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->∞