Limits

[The_ArtofScience]

1. The problem statement, all variables and given/known data

Prove that lim n--> infinity a_n{b}_n=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

[HallsofIvy]

That sounds like a pretty good way! You will want to use the fact that
|a_nb_n- AB|= |a_nb_n- a_nB+ a_nB- AB|\le |a_n||b_n-B|+ |B||a_n-B|
That first term is a little tricky!

[The_ArtofScience]

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 \left|A\right|. I'm thinking that its limit should be between \left|A\right|+e and \left|A\right|. If that's true then b_subn should also have that condition....Please correct me if I'm wrong

[HallsofIvy]

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

To show that sum is "< \epsilon" you want to be able to make each part less than \epsilon/2". Since |B| is a fixed number, you just need to make |a_n- A|< \epsilon/2|B|. Since |a_n depends on n, you cannot just make |b_n-B|< \epsilon/2a_n. Do just as you say: use the fact that {a_n} is bounded: for large enough n, A-1< a_n< A+ 1.

[The_ArtofScience]

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(\left|A\right|+1)

[HallsofIvy]

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

[matt grime]

From your answer, the absolute value of b_subn - B must then be less than e/2(\left|A\right|+1)

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.