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# Proof of the Limit Theorems

For each of the four proofs let $\epsilon >0$ be arbitrary.

1.

As ${a}_{n}\to a$  we find an  ${n}_{1}\in {ℕ}^{\ast }$ for $\frac{\epsilon }{2}>0$ such that . Similar we get an ${n}_{2}\in {ℕ}^{\ast }$ with .

Thus for all $n\ge {n}_{0}≔\mathrm{max}\left\{{n}_{1},{n}_{2}\right\}$  the following holds due to the triangle inequality:

$|{a}_{n}+{b}_{n}-\left(a+b\right)|=|{a}_{n}-a+{b}_{n}-b|\le |{a}_{n}-a|+|{b}_{n}-b|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \text{.}$

2.

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

3.

Proving the third limit theorem is a more elaborate task. At first we note that $\left({a}_{n}\right)$ is bounded (cf. [5.5.1]), so that there is an $s\in {ℝ}^{>0}$ with .

As ${a}_{n}\to a$  we find an  ${n}_{1}\in {ℕ}^{\ast }$ for $\frac{\epsilon }{2\left(|b|+1\right)}>0$ such that

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

With these inequlities the following estimate holds for all $n\ge {n}_{0}≔\mathrm{max}\left\{{n}_{1},{n}_{2}\right\}$ :

$\begin{array}{ll}|{a}_{n}{b}_{n}-ab|=|{a}_{n}{b}_{n}-{a}_{n}b+{a}_{n}b-ab|\hfill & \le |{a}_{n}\left({b}_{n}-b\right)|+|b\left({a}_{n}-a\right)|\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 &

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

4.

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

 ${b}_{n}\to b⇒\frac{1}{{b}_{n}}\to \frac{1}{b}\text{,}$ 

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

Now let's prove . At first we get an ${n}_{1}\in {ℕ}^{\ast }$ such that

 , 

because: As ${b}_{n}\to b$ there is an  ${n}_{1}\in {ℕ}^{\ast }$ for $\frac{|b|}{2}>0$ with . Using the second triangle inequality we get for these n$|b|-|{b}_{n}|\le |b-{b}_{n}|<\frac{|b|}{2}$  and thus

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

 

Combining  and  now allows the following estimate for all  $n\ge {n}_{0}≔\mathrm{max}\left\{{n}_{1},{n}_{2}\right\}$ :

$|\frac{1}{{b}_{n}}-\frac{1}{b}|=|\frac{b-{b}_{n}}{{b}_{n}\cdot b}|\le \frac{|b-{b}_{n}|}{\frac{|b|}{2}\cdot |b|}=\frac{2|b-{b}_{n}|}{{b}^{2}}<\frac{2\epsilon {b}^{2}}{2{b}^{2}}=\epsilon$.