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Proving The Continuous Theorem for Sequences

  1. May 10, 2007 #1
    1. The problem statement, all variables and given/known data
    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].

    2. Relevant equations

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

    3. 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!
  2. jcsd
  3. May 10, 2007 #2


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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.
  4. May 10, 2007 #3
    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).
  5. May 11, 2007 #4


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    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]?
  6. May 11, 2007 #5
    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 =(
  7. May 11, 2007 #6


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    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]
  8. May 11, 2007 #7
    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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