# How to begin this limit problem

1. Jun 19, 2008

### 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

2. Jun 20, 2008

### HallsofIvy

Staff Emeritus
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!

3. Jun 20, 2008

### 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

4. Jun 21, 2008

### HallsofIvy

Staff Emeritus
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, $A-1< a_n< A+ 1$.

5. Jun 22, 2008

### 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)

6. Jun 22, 2008

### HallsofIvy

Staff Emeritus
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, $|a_n|< A+1$.

7. Jun 22, 2008

### matt grime

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.