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

Question

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 $

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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 $lim_{n\to\infty}(\frac{x_n}{y_n})=\frac{X}{Y}$

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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 $0$.

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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 $\{(-1)^n\}$).

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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 $\lim_{A_2\ni n\to \infty}y_n = L$

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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 $lim_{A\ni n\to\infty}(\frac{x_n}{y_n})=\frac{X}{Y}$

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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.

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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.

Answer 1

I was able to prove that my definitions above are actually equivalent to the original definition of limit of a sequence, And that in order for the original definition of limit of a sequence to actully mean something, The sequences on which the limit is defined must be defined on sets of the form $\mathscr{A}(\mathbb{N})=\{A\in\mathscr{P}(\mathbb{N})|\exists K\in\mathbb{N},\forall K<n\in\mathbb{N},n\in A\}$ (i.e. $\mathscr{A}(\mathbb{N})$ is the set of all subsets of $\mathbb{N}$ that contain a neighborhood of $\infty$ that consists only of natural numbers, which we may call from now on succinctly a natural neighborhood of $\infty$).

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NEW DEFINITION (1): Limit of a Sequence

Let $A\in\mathscr{A}(\mathbb{N})$ be some subset of $\mathbb{N}$ that contains a natural neighborhood of $\infty$,

(i.e. $A\in\mathscr{P}(\mathbb{N})$ is some subset of the natural numbers for which $\exists K\in \mathbb{N}, \forall K<n\in \mathbb{N}, n \in A$)

And let $\{x_n\}_{n\in A}$ be some sequence of real numbers defined on $A$,

(i.e. $x$ is a function $x:A\to\mathbb{R}$),

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

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 $

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Now we'll prove the following theorem:

THEOREM (2): The limit (if exists) is independent of the underlying set

Let $A_1\in\mathscr{A}(\mathbb{N})$ be some subset of $\mathbb{N}$ that contains a natural neighborhood of $\infty$,

(i.e. $A_1\in\mathscr{P}(\mathbb{N})$ is some subset of the natural numbers for which $\exists K_1\in \mathbb{N}, \forall K_1<n\in\mathbb{N}, n\in A_1$)

Let $A_2\in\mathscr{A}(\mathbb{N})$ be some subset of $\mathbb{N}$ that contains a natural neighborhood of $\infty$,

(i.e. $A_2\in\mathscr{P}(\mathbb{N})$ is some subset of the natural numbers for which $\exists K_2\in \mathbb{N}, \forall K_2<n\in\mathbb{N},n\in A_2$)

Let $\{x^1_n\}_{n\in A_1}$ and $\{x^2_n\}_{n\in A_2}$ be two sequences defined on $A_1$ and $A_2$ respectively such that $\forall n\in A_1\cap A_2,x^1_n = x^2_n$.

(i.e. $x^1$ and $x^2$ are two functions: $x^1:A_1\to\mathbb{R}$ and $x^2:A_2\to\mathbb{R}$),

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

Then:

$\lim_{A_1\ni n\to\infty}x^1_n = L $ if and only if $\lim_{A_2\ni n\to\infty}x^2_n = L$

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Proof:

Suppose that $\lim_{A_1\ni n\to\infty}x^1_n = L $ and we'll prove that $\lim_{A_2\ni n\to \infty}x^2_n = L$:

Let $0<\epsilon \in \mathbb{R}$ be some positive real number, Then because $\lim_{A_1\ni n\to\infty}x^1_n = L $, We get by Definition (1) that $\exists N_1\in\mathbb{N},\forall N_1<n\in A_1,|x^1_n-L|<\epsilon$.

Now take $N_2 = max(N_1,K_1)$, Because $N_1,K_1\in\mathbb{N}$ we get that $N_2\in\mathbb{N}$ as $N_2$ is the maximum of two natural numbers and thus must be also a natural number.

Now we'll prove that: $\forall N_2<n\in\ A_2,|x^2_n-L|<\epsilon$

Let $N_2 < n\in A_2$, Now since $N_1,K_1 \leq N_2$ (as $N_2$ is the maximum of both of those two numbers), We get that $N_1, K_1<n$, Also since $A_2\subseteq \mathbb{N}$ we get that $n\in\mathbb{N}$, And so $N_1, K_1<n\in \mathbb{N}$, Now since it known that $\forall K_1<n\in \mathbb{N}, n \in A_1$, we get that for this particular $n$, $n\in\ A_1$ and thus $n\in A_1\cap A_2$, And we can conclude by the fact $\forall n\in A_1 \cap A_2, x^1_n = x^2_n$ that for this particular $n$, (I) $x^1_n=x^2_n$ .

Also since it is known that $\forall N_1<n\in A_1,|x^1_n-L|<\epsilon$ we can conclude that for this particular $n$, (II) $|x^1_n-L|<\epsilon$, And, Now by substituting (I) into (II) we get that $|x^2_n-L|<\epsilon$.

Thus we have shown that $\forall 0<\epsilon\in\mathbb{R},\exists N_2\in\mathbb{N},\forall N_2<n\in A_2,|x^2_n-L|<\epsilon$ and we can concluded by Definition (1) that $\lim_{A_2\ni n\to \infty}x^2_n = L$ as was to be shown.

The proof of the other direction (i.e. that if $\lim_{A_2\ni n\to \infty}x^2_n = L$ then $\lim_{A_1\ni n\to\infty}x^1_n = L $) is similar and thus will not be shown.

Q.E.D.

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Now we'll prove the following immediate corollary of the theorem just proved:

COROLLARY (3): The limit (if exists) for one subset of $\mathbb{N}$ that contains a natural neighborhood of $\infty$, is the same for all subsets of the subset that contain a natural neighborhood of $\infty$, And the limit (if exists) for one subset of $\mathbb{N}$ that contains a natural neighborhood of $\infty$ is the same for all subsets of $\mathbb{N}$ that include the subset

Let $A_1,A_2\in\mathscr{A}(\mathbb{N})$ be some subsets of $\mathbb{N}$ that contain a natural neighborhood of $\infty$,

such that $A_2\subseteq A_1$.

Let $\{x^1_n\}_{n\in A_1}$ and $\{x^2_n\}_{n\in A_2}$ be two sequences defined on $A_1$ and $A_2$ respectively such that $\forall n\in A_2, x^1_n = x^2_n$.

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

$\lim_{A_1\ni n\to\infty}x^1_n = L $ if and only if $\lim_{A_2\ni n\to\infty}x^2_n = L$

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Proof:

Since $A_2\subseteq A_1$, We have that $A_1\cap A_2 = A_2$, And thus we can conclude from the fact $\forall n\in A_2, x^1_n = x^2_n$ that $\forall n\in A_1\cap A_2, x^1_n = x^2_n$, And therfore all the conditions of THEOREM (2) are satisfied and we can conclude that $\lim_{A_1\ni n\to\infty}x^1_n = L $ if and only if $\lim_{A_2\ni n\to \infty}x^2_n = L$ as was to be shown.

Q.E.D.

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Now we can state the following theorem which can be proved in a similar way to the original theorem as given in standard anlysis books:

THEOREM (4): Limits of Sequences under Arithmetic Operations:

Let $A_1,A_2\in\mathscr{A}(\mathbb{N})$ be some subsets of $\mathbb{N}$ that contain a natural neighborhood of $\infty$,

Let $\{x_n\}_{n\in A_1}$ and $\{y_n\}_{n\in A_2}$ be two sequences defined on $A_1$ and $A_2$ respectively.

And let $X_1,X_2\in\mathbb{R}$ be two real numbers such that $lim_{A_1\ni n\to\infty}x^1_n=X_1$ and $lim_{A_2\ni n\to\infty}x^2_n=X_2$ then the following are true:

1. $lim_{A_1\cap A_2\ni n\to\infty}(x^1_n+x^2_n)=X_1+X_2$
2. $lim_{A_1\cap A_2\ni n\to\infty}(x^1_n-x^2_n)=X_1-X_2$
3. $lim_{A_1\cap A_2\ni n\to\infty}(x^1_n\times x^2_n)=X_1\times X_2$
4. If in addition $\forall n\in A_2, x^2_n \neq 0$ and $X_2\neq 0$ then $lim_{A_1\cap A_2\ni n\to\infty}(\frac{x^1_n}{x^2_n})=\frac{X_1}{X_2}$

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Now we'll prove a trivial theorem that claims that if the limit in the definition above exists, then the limit in the original definition given in analysis books exists:

THEOREM (5): If the limit of seqeunce according to DEFNITION (1) exist then the limit of sequence according to the standard definition given in standard analisys books also exist and both limits agree

Let $A\in\mathscr{A}(\mathbb{N})$ be some subset of $\mathbb{N}$ that contains a natural neighborhood of $\infty$,

And let $\{x_n\}_{n\in A}$ be some sequence of real numbers defined on $A$,

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

Then:

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

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

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Proof:

We must prove that: $\forall 0 < \epsilon \in\mathbb{R} ,\exists N\in\mathbb{N}, \forall N<n\in \mathbb{N}, |x_n - L| < \epsilon $:

Let $ 0 < \epsilon \in\mathbb{R}$ be some positive real number, Because $A\in\mathscr{A}(\mathbb{N})$ we get that:

(I) $\exists K\in\mathbb{N},\forall K<n\in\mathbb{N},n\in A$,

Now since $lim_{A\ni n\to\infty}x_n=L$ we get by DEFINITION (1) that

(II) $\exists \hat{N}\in\mathbb{N},\forall \hat{N}<n\in\ A, |x_n-L|<\epsilon$,

Now let $N=max(K,\hat{N})$, Then clearly $N\in\mathbb{N}$ and $K,\hat{N}\leq N$,

Now we'll prove that $\forall N<n\in\mathbb{N},|x_n-L|<\epsilon$:

Let $N<n\in\mathbb{N}$, Then $K<n$ and thus by (I) we get that $n\in A$, And, Now since $\hat{N}<n$ and since we've shown that $n\in A$, By (II) we get that $|x_n-L|<\epsilon$.

Thus we've shown that $\forall 0 < \epsilon \in\mathbb{R} ,\exists N\in\mathbb{N}, \forall N<n\in \mathbb{N}, |x_n - L| < \epsilon $ and we can conclude that $lim_{n\to\infty}x_n = L$ as was to be shown.

Q.E.D.

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Now, After all this theory, We can by using THEOREM (4) calculate limits like the ones below more rigorously:

EXAMPLE 1:

Show that $lim_{ n\to\infty}(\frac{n + 1}{n - 1}) = 1$

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ANSWER:

By observing that:

$lim_{\mathbb{N}\ni n\to\infty}(1+\frac{1}{n}) = 1$

And that:

$lim_{\mathbb{N}\ni n\to\infty}(1-\frac{1}{n}) = 1$

Since $\mathbb{N}_{\geq 2} \subset \mathbb{N}$, We can conclude by COROLLARY (3) that:

$lim_{\mathbb{N}_{\geq 2}\ni n\to\infty}(1-\frac{1}{n}) = 1 \neq 0$,

Also since $\forall n\in\mathbb{N}_{\geq 2}, 1-\frac{1}{n} \neq 0$, We can conclude by PART 4. OF THEOREM (4) that $lim_{ n\in\mathbb{N}\cap\mathbb{N}_{\geq 2}\ni n\to\infty} \frac{1+\frac{1}{n}}{1-\frac{1}{n}} = \frac{1}{1} = 1$, But since $\mathbb{N}_{\geq 2} \subset \mathbb{N}$ we can conclude that $\mathbb{N}\cap\mathbb{N}_{\geq 2}=\mathbb{N}_{\geq 2}$, And thus:

$lim_{\mathbb{N}_{\geq 2}\ni n\to\infty} \frac{1+\frac{1}{n}}{1-\frac{1}{n}} = 1$,

But since $\forall n\in\mathbb{N}_{\geq 2}, \frac{1+\frac{1}{n}}{1-\frac{1}{n}} = \frac{n+1}{n-1}$ We can conclude that:

$lim_{\mathbb{N}_{\geq 2}\ni n\to\infty} \frac{n+1}{n-1} = 1$, And now by using THEOREM (5) we get that $lim_{n\to\infty} \frac{n+1}{n-1} = 1$ as was to be shown.

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EXAMPLE 2:

Show that $lim_{n\to\infty}\frac{3n^2+n-1}{5n^3-2n^2+n-4} = 0$

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ANSWER:

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 is clear that $\exists K\in \mathbb{N},\forall K<n\in \mathbb{N}, n\in A$, since $B$ includes only a finite number of elements or none at all. (Just take $K=1$ if $B=\emptyset$, or choose any $K\in\mathbb{N}$ that satisfy $max(B)<K$ if $B\neq \emptyset$), And thus $A\in\mathscr{A}(\mathbb{N})$.

Now, Define two sequences $\{x_n\}_{n\in A}$ 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 $lim_{A \ni n\to\infty} y_n = 5 \neq 0$.

Also, by definition of the set $A$, we have that $\forall n\in A, y_n\neq 0$.

And thus, By using PART 4. OF THEOREM (4), We can conclude that $lim_{A\cap A \ni n\to\infty}\frac{x_n}{y_n}=0$, But since $A \cap A = A$ 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 $lim_{A \ni n\to\infty}\frac{3n^2+n-1}{5n^3-2n^2+n-4}=0$ and by using THEOREM (5) we can conclude that $lim_{n\to\infty}\frac{3n^2+n-1}{5n^3-2n^2+n-4}=0$ as was to be shown.

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Answer 2

When we speak about the limit of a sequence, only terms $u_n $ with $n $ enough great will be taken into account. The nature of a sequence and its limit (if it exists) do not change if we modify a finite number of terms. In the first theorem, the property 4 should be

If for $n $ large enough, $y_n\ne 0$ and $y_\to Y\ne 0$ then $x_n/y_n\to X/Y $.