that:
[tex]|a_n-L|<\delta \Rightarrow |f(a_n)-f(L)| < \epsilon[/tex]
 
I'm following you and it looks like I'll be proving a limit at a point, but I'm still stuck with how we can relate a_n to f(a_n), and L to f(L). If I can somehow extract the a_n and the L from the f, I could do it...again, thanks for the help so far and sorry I'm still not quite getting what to do...
 

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