User:Corkgkagj/draft

Addition
If
 * $$\lim_{n\to\infty}a_n=A\qquad (1),$$
 * $$\lim_{n\to\infty}b_n=B\qquad (2),$$

then,
 * $$\lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n=A+B$$

Proof

 * $$|a_n+b_n-(A+B)|\leq|a_n-A|+|b_n-B|$$

Assume
 * $$\varepsilon\in\mathbb{R}^+$$

By (1),
 * $$\exists N_1\in\mathbb{N}:\forall n\geq N_1, |a_n-A|<\frac{\varepsilon}{2}.$$

By (2),
 * $$\exists N_2\in\mathbb{N}:\forall n\geq N_2, |b_n-B|<\frac{\varepsilon}{2}.$$

Now,
 * $$N:=\max{\{N_1,N_2\}}\,$$
 * $$\forall n\geq N,|a_n+b_n-(A+B)|\leq|a_n-A|+|b_n-B|<\varepsilon$$
 * $$\Rightarrow\lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n=A+B$$

Q.E.D.

By a Constant Scalar
If
 * $$\lim_{n\to\infty}a_n=A\qquad (1),$$

then
 * $$\forall c\in\mathbb{R},\lim_{n\to\infty}c\,a_n=c\lim_{n\to\infty}a_n=cA$$

Proof

 * $$|c\,a_n-cA|=|c||a_n-A|$$

By (1),
 * $$\exists N\in\mathbb{N}:\forall n\geq N\land c\neq 0, |a_n-A|<\frac{\varepsilon}{|c|}$$
 * $$\Rightarrow |c\,a_n-cA|=|c||a_n-A|<\varepsilon$$
 * $$\Rightarrow \lim_{n\to\infty}c\,a_n=c\lim_{n\to\infty}a_n=cA$$

If $$c=0$$, then the sequence has constant value zero, and the proof is trivial. Q.E.D.

By a Sequence
If
 * $$\lim_{n\to\infty}a_n=A\qquad (1),$$
 * $$\lim_{n\to\infty}b_n=B\qquad (2),$$

then,
 * $$\lim_{n\to\infty}(a_nb_n)=\left(\lim_{n\to\infty}a_n\right)\left(\lim_{n\to\infty}b_n\right)=AB$$

Proof

 * $$\forall n\in\mathbb{N},|a_nb_n-AB|=|a_n(b_n-B)+B(a_n-A)|\leq|a_n||b_n-B|+|B||a_n-A|$$

Assume
 * $$\varepsilon\in\mathbb{R}^+$$

By (1),
 * $$\exists N_1\in\mathbb{N}:\forall n\geq N_1, |a_n-A|<

\begin{cases} \varepsilon/(2B), & B\neq 0 \\ \varepsilon/(2B+1), & B=0 \end{cases}$$ Because $$\{a_n\}$$ is a convergent sequence, it is also bounded.
 * $$M:=\sup{\{a_n\}}$$

By (2),
 * $$\exists N_2\in\mathbb{N}:\forall n\geq N_2, |b_n-B|<

\begin{cases} \varepsilon/(2M), & M\neq 0 \\ \varepsilon/(2M+1), & M=0 \end{cases}$$ Using the above conclusions,
 * $$N:=\max{\{N_1,N_2\}}\,$$
 * $$\forall n\geq N,|a_nb_n-AB|\leq|a_n||b_n-B|+|B||a_n-A|<

\begin{cases} B\left(\frac{\varepsilon}{2B}\right)+M\left(\frac{\varepsilon}{2M}\right)=\varepsilon, & B,M\neq 0 \\ B\left(\frac{\varepsilon}{2B+1}\right)+M\left(\frac{\varepsilon}{2M+1}\right)<\varepsilon, & B,M=0 \end{cases}$$
 * $$|a_nb_n-AB|<\varepsilon$$
 * $$\Rightarrow \lim_{n\to\infty}a_nb_n=\left(\lim_{n\to\infty}a_n\right)\left(\lim_{n\to\infty}b_n\right)=AB$$

Q.E.D.

Subtraction and Division
The structures of limit subtraction and division follow from those of limit addition and multiplication by considering additive and multiplicative inverses.

A conclusion considering multiplicative inverses is proven below.

Multiplicative Inverse
If
 * $$\lim_{n\to\infty}a_n=A\land a_n,A\neq 0\qquad (1),$$

then
 * $$\lim_{n\to\infty}\frac{1}{a_n}=\frac{1}{A}$$

Proof

 * $$\left|\frac{1}{a_n}-\frac{1}{A}\right|=\frac{|a_n-A|}{|A||a_n|}$$

Assume
 * $$\varepsilon\in\mathbb{R}^+$$

Because $$\{a_n\}$$ is a convergent sequence, so too is it bounded.
 * $$M:=\sup{\{a_n\}}$$

By (1),
 * $$\exists N\in\mathbb{N}:\forall n\geq N, |a_n-A|<AM\varepsilon$$
 * $$\forall n\geq N, \left|\frac{1}{a_n}-\frac{1}{A}\right|=\frac{|a_n-A|}{|A||a_n|}<\frac{AM\varepsilon}{AM}=\varepsilon$$

Q.E.D.

Main Idea
Assume
 * $$\lim_{n\to\infty}a_n=A\qquad (1),$$
 * $$\lim_{n\to\infty}b_n=B\qquad (2).$$

If
 * $$\forall n\in\mathbb{N}, a_n\leq b_n,$$

then
 * $$\lim_{n\to\infty}a_n\leq\lim_{n\to\infty}b_n.$$

Proof

 * $$0\leq b_n-a_n=(b_n-B)+(A-a_n)+B-A$$

By (1),
 * $$\forall\varepsilon_1\in\mathbb{R}^+,\exists N_1\in\mathbb{N}:\forall n\geq N_1, |a_n-A|<\varepsilon_1.$$

By (2),
 * $$\forall\varepsilon_2\in\mathbb{R}^+,\exists N_2\in\mathbb{N}:\forall n\geq N_2, |b_n-B|<\varepsilon_2.$$
 * $$\varepsilon:=\varepsilon_1+\varepsilon_2$$
 * $$N:=\max{\{N_1,N_2\}}\,$$
 * $$\forall n\geq N,0\leq b_n-a_n<\varepsilon+B-A$$
 * $$\Rightarrow-\varepsilon\leq B-A.$$

Since $$\varepsilon$$ can be any positive number, it follows that $$B-A$$ is positive, and thus
 * $$A\leq B.$$

Q.E.D.

Corollary
If
 * $$\alpha,\beta\in\mathbb{R}\land\forall n\in\mathbb{N},\alpha0$$.
 * $$\exists N_1: \forall n\geq N_1,c_n-b_n<\varepsilon_1+\varepsilon_2+L-B$$
 * $$-(\varepsilon_1+\varepsilon_2)\leq L-B$$