Proof of the Algebraic Limit Theorem

Statement can be found here.

1. ${\lim }_{n\to \infty }(c{a}_n)=ca$

Assume $c$ is not 0 (constant sequences converge). We know that ${\lim }_{n\to \infty }{a}_n=a$. Thus, we can choose $N\in \mathbb{N}$ such that for all $n\ge N$, $|a_n-a|<\frac{ϵ}{|c|}$, for some arbitrary positive number $ϵ$. Then we have

$$\begin{array}{rl}& |c||a_n-a|<ϵ\\ ⟹& |c{a}_n-ca|<ϵ\forall n\ge N\end{array}$$

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2. ${\lim }_{n\to \infty }(a_n+b_n)=a+b$

Let $ϵ>0$ be arbitrary. Since $(a_n)\to a$ and $(b_n)\to b$, we can choose $N_1\in \mathbb{N}$ and $N_2\in \mathbb{N}$ such that

$$\begin{array}{rl}|a_n-a|<\frac{ϵ}{2}& \forall n\ge N_1,\text{ and }\\ |b_n-b|<\frac{ϵ}{2}& \forall n\ge N_2\end{array}$$

Let $N=\max \{N_1,N_2\}$. From the triangle inequality, we have

$$\begin{array}{rl}|(a_n+b_n)-(a+b)|& =|(a_n-a)+(b_n-b)|\\ & \le |a_n-a|+|b_n-b|\\ & <\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ\end{array}$$

for all $n\ge N$. ❏

Corollary

${\lim }_{n\to \infty }{a}_n=a$ is the same as ${\lim }_{n\to \infty }(a_n-a)=0$.

3. ${\lim }_{n\to \infty }(a_n{b}_n)=ab$

We have to show that ${\lim }_{n\to \infty }(a_n{b}_n-ab)=0$.

$$a_n{b}_n-ab=(a_n-a)(b_n-b)+b(a_n-a)+a(b_n-b)$$

Using parts (1) and (2) and the above equality, we have

$$\begin{array}{rl}& \underset{n\to \infty }{\lim }(a_n{b}_n-ab)\\ =& \underset{n\to \infty }{\lim }(a_n-a)(b_n-b)\end{array}$$

Let $ϵ>0$ be arbitrary. Since $(a_n)\to a$ and $(b_n)\to b$, we can choose $N_1\in \mathbb{N}$ and $N_2\in \mathbb{N}$ such that

$$\begin{array}{rl}|a_n-a|<\sqrt{ϵ}& \forall n\ge N_1,\text{ and }\\ |b_n-b|<\sqrt{ϵ}& \forall n\ge N_2\end{array}$$

Let $N=\max \{N_1,N_2\}$. Multiplying the two inequalities gives us

$$\begin{array}{rl}& |(a_n-a)(b_n-b)-0|<ϵ\forall n\ge N\\ & \therefore \underset{n\to \infty }{\lim }(a_n-a)(b_n-b)=0\end{array}$$

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4. ${\lim }_{n\to \infty }\left(\frac{a_n}{b_n}\right)=\frac{a}{b}$, provided $b\ne 0$.

This will follow from (3) if we can prove that

$$(b_n)\to b⟹\left(\frac{1}{b_n}\right)\to \frac{1}{b}$$

whenever $b\ne 0$.
Observe that

$$\left|\frac{1}{b_n}-\frac{1}{b}\right|=\frac{|b-b_n|}{|b||b_n|}.$$

Because $(b_n)\to b$, we can make the numerator arbitrarily small. Now, it would be great if we could come up with a number $M$ such that

$$\frac{1}{|b||b_n|}<M$$

for all $n>N_1$, for some $N_1$. In this vein, let $N_1$ be such that for all $n\ge N_1$,

$$|b_n-b|<\frac{|b|}{2}.$$

From the triangle inequality,

$$\begin{array}{rl}|b|& \le |b-b_n|+|b_n|\\ & <\frac{|b|}{2}+|b_n|\\ & \\ ⟹& |b_n|>\frac{|b|}{2}.\end{array}$$

Thus, we have

$$\frac{1}{|b||b_n|}<\frac{2}{|b{|}^2}$$

for all $n\ge N_1$.
Now, let $ϵ>0$ be arbitrary. Let $N_2$ be such that for all $n\ge N_2$,

$$|b-b_n|<\frac{ϵ|b{|}^2}{2}.$$

Let $N=\max \{N_1,N_2\}$. Multiplying the preceding inequalities, we get

$$\frac{|b-b_n|}{|b||b_n|}<ϵ$$

for all $n\ge N$. ❏