# Theorem: If lim a n = A and lim b n = B, Then lim anbn = AB

• Mathos
In summary: Now do the same for |bn - B| and you get the right hand side of the inequality to be < epsilon/2. So you have a number N such that for all n>N you have |an-A||bn| + |bn-B||A| < epsilon/2 + epsilon/2 = epsilon. This is exactly what is needed to show that the limit is AB.
Mathos

## Homework Statement

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

|an - A| < ε

|bn - 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 bn converges it is bounded and there exists an M1 such that for all n $\in$ N we have |bn| < M1. Define M1 $\geq$ 1. Then |bn| / M1 ≤ 1.

Skipping ahead, do the same for an and let |an - A| < $\frac{ε}{2M1}$

and let

|bn - B| < $\frac{ε}{2M2}$.

|anbn - AB| = |anbn -Abn + Abn - AB|

So are we trying to manipulate the limit anbn into the familiar definition of |an - L| < ε so we can compare them?

So |anbn -Abn + Abn - AB|

|anbn -Abn| + |Abn - AB|

=

|an - A||bn| + |bn - B||A|

<

(ε/2M1)|bn|

+

(ε/@M2)|A|

Stupid question, but how are we multiplying each ε by |bn| and |A|? I know that it gives us the two original lim an and lim bn, but how can we just stick them on the other side of the "<" symbol? And how do |bn|/M1 = 1 and |A|/M2 = 1 if M is just defined as being larger than both? Can we just take the max of |bn| and |A|?

Mathos said:

## Homework Statement

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

|an - A| < ε

|bn - 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 bn converges it is bounded and there exists an M1 such that for all n $\in$ N we have |bn| < M1. Define M1 $\geq$ 1. Then |bn| / M1 ≤ 1.

Skipping ahead, do the same for an and let |an - A| < $\frac{\varepsilon}{2M_1}$

and let

|bn - B| < $\frac{\varepsilon}{2M_2}$.

|anbn - AB| = |anbn -Abn + Abn - AB|

So are we trying to manipulate the limit anbn into the familiar definition of |an - L| < ε so we can compare them?

So |anbn -Abn + Abn - AB|

|anbn -Abn| + |Abn - AB|

=

|an - A||bn| + |bn - B||A|

<

(ε/2M1)|bn|

+

(ε/@M2)|A|

Stupid question, but how are we multiplying each ε by |bn| and |A|? I know that it gives us the two original lim an and lim bn, but how can we just stick them on the other side of the "<" symbol? And how do |bn|/M1 = 1 and |A|/M2 = 1 if M is just defined as being larger than both? Can we just take the max of |bn| and |A|?
(In LaTeX, don't use the special characters at the right, like "ε" . Use \varepsilon . For subscripts use M_1, etc.)

First it will help to state what is required to show that ##\displaystyle \lim\ \left(a_n\ b_n\right) = AB \ ## .

You know that the sequences an and Bn converge, so you can use any positive quantity to represent "ε" for each of these sequences.

|an - A||bn| + |bn - B||A|

You know that bn converges thus there is a M such that bn<M for all n. Since |an-A| can be made really small there is a n such that for all n> N there holds

|an - A| < epsilon/(2*(M+1)).

but then for all n>N:

|an - A||bn| < epsilon*M/(2*(M+1)) < epsilon /2.

## 1. What is the theorem stating the relationship between limits and multiplication?

The theorem states that if the limit of a sequence an is equal to A and the limit of a sequence bn is equal to B, then the limit of the product of the two sequences anbn is equal to the product of their respective limits, AB.

## 2. What is the significance of this theorem in mathematics?

This theorem is significant because it allows us to simplify the process of finding limits of products of sequences. Instead of evaluating the limit of the product directly, we can evaluate the limits of the individual sequences and then multiply them together.

## 3. Can this theorem be used for any two sequences?

No, this theorem only applies to sequences where the limits of both sequences an and bn exist. If either of the limits does not exist, then this theorem cannot be applied.

## 4. Are there any other theorems related to limits and operations?

Yes, there are several other theorems that relate to limits and operations, such as the Sum/Difference Rule, the Product Rule, and the Quotient Rule. These theorems provide rules for finding limits of sums, differences, products, and quotients of sequences.

## 5. Can this theorem be extended to limits of functions?

Yes, this theorem can be extended to limits of functions. It is known as the Limit Product Rule and states that if the limits of two functions f(x) and g(x) exist as x approaches a, then the limit of the product f(x)g(x) is equal to the product of their respective limits, limx→af(x)·limx→ag(x).

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