- #1

Mathos

- 26

- 3

## Homework Statement

Theorem. If lim a n = A and lim b n = B, then lim a n b n = AB.

## Homework Equations

|a

_{n}- A| < ε

|b

_{n}- B| > ε

## The Attempt at a Solution

I have the solution to this, but I'm unclear on why one part is done.

Let ε > 0. Since b

_{n}converges it is bounded and there exists an M

_{1}such that for all n [itex]\in[/itex] N we have |b

_{n}| < M

_{1}. Define M

_{1}[itex]\geq[/itex] 1. Then |b

_{n}| / M

_{1}≤ 1.

Skipping ahead, do the same for a

_{n}and let |a

_{n}- A| < [itex]\frac{ε}{2M

_{1}}[/itex]

and let

|b

_{n}- B| < [itex]\frac{ε}{2M

_{2}}[/itex].

|a

_{n}b

_{n}- AB| = |a

_{n}b

_{n}-Ab

_{n}+ Ab

_{n}- AB|

**So are we trying to manipulate the limit a**

_{n}b_{n}into the familiar definition of |a_{n}- L| < ε so we can compare them?So |a

_{n}b

_{n}-Ab

_{n}+ Ab

_{n}- AB|

≤

|a

_{n}b

_{n}-Ab

_{n}| + |Ab

_{n}- AB|

=

|a

_{n}- A||b

_{n}| + |b

_{n}- B||A|

<

(ε/2M

_{1})|b

_{n}|

+

(ε/@M

_{2})|A|

**Stupid question, but how are we multiplying each ε by |b**

_{n}| and |A|? I know that it gives us the two original lim a_{n}and lim b_{n}, but how can we just stick them on the other side of the "<" symbol? And how do |bn|/M_{1}= 1 and |A|/M_{2}= 1 if M is just defined as being larger than both? Can we just take the max of |bn| and |A|?