How to begin this limit problem

[The_ArtofScience]

Homework Statement

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

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.