Proof of the Limit Theorems

As $a_n\to a$  we find an  $n_1\in {\mathbb{N}}^{\ast }$ for ${\displaystyle \frac{\epsilon }{2}>0}$ such that ${\displaystyle |a_n-a|<\frac{\epsilon }{2}\text{ for all }n\ge n_1}$. Similar we get an $n_2\in {\mathbb{N}}^{\ast }$ with ${\displaystyle |b_n-b|<\frac{\epsilon }{2}\text{ for all }n\ge n_2}$.

Thus for all $n\ge n_0≔\max \{n_1,n_2\}$  the following holds due to the triangle inequality:

${\displaystyle |a_n+b_n-(a+b)|=|a_n-a+b_n-b|\le |a_n-a|+|b_n-b|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \text{.}}$
 

The proof is just a slight modification of the above arguments.
 

Proving the third limit theorem is a more elaborate task. At first we note that $(a_n)$ is bounded (cf. [5.5.1]), so that there is an $s\in {ℝ}^{>0}$ with $|a_n|\le s\text{ for all }n\in {\mathbb{N}}^{\ast }$.

As $a_n\to a$  we find an  $n_1\in {\mathbb{N}}^{\ast }$ for ${\displaystyle \frac{\epsilon }{2(|b|+1)}>0}$ such that

Furthermore there is an  $n_2\in {\mathbb{N}}^{\ast }$ for ${\displaystyle \frac{\epsilon }{2s}>0}$ so that

With these inequlities the following estimate holds for all $n\ge n_0≔\max \{n_1,n_2\}$ :

${\displaystyle \begin{array}{ll}|a_n{b}_n-ab|=|a_n{b}_n-a_{n}b+a_{n}b-ab|\hfill & \le |a_n(b_n-b)|+|b(a_n-a)|\hfill \\ \hfill & =|a_n|\cdot |b_n-b|+|b|\cdot |a_n-a|\hfill \\ \hfill & \le s\cdot |b_n-b|+|b|\cdot |a_n-a|\hfill \\ \hfill & <s\cdot \frac{\epsilon }{2s}+|b|\cdot \frac{\epsilon }{2(|b|+1)}\hfill \\ \hfill & <\frac{\epsilon }{2}+\frac{\epsilon }{2}\hfill \\ \hfill & =\epsilon \text{.}\hfill \end{array}}$

Note that the last estimate is correct due to ${\displaystyle \frac{|b|}{|b|+1}<1}$. (By the way: The strange addend 1 in the denominator only serves to cope with possibility $b=0$.)
 

According to a remark on the main page we suppose without restriction that $b_n\ne 0$ holds for all n. Furthermore it is sufficient to prove only the special case

namely because the third limit theorem will then provide:   ${\displaystyle \frac{a_n}{b_n}=\frac{1}{b_n}\cdot a_n\to \frac{1}{b}\cdot a=\frac{a}{b}}$ .

Now let's prove [1]. At first we get an $n_1\in {\mathbb{N}}^{\ast }$ such that

because: As $b_n\to b$ there is an  $n_1\in {\mathbb{N}}^{\ast }$ for ${\displaystyle \frac{|b|}{2}>0}$ with ${\displaystyle |b-b_n|<\frac{|b|}{2}\text{ for all }n\ge n_1}$. Using the second triangle inequality we get for these n:  ${\displaystyle |b|-|b_n|\le |b-b_n|<\frac{|b|}{2}}$  and thus

At second the convergence $b_n\to b$ provides us with an  $n_2\in {\mathbb{N}}^{\ast }$ for ${\displaystyle \frac{\epsilon b^2}{2}>0}$ such that

Combining [2] and [3] now allows the following estimate for all  $n\ge n_0≔\max \{n_1,n_2\}$ :