Real Analysis – Sequences II
After having introduced the notion of neighborhood of a real number in the previous post we are going to introduce a further notion that will permit us to unify a few results.
Definition 16 The set formed by will be the set of affinely extended real numbers and denote it by . The elements of this new set have the following properties:

After the introduction of this new set and the new elements we can define two new neighborhoods:
Definition 17

It is very important that the reader can understand the reasons of these two definitions. I know that the first time I looked at them I was puzzled.
The notion of neighborhood of a real number is straightforward. Essentially, it is the open interval centered at a given number.
If one wants to extend the notion of neighborhood to the new elements and one has to realize that one has to abandon the hope of getting a centered interval. This can’t be done because those two new elements are the edges of .
One thing that one can, and should keep, though, is the fact that the bigger the the bigger the interval one gets. But with one gets smaller and smaller ratios as gets bigger!!! And that’s just want we want! If gets smaller this means that the left edge of the neighborhood (I’m assuming we are talking about the case) is moving even more to the left, thus making the interval bigger!
The neighborhood for has the same reasoning behind it and I hope that now any doubts that you might have had regarding the definitions of the neighborhoods of and are gone.
Let us now consider a sequence, , that is bounded below but not bounded above. That is to say that in we have that . This is equivalent to the following:
One can do the analogous derivation for . Hence one can write with full generality:
Definition 18 Given it is iff . 
— 5.2. Limits and Inequalities —
Theorem 14 Given and let us suppose that there exists an order such as . If and exist, it follows
Proof: Omitted. 
After this theorem we may ask ourselves is also a theorem? Well, it isn't. To prove so a single counterexample suffices.
For instance and nevertheless .
Corollary 15 Let be a sequence and . Let us also suppose that from a certain order we have (), then if exists we have ( ).
Proof: Take in theorem 14. 
Theorem 16
Given and such as for some order . Then: Proof: Omitted. 
Theorem 17 (Squeezed sequence theorem) Given , , and such as, for some order , . Then, if , has a limit and it is .
Proof: Omitted. 
As an application of theorem 17 let us look at the following example: and we wish to compute .
Well, .
We know that , hence .
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