# Real numbers/Convergent sequences/Rules/Section
## Lemma

Let ${\left(x_{n}\right)}_{n\in \mathbb {N} }$ and ${\left(y_{n}\right)}_{n\in \mathbb {N} }$ be
* The sequence ${\left(x_{n}+y_{n}\right)}_{n\in \mathbb {N} }$ is convergent, and holds.
    $\lim _{n\rightarrow \infty }{\left(x_{n}+y_{n}\right)}={\left(\lim _{n\rightarrow \infty }x_{n}\right)}+{\left(\lim _{n\rightarrow \infty }y_{n}\right)}\,$
* The sequence ${\left(x_{n}\cdot y_{n}\right)}_{n\in \mathbb {N} }$ is convergent, and holds.
    $\lim _{n\rightarrow \infty }{\left(x_{n}\cdot y_{n}\right)}={\left(\lim _{n\rightarrow \infty }x_{n}\right)}\cdot {\left(\lim _{n\rightarrow \infty }y_{n}\right)}\,$
* For $c\in \mathbb {R}$, we have
    $\lim _{n\rightarrow \infty }cx_{n}=c{\left(\lim _{n\rightarrow \infty }x_{n}\right)}\,.$
* Suppose that $\lim _{n\rightarrow \infty }x_{n}=x\neq 0$ and $x_{n}\neq 0$ for all $n\in \mathbb {N}$. Then $\left({\frac {1}{x_{n}}}\right)_{n\in \mathbb {N} }$ is also convergent, and holds.
    $\lim _{n\rightarrow \infty }{\frac {1}{x_{n}}}={\frac {1}{x}}\,$
* Suppose that $\lim _{n\rightarrow \infty }x_{n}=x\neq 0$ and that $x_{n}\neq 0$ for all $n\in \mathbb {N}$. Then $\left({\frac {y_{n}}{x_{n}}}\right)_{n\in \mathbb {N} }$ is also convergent, and holds.
    $\lim _{n\rightarrow \infty }{\frac {y_{n}}{x_{n}}}={\frac {\lim _{n\rightarrow \infty }y_{n}}{x}}\,$

### Proof

(1). Denote the limits of the sequences by $x$ and $y$, respectively. Let $\epsilon >0$ be given. Due to the convergence of the first sequence, there exists for
    $\epsilon '={\frac {\epsilon }{2}}\,$
some $n_{0}$ such that for all $n\geq n_{0}$ the estimate
    $\vert {x_{n}-x}\vert \leq \epsilon '\,$
holds. In the same way there exists due to the convergence of the second sequence for $\epsilon '={\frac {\epsilon }{2}}$ some $n_{0}'$ such that for all $n\geq n_{0}'$ the estimate
    $\vert {y_{n}-y}\vert \leq \epsilon '\,$
holds. Set
    $N={\max {\left(n_{0},n_{0}'\right)}}\,.$
Then for all $n\geq N$ the estimate
    ${\begin{aligned}\vert {x_{n}+y_{n}-(x+y)}\vert &=\vert {x_{n}+y_{n}-x-y}\vert \\&=\vert {x_{n}-x+y_{n}-y}\vert \\&\leq \vert {x_{n}-x}\vert +\vert {y_{n}-y}\vert \\&\leq \epsilon '+\epsilon '\\&=\epsilon \end{aligned}}$
holds.
(2). Let $\epsilon >0$ be given. The convergent sequence ${\left(x_{n}\right)}_{n\in \mathbb {N} }$ is bounded, due to fact, and therefore there exists a $D>0$ such that $\vert {x_{n}}\vert \leq D$ for all $n\in \mathbb {N}$. Set $x:=\lim _{n\rightarrow \infty }x_{n}$ and $y:=\lim _{n\rightarrow \infty }y_{n}$. We put $C:={\max {\left(D,\vert {y}\vert \right)}}$. Because of the convergence, there are natural numbers $N_{1}$ and $N_{2}$ such that
    $\vert {x_{n}-x}\vert \leq {\frac {\epsilon }{2C}}{\text{ for }}n\geq N_{1}{\text{ and }}\vert {y_{n}-y}\vert \leq {\frac {\epsilon }{2C}}{\text{ for }}n\geq N_{2}.$
These estimates hold also for all $n\geq N:={\max {\left(N_{1},N_{2}\right)}}$. For these numbers, the estimates
    ${\begin{aligned}\vert {x_{n}y_{n}-xy}\vert &=\vert {x_{n}y_{n}-x_{n}y+x_{n}y-xy}\vert \\&\leq \vert {x_{n}y_{n}-x_{n}y}\vert +\vert {x_{n}y-xy}\vert \\&=\vert {x_{n}}\vert \vert {y_{n}-y}\vert +\vert {y}\vert \vert {x_{n}-x}\vert \\&\leq C{\frac {\epsilon }{2C}}+C{\frac {\epsilon }{2C}}\\&=\epsilon \end{aligned}}$
hold.
For the other parts, see exercise, exercise and exercise.
$\Box$
We give a typical application of this statement.

## Example

We consider the sequence given by
    $x_{n}={\frac {-5n^{3}+6n^{2}-n+8}{11n^{3}+7n^{2}+3n-1}}\,,$
and want to know whether it converges and if so, what the limit is. We can not use fact immediately, as neither the numerator nor the denominator converges. However, we can use the following trick. We write
    $x_{n}={\frac {-5n^{3}+6n^{2}-n+8}{11n^{3}+7n^{2}+3n-1}}={\frac {{\left(-5n^{3}+6n^{2}-n+8\right)}{\frac {1}{n^{3}}}}{{\left(11n^{3}+7n^{2}+3n-1\right)}{\frac {1}{n^{3}}}}}={\frac {-5+{\frac {6}{n}}-{\frac {1}{n^{2}}}+{\frac {8}{n^{3}}}}{11+{\frac {7}{n}}+{\frac {3}{n^{2}}}-{\frac {1}{n^{3}}}}}\,.$
In this form, the numerator and the denominator converges, and the limits are $-5$ and $11$ respectively. Therefore, the sequence converges to $-{\frac {5}{11}}$.