Proving Convergence of \{b_n\} when \{a_n\}\to A, \{a_nb_n\} Converge

[Dustinsfl]

If \{a_n\}\to A, \ \{a_nb_n\} converge, and A\neq 0, then prove \{b_n\} converges.

Let \epsilon>0. Then \exists N_1,N_2\in\mathbb{N}, \ n\geq N_1,N_2

|a_n-A|<\frac{\epsilon}{2}

And let \{a_nb_n\}\to AB

So, |a_nb_n-AB|<\epsilon

I don't know how to show b_n is < epsilon.

 

[tiny-tim]

Hi Dustinsfl!

Hint: a_n(b_n - B) ​

 

[Dustinsfl]

Hi Dustinsfl!

Hint: a_n(b_n - B) ​

I am don't understand, so we have:

(a_nb_n-a_nB)

Ok, now what?

 

[tiny-tim]

lim_n->∞