# Proving a limit and the Order Limit Theorem

• rbzima
In summary, the conversation is discussing how to prove that if a sequence (a_n) converges to 0 and the absolute difference between another sequence (b_n) and a constant b is less than or equal to the terms of the first sequence, then (b_n) also converges to b. The conversation includes discussions on the Order Limit Theorem and the definition of convergence, and ultimately concludes that the additional restriction of {an} being positive or the inequality being modified is necessary for the proof to make sense.
rbzima
If $$(a_n) \rightarrow 0$$ and $$\left|b_n - b\right| \leq a_n$$, then $$(b_n) \rightarrow b$$.

I'm a little bit stuck on proving this! Part (iii) of the Order Limit Theorem states:

Assume:$$lim a_n = a$$ and $$lim b_n = b$$
If there exists $$c \in$$ R for which $$c \leq b_n$$ for all $$n \in$$ N, then $$c \leq b$$. Similarly if $$a_n \leq c$$ for all $$n \in$$ N, then $$a \leq c$$.

I can then take this to a point where $$b_n = b$$, but somehow that doesn't seem right because $$\left|b_n - b\right|<\epsilon$$, where $$\epsilon > 0$$. Where am I going wrong here!

What happens if you choose n so large that |an| is less than epsilon? Why can you choose such an n?

That's exactly what I want. I want |an| to be less than epsilon because it implies convergence for any n, which means |an| is practically nothing because it approaches 0.

I'm not really sure what you are getting at here.

You know {an} converges to 0, what does that tell you?

d_leet said:
You know {an} converges to 0, what does that tell you?

$$b_n - b = 0$$

So, basically, $$b_n = b$$

But I'm wondering if that's really all that's necessary?

rbzima said:
$$b_n - b = 0$$

So, basically, $$b_n = b$$

But I'm wondering if that's really all that's necessary?

I don't understand how you get that from what I wrote.

Now IGNORE {bn} for a minute you know that {an} converges to zero, what does that mean?

d_leet said:
I don't understand how you get that from what I wrote.

Now IGNORE {bn} for a minute you know that {an} converges to zero, what does that mean?

We had $$a_n$$ in the inequality, that's what I thought you were referring to.

Basically, the sequence approaches zero! Convergent series are bounded above and below, and a sequence converges when, for every positive integer $$\epsilon$$, there exists and $$N \in$$ N such that whenever $$n \leq$$ N, it follows that |an - a| < $$\epsilon$$.

I know the definition of convergence, in this case: |an| < $$\epsilon$$

Shall I list the properties of convergent sequences while I'm at it too! LOL!

rbzima said:
If $$(a_n) \rightarrow 0$$ and $$\left|b_n - b\right| \leq a_n$$, then $$(b_n) \rightarrow b$$.

I'm a little bit stuck on proving this!
If you are trying to prove just this part, than here you go:
since $$(a_n) \rightarrow 0$$ it means that for every e>o there exists some N(e)>0 such that whenever n>N we have Ia_nI<e, which actually is

-e<a_n<e
Now going back to $$\left|b_n - b\right| \leq a_n<e$$ which actually means that b_n converges to b.

what else are u looking for?
P.S. This is what d_leet also suggested, i guess!

sutupidmath said:
If you are trying to prove just this part, than here you go:
since $$(a_n) \rightarrow 0$$ it means that for every e>o there exists some N(e)>0 such that whenever n>N we have Ia_nI<e, which actually is

-e<a_n<e
Now going back to $$\left|b_n - b\right| \leq a_n<e$$ which actually means that b_n converges to b.

what else are u looking for?
P.S. This is what d_leet also suggested, i guess!

Two things. One please don't post entire solutions to questions such as this. And two, there actually needs to be an additional restriction to {an} or to the inequality of |b{sub]n[/sub]-b|<an in order for it to make sense; that restriction being either {an} is a sequence of positive terms or the inequality should be |bn-b|<|an|

## 1. What is a limit in mathematics?

A limit in mathematics is a fundamental concept that defines the behavior of a function as its input approaches a certain value. It represents the value that a function approaches as its input gets closer and closer to a specific value, without ever actually reaching it.

## 2. How do you prove a limit in mathematics?

In order to prove a limit in mathematics, you must show that the function approaches a specific value as its input approaches a certain value. This can be done by using the formal definition of a limit, which involves using the epsilon-delta method to show that for any epsilon (a small value) there exists a delta (a small distance) such that the function's output is within epsilon of the limit value for all input values within delta of the limit value.

## 3. What is the Order Limit Theorem?

The Order Limit Theorem is a mathematical theorem that states that if a function f(x) is less than or equal to another function g(x) for all values of x in a certain interval, and both functions have the same limit as x approaches a certain value, then the limit of f(x) is less than or equal to the limit of g(x).

## 4. How is the Order Limit Theorem used in proving limits?

The Order Limit Theorem is used in proving limits by providing a convenient way to compare the limits of two functions. By showing that one function is always less than or equal to another function and that both have the same limit, the theorem allows us to conclude that the limit of the first function must also be less than or equal to the limit of the second function.

## 5. Can the Order Limit Theorem be applied to all functions?

No, the Order Limit Theorem can only be applied to functions that are continuous within a certain interval. This means that the function must be defined and have a limit at every point within the interval. Additionally, both functions being compared must have a limit as x approaches the same value within that interval.

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