# Updating the Definition of Limit of a Sequence to take in consideration sequences that are not defined for some finite number of natural numbers

The definition of limit of a sequence I always encounterd was of the form:

DEFINITION: Limit of a Sequence:
Let $\{x_n\}_{n\in\mathbb{N}}$ be some sequence of real numbers, (i.e. $x:\mathbb{N}\to\mathbb{R}$)
We say that:

$lim_{n\to\infty}x_n=L$

if and only if

$\forall 0 < \epsilon \in\mathbb{R} ,\exists N\in\mathbb{N}, \forall N<n\in\mathbb{N}, |x_n - L| < \epsilon$

And the theorem of arithmetic laws for limits of sequences was always of the form:

THEOREM: Limits of Sequences under Arithmetic Operations:
If $\{x_n\}_{n\in\mathbb{N}}$ and $\{y_n\}_{n\in\mathbb{N}}$ are two sequences
(i.e. $x$ and $y$ are functions: $x:\mathbb{N}\to\mathbb{R}$ and $y:\mathbb{N}\to\mathbb{R}$)
and $X,Y\in\mathbb{R}$ are two real numbers such that $lim_{n\to\infty}x_n=X$ and $lim_{n\to\infty}y_n=Y$ then the following are true:

1. $lim_{n\to\infty}(x_n+y_n)=X+Y$
2. $lim_{n\to\infty}(x_n-y_n)=X-Y$
3. $lim_{n\to\infty}(x_n\times y_n)=X\times Y$
4. If in addition $\forall n\in\mathbb{N}, y_n \neq 0$ and $Y\neq 0$ then $lim_{n\to\infty}(\frac{x_n}{y_n})=\frac{X}{Y}$

Now the problem is that this definition and theorem does not allow to handle limits such as the one below in a truly rigorous manner:

$lim_{n\to\infty}\frac{n-1}{2n-6} = \frac{1}{2}$

Suppose that we try to solve this limit by using the standard definition and theorem as given above:

Define two sequences (i.e. two functions from $\mathbb{N}$ to $\mathbb{R}$) $\{x_n\}_{n\in\mathbb{N}}$ and $\{y_n\}_{n\in\mathbb{N}}$

such that $\forall n\in\mathbb{N}, x_n \triangleq 1-\frac{1}{n}$ and $\forall n\in\mathbb{N}, y_n \triangleq 2-\frac{6}{n}$ where the symbol $\triangleq$ means "equal by definition". (The sequences are well defined since for all $n\in\mathbb{N}$, the denominator in each of those two sequences does not match $0$).

Now, We can show that for $n = 3$ we have $y_3 = 2-\frac{6}{3}=0$ and that $\forall 4\leq n\in\mathbb{N}, y_n \neq 0$, Also it is clear that $lim_{n\to\infty}x_n=1$ and that $lim_{n\to\infty} y_n = 2 \neq 0$.

Now inorder to claim that $lim_{n\to\infty}\frac{x_n}{y_n}=\frac{1}{2}$ we have to change the condition of part 4. of the theorem (that is $\forall n\in\mathbb{N},y_n\neq 0$),
and say that it is enough that almost for all $n\in\mathbb{N}$, we have $y_n \neq 0$,
I.e. that is enough that $\exists N\in\mathbb{N},\forall N<n\in\mathbb{N}, y_n \neq 0$ inorder to conclude that
(*) $lim_{n\to\infty}\frac{x_n}{y_n}=\frac{1}{2}$.

Now since it can be shown that $\forall 4\leq n\in\mathbb{N}, \frac{n-1}{2n-6} = \frac{x_n}{y_n}$, We can conclude by (*) that $lim_{n\to\infty}\frac{n-1}{2n-6} = \frac{1}{2}$ as was to be shown.

Now the problem is that the limit $lim_{n\to\infty}\frac{n-1}{2n-6} = \frac{1}{2}$ itself, also does not confine to the original definition of limit of a sequence that requires the sequence to be defined for every natural number $n\in\mathbb{N}$ because for $n=3$ the denominator is $0$.

Now the problem is, in order to tackle limits like this we have to "update" the definition of limit of a sequence so that it will be able to handle rigorously limits of sequences like the one above or limits like this one $lim_{n\to\infty}\frac{3n^2+n-1}{5n^3-2n^2+n-4}$ which may have denominators that are equal to zero for a finite number of natural numbers, and thus the sequence may not be defined for all $n\in\mathbb{N}$.

What definition of limit we should take? Maybe this one?

UPDATED DEFINITION: Limit of a Sequence:
Let $A\in\mathscr{P}(\mathbb{N})$ (the power set of $\mathbb{N}$) be some subset of the natural numbers with cardinality $|A|=\aleph_0$ that satisfies $\exists K\in\mathbb{N},\forall K<n\in\mathbb{N},n\in A$, And let $\{x_n\}_{n\in A}$ be some sequence of real numbers (i.e. $x:A\to\mathbb{R}$),
We say that:

$lim_{A\ni n\to\infty}x_n=L$

if and only if

$\forall 0 < \epsilon \in\mathbb{R} ,\exists N\in\mathbb{N}, \forall N<n\in A, |x_n - L| < \epsilon$

IMPORTANT NOTE: The condition $\exists K\in\mathbb{N},\forall K<n\in\mathbb{N},n\in A$ must be included, Otherwise we may get a subsequential limit (i.e. limit of some subsequence), and the limit will not be uniquely defined. (take for example the sequence $\{(-1)^n\}$).

This new definition probably will able to tackle this kind of limits rigorously, The only downside is that many theorems that were already proven for the original defintion, Now must be reproved again. Also we must prove that the sets on which the limit will be taken does not matter, In other words, We must prove that the limit is independent on the underlying set. I will state it formally as a theorem which I haven't tried to prove yet:

NEW THEOREM: The limit (if exists) is independent of the underlying set
Let $A_1\in\mathscr{P}(\mathbb{N})$ be some subset of the natural numbers with cardinality $|A_1|=\aleph_0$ that satisfies $\exists K_1\in \mathbb{N}, \forall K_1<n\in \mathbb{N}, n \in A_1$,

Let $A_2\in\mathscr{P}(\mathbb{N})$ be some subset of the natural numbers with cardinality $|A_2|=\aleph_0$ that satisfies $\exists K_2\in \mathbb{N}, \forall K_2<n\in \mathbb{N}, n \in A_2$,

Let $\{x_n\}_{n\in A_1}$ and $\{y_n\}_{n\in A_2}$ be two sequences such that $\forall n\in A_1 \cap A_2, x_n = y_n$.

And let $L\in\mathbb{R}$ be some real number, Then:

$lim_{A_1\ni n\to\infty}x_n = L$ if and only if $\lim_{A_2\ni n\to \infty}y_n = L$

We can also restate and prove (I haven't proved it yet, But the proof is supposed to be straitforward) the thoerem of arithmetic laws of limits of sequences:

UPDATED THEOREM: Limits of Sequences under Arithmetic Operations:
Let $A\in\mathscr{P}(\mathbb{N})$ be some subset of the natural numbers with cardinality $|A|=\aleph_0$ that satisfies $\exists K\in \mathbb{N}, \forall K<n\in \mathbb{N}, n \in A$,

Let $\{x_n\}_{n\in A}$ and $\{y_n\}_{n\in A}$ be two sequences

(i.e. $x$ and $y$ are functions: $x:A\to\mathbb{R}$ and $y:A\to\mathbb{R}$),

And let $X,Y\in\mathbb{R}$ be two real numbers such that $lim_{A\ni n\to\infty}x_n=X$ and $lim_{A\ni n\to\infty}y_n=Y$ then the following are true:

1. $lim_{A\ni n\to\infty}(x_n+y_n)=X+Y$
2. $lim_{A\ni n\to\infty}(x_n-y_n)=X-Y$
3. $lim_{A\ni n\to\infty}(x_n\times y_n)=X\times Y$
4. If in addition $\forall n\in A, y_n \neq 0$ and $Y\neq 0$ then $lim_{A\ni n\to\infty}(\frac{x_n}{y_n})=\frac{X}{Y}$

Now, For example, Let's try to prove that $lim_{n\to\infty}\frac{3n^2+n-1}{5n^3-2n^2+n-4} = 0$ by using the new definitions and theorems:

Let $B\in\mathscr{P}(\mathbb{R})$ be defined as $B=\{x\in\mathbb{R}|5x^3-2x^2+x-4=0\}$, By the Fundamental Theorem of Algebra, This set have at most $3$ elements (i.e. $0\leq |B| \leq 3$).

Now, Define $A\in\mathscr{P}(\mathbb{R})$ as $A = \mathbb{N}-B$, It clear that $|A| = \aleph_0$ since $B$ includes only a finite number of elements or none at all.

Also, It is clear that $\exists K\in \mathbb{N},\forall K<n\in \mathbb{N}, n\in A$, (Just take $K=1$ if $B=\emptyset$, or choose any $K\in\mathbb{N}$ that satisfy $max(B)<K$ if $B\neq \emptyset$).

Now, Define two sequences (i.e. two functions from $A$ to $\mathbb{R}$) $\{x_n\}_{n\in A}$ and $\{y_n\}_{n\in A}$

such that $\forall n\in A, x_n \triangleq \frac{3}{n}+\frac{1}{n^2}-\frac{1}{n^3}$ and $\forall n\in A, y_n \triangleq 5 - \frac{2}{n}+\frac{1}{n^2}-\frac{4}{n^3}$. (The sequences are well defined since for all $n\in\mathbb{N}$, the denominator in each of those two sequences does not match $0$).

Now, It is clear that $lim_{A\ni n\to\infty}x_n=0$ and that $lim_{A \ni n\to\infty} y_n = 5 \neq 0$.

Now by using the new form of the theorem of arithmetic laws for limits of sequences, We can conclude that (*) $lim_{A \ni n\to\infty}\frac{x_n}{y_n}=0$.

Now since it can be shown that $\forall n\in A, \frac{3n^2+n-1}{5n^3-2n^2+n-4} = \frac{x_n}{y_n}$, We can conclude by (*) that $lim_{A \ni n\to\infty}\frac{3n^2+n-1}{5n^3-2n^2+n-4}=0$ as was to be shown. Q.E.D.

Thanks for any ideas, Also suggestions/references for books (or papers/articles) that treat limits in such a manner will be appreciated. Thanks a lot.

• That's a lot of text to deal with a problem whose solution, I think, is hardly worthy of more than a short remark (i.e. "Convergence of sequences is completely determined by any tail-end.") But yes, what you've developed here in detail is one way to formalize this remark. I haven't read all of it but skimming through some of the details, it seems perfectly fine to me. +1 – Stefan Mesken Feb 23 '18 at 19:55