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Proof about a limit property clarification

  1. Jun 3, 2013 #1
    The book proves this limit and I am a bit confused how all the pieces fit together.
    So the book proves "If [itex](s_n)[/itex] converges to [itex]s[/itex] and [itex](t_n)[/itex] converges to [itex]t[/itex], then [itex](s_nt_n)[/itex] converges to [itex]st [/itex]. That is, [itex] lim(s_nt_n) = (lim s_n)(lim t_n)[/itex].

    The proof goes like this
    Let [itex] \epsilon> 0 [/itex] . By Theorem 9.1 there is a constant [itex] M > 0[/itex] such that
    [itex]|s_n| ≤ M [/itex]for all [itex]n[/itex]. Since [itex]lim t_n = t[/itex] there exists [itex]N_1 [/itex] such that [itex]n > N1 [/itex] implies [itex]|t_n − t| <\epsilon/(2M) [/itex] Also, since [itex] lim s_n = s [/itex] there exists [itex]N_2 [/itex] such that [itex]n > N_2 [/itex]implies [itex]|s_n − s| < \epsilon/(2(|t| + 1)) [/itex] Then [itex]|s_nt_n − st| ≤ |s_n| · |t_n − t| + |t| · |s_n − s|
    ≤ M · (\epsilon/2M)+ |t| · (\epsilon/(2(|t| + 1))<\epsilon/2+\epsilon/2=\epsilon [/itex].

    The part I do not understand about the proof is this jump in the inequality in the last step that is how is [itex]|s_n| · |t_n − t| + |t| · |s_n − s|≤ M · (\epsilon/(2M))+ |t| · (\epsilon/(2(|t| + 1))[/itex] instead of just less than?
     
    Last edited: Jun 3, 2013
  2. jcsd
  3. Jun 3, 2013 #2
    Sorry I had accidentally posted this post before I had completed it. So I made several changes to it but now its completed.
     
  4. Jun 3, 2013 #3

    Office_Shredder

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    If your confusion is just in that you expect a [itex] < [/itex] and the book uses a [itex] \leq [/itex], then it's a simple observation that if [itex] a < b[/itex], it is certainly true that [itex] a\leq b[/itex], but maybe it would have been more consistent on the part of the author to use a strict inequality there
     
  5. Jun 3, 2013 #4
    Oh yes I did forget to ask I also noticed that the author selects two different epsilon values that each limit that is the [itex]\epsilon/2M[/itex] and [itex] \epsilon/(2|t|+1).[/itex] that seem to be completely arbitrary. If I redid the proof and could I chose different values than the author? And also how does the author verify that [itex]M⋅\epsilon/2M+|t|⋅(ϵ/(2(|t|+1))<ϵ /(2)+ϵ /(2)[/itex]?. It seems that should require another proof on its own or is it just possible just by observation?
     
    Last edited: Jun 3, 2013
  6. Jun 3, 2013 #5
    Well anyways I think I figured out a simpler way of doing this proof. By definition of convergence let [itex] \epsilon>0[/itex]. Since all convergent sequences are bounded there exists a real number call it [itex]M [/itex] such that [itex] M>0 \text{ and }\ |s_n| < M \text{ for all n } [/itex]. Since
    [itex] lim t_n=t \text{ there exists }\ N_1 \text{ such that }\ n>N_1 \text{ implies } |t_n-t|< \frac{\epsilon}{2M}[/itex]. Similarly for [itex] lim s_n=s \text{ there exists} \ N_2 \text{ such that }\ n>N_2 \text{ implies } |s_n-s|< \frac{\epsilon}{2M}[/itex]. Thus we get [itex] |s_nt_n- st|≤|s_n|*|t_n-t|+|t|*|s_n-s|< M*\frac{\epsilon}{2M} + M*\frac{\epsilon}{2M}=\epsilon [/itex]. Hence [itex] lim s_nt_n=(lim s_n)(lim t_n)[/itex].
     
    Last edited: Jun 3, 2013
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