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How to begin this limit problem

  1. Jun 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that lim n--> infinity [tex]a_n{b}_n[/tex]=AB from the definition of a limit.



    3. The attempt at a solution

    I'm not sure how to begin this other than using the identity given that the absolute value of a function minus its L is less than some value e. I'm thinking that e would eventually have to split into halves and then added to e. Other than that, my attempts at trying to prove it (or understand it) haven't really lead me anywhere
     
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  3. Jun 20, 2008 #2

    HallsofIvy

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    That sounds like a pretty good way! You will want to use the fact that
    [tex]|a_nb_n- AB|= |a_nb_n- a_nB+ a_nB- AB|\le |a_n||b_n-B|+ |B||a_n-B|[/tex]
    That first term is a little tricky!
     
  4. Jun 20, 2008 #3
    Thanks for the help HallsofIvy,

    I understand what you did there but I'm still curious if I also need to do another step in the proof where I have to show that its bounded? a_subn is less than some value epsilon plus [tex]\left|A\right|[/tex]. I'm thinking that its limit should be between [tex]\left|A\right|[/tex]+e and [tex]\left|A\right|[/tex]. If that's true then b_subn should also have that condition....Please correct me if I'm wrong
     
  5. Jun 21, 2008 #4

    HallsofIvy

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    Yes, that was my point in "That first term is a little tricky!"

    To show that sum is "[itex]< \epsilon[/itex]" you want to be able to make each part less than [itex]\epsilon/2[/itex]". Since |B| is a fixed number, you just need to make [itex]|a_n- A|< \epsilon/2|B|[/itex]. Since [itex]|a_n[/itex] depends on n, you cannot just make [itex]|b_n-B|< \epsilon/2a_n[/itex]. Do just as you say: use the fact that [itex]{a_n}[/itex] is bounded: for large enough n, [itex]A-1< a_n< A+ 1[/itex].
     
  6. Jun 22, 2008 #5
    Where did you get the 1 from? Sorry I'm not understanding that part

    From your answer, the absolute value of b_subn - B must then be less than e/2([tex]\left|A\right|[/tex]+1)
     
  7. Jun 22, 2008 #6

    HallsofIvy

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    I used "1" because it is easy! Any specific number would do. Since [itex]a_n[/itex] goes to A, given any [itex]\epsilon>0[/itex], for "large enough n" (i.e, there exist N such that if n> N...) [itex]|a_n- A|< \epsilon[/itex]. Just take [itex]\epsilon= 1[/itex] and that [itex]|a_n- A|< 1[/itex] so [itex]-1< a_n-A< 1[/itex] or [itex]A-1< a_n< A+1[/itex]. Since both of those are less than or equal to |A|+ 1, you no know that for n> N, [itex]|a_n|< A+1[/itex].
     
  8. Jun 22, 2008 #7

    matt grime

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    Yep, that's fine. Getting something precisely less than epsilon is a complete red herring. You just need to make it arbtirarily small.

    If, given any e>0 I can make something less than 3e or e^2, or e(1 + 1/(|A|+1))/2 for some fixed number A, then I've done the job.
     
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