editar
Toda sequência convergente é limitada.
Seja
x
n
{\displaystyle x_{n}}
c
∈
Q
⇒
∀
ϵ
>
0
∃
n
0
∈
N
{\displaystyle c\in \mathbb {Q} \Rightarrow \forall \;\epsilon >0\;\exists \;n_{0}\in \mathbb {N} }
n
>
n
0
⇒
x
n
∈
(
c
−
ϵ
,
c
+
ϵ
)
{\displaystyle n>n_{0}\Rightarrow x_{n}\in (c-\epsilon ,c+\epsilon )}
Tomando o conjunto
A
=
{
x
1
,
x
2
,
.
.
.
,
x
n
0
,
c
−
ϵ
,
c
+
ϵ
}
{\displaystyle A=\{x_{1},x_{2},...,x_{n_{0}},c-\epsilon ,c+\epsilon \}}
a
=
m
i
n
A
{\displaystyle a=minA}
b
=
m
a
x
A
{\displaystyle b=maxA}
Temos que
∀
n
∈
N
,
x
n
∈
(
a
,
b
)
{\displaystyle \forall n\in \mathbb {N} ,x_{n}\in (a,b)}
Se
(
x
n
)
{\displaystyle (x_{n})}
c
>
0
,
c
∈
Q
{\displaystyle c>0,c\in \mathbb {Q} }
|
x
n
|
<
c
{\displaystyle |x_{n}|<c}
Tome
A
=
{
x
n
∈
(
x
n
)
:
n
∈
N
}
{\displaystyle A=\{x_{n}\in (x_{n}):n\in \mathbb {N} \}}
a
,
b
∈
Q
{\displaystyle a,b\in \mathbb {Q} }
a
<
x
n
<
b
,
∀
n
∈
N
{\displaystyle a<x_{n}<b,\forall \;n\in \mathbb {N} }
Tome
c
=
max
{
|
a
|
,
|
b
|
}
⇒
−
c
<
x
n
<
c
⇒
|
x
n
|
<
c
{\displaystyle c=\max\{|a|,|b|\}\Rightarrow -c<x_{n}<c\Rightarrow |x_{n}|<c}
Toda sequência de Cauchy é limitada.
Seja
x
n
{\displaystyle x_{n}}
∀
ϵ
>
0
∃
n
0
∈
N
{\displaystyle \forall \;\epsilon >0\;\exists \;n_{0}\in \mathbb {N} }
m
,
n
≥
n
0
⇒
|
x
m
−
x
n
|
<
ϵ
⇒
x
n
∈
(
x
n
0
−
ϵ
,
x
n
0
+
ϵ
)
{\displaystyle m,n\geq n_{0}\Rightarrow |x_{m}-x_{n}|<\epsilon \Rightarrow x_{n}\in (x_{n_{0}}-\epsilon ,x_{n_{0}}+\epsilon )}
Tomando
A
=
{
x
1
,
x
2
,
.
.
.
,
x
n
0
,
x
n
0
−
ϵ
,
x
n
0
+
ϵ
}
{\displaystyle A=\{x_{1},x_{2},...,x_{n_{0}},x_{n_{0}}-\epsilon ,x_{n_{0}}+\epsilon \}}
a
=
m
i
n
A
,
b
=
m
a
x
A
{\displaystyle a=minA,b=maxA}
Tome
c
=
max
{
a
,
b
}
,
⇒
−
c
<
x
n
<
c
⇒
|
x
n
|
<
c
,
∀
n
∈
N
{\displaystyle c=\max\{a,b\},\Rightarrow -c<x_{n}<c\Rightarrow |x_{n}|<c,\forall \;n\in \mathbb {N} }
Se
lim
x
n
=
0
{\displaystyle \lim x_{n}=0}
(
y
n
)
{\displaystyle (y_{n})}
lim
x
n
⋅
y
n
=
0
{\displaystyle \lim x_{n}\cdot y_{n}=0}
lim
y
n
{\displaystyle \lim y_{n}}
Como
(
y
n
)
{\displaystyle (y_{n})}
c
>
0
{\displaystyle c>0}
|
y
n
|
<
c
{\displaystyle |y_{n}|<c}
n
∈
N
{\displaystyle n\in \mathbb {N} }
ϵ
>
0
{\displaystyle \epsilon >0}
lim
x
n
=
0
{\displaystyle \lim x_{n}=0}
n
0
∈
N
{\displaystyle n_{0}\in \mathbb {N} }
n
>
n
0
⇒
|
x
n
|
<
ϵ
c
{\displaystyle n>n_{0}\Rightarrow |x_{n}|<{\epsilon \over c}}
Logo,
n
>
n
0
⇒
|
x
n
⋅
y
n
|
=
|
x
n
|
⋅
|
y
n
|
<
ϵ
c
⋅
c
=
ϵ
{\displaystyle n>n_{0}\Rightarrow |x_{n}\cdot y_{n}|=|x_{n}|\cdot |y_{n}|<{\epsilon \over c}\cdot c=\epsilon }
lim
x
n
⋅
y
n
=
0
{\displaystyle \lim x_{n}\cdot y_{n}=0}
lim
x
n
=
a
⇔
lim
x
n
−
a
=
0
{\displaystyle \lim x_{n}=a\Leftrightarrow \lim x_{n}-a=0}
(
⇒
{\displaystyle \Rightarrow }
lim
x
n
=
a
⇒
∀
ϵ
>
0
,
∃
n
0
∈
N
{\displaystyle \lim x_{n}=a\Rightarrow \forall \;\epsilon >0,\exists \;n_{0}\in \mathbb {N} }
n
>
n
0
⇒
|
x
n
−
a
|
<
ϵ
⇒
|
(
x
n
−
a
)
−
0
|
<
ϵ
⇒
lim
(
x
n
−
a
)
=
0
{\displaystyle n>n_{0}\Rightarrow |x_{n}-a|<\epsilon \Rightarrow |(x_{n}-a)-0|<\epsilon \Rightarrow \lim(x_{n}-a)=0}
(
⇐
{\displaystyle \Leftarrow }
lim
x
n
−
a
=
0
⇒
∀
ϵ
>
0
,
∃
n
0
∈
N
{\displaystyle \lim x_{n}-a=0\Rightarrow \forall \;\epsilon >0,\exists \;n_{0}\in \mathbb {N} }
n
>
n
0
⇒
|
(
x
n
−
a
)
−
0
|
<
ϵ
⇒
|
x
n
−
a
|
<
ϵ
⇒
lim
x
n
=
a
{\displaystyle n>n_{0}\Rightarrow |(x_{n}-a)-0|<\epsilon \Rightarrow |x_{n}-a|<\epsilon \Rightarrow \lim x_{n}=a}
Mostre que a sequência
x
n
=
1
+
1
2
+
1
3
+
.
.
.
+
1
n
{\displaystyle x_{n}=1+{1 \over 2}+{1 \over 3}+...+{1 \over n}}
Q
{\displaystyle \mathbb {Q} }
Primeiro vamos exibir o valor de
x
n
{\displaystyle x_{n}}
x
n
=
1
+
1
2
+
(
1
3
+
1
4
)
+
(
1
5
+
1
6
+
1
7
+
1
8
)
+
.
.
.
+
(
1
2
j
−
1
+
1
+
1
2
j
−
1
+
2
+
.
.
.
+
1
2
j
)
+
.
.
.
+
1
n
{\displaystyle x_{n}=1+{1 \over 2}+{\bigg (}{1 \over 3}+{1 \over 4}{\bigg )}+{\bigg (}{1 \over 5}+{1 \over 6}+{1 \over 7}+{1 \over 8}{\bigg )}+...+{\bigg (}{1 \over 2^{j-1}+1}+{1 \over 2^{j-1}+2}+...+{1 \over 2^{j}}{\bigg )}+...+{1 \over n}}
x
n
>
1
+
1
2
+
(
1
4
+
1
4
)
+
(
1
8
+
1
8
+
1
8
+
1
8
)
+
.
.
.
+
(
1
2
j
+
1
2
j
+
.
.
.
+
1
2
j
)
+
.
.
.
+
1
n
{\displaystyle x_{n}>1+{1 \over 2}+{\bigg (}{1 \over 4}+{1 \over 4}{\bigg )}+{\bigg (}{1 \over 8}+{1 \over 8}+{1 \over 8}+{1 \over 8}{\bigg )}+...+{\bigg (}{1 \over 2^{j}}+{1 \over 2^{j}}+...+{1 \over 2^{j}}{\bigg )}+...+{1 \over n}}
x
n
>
1
+
1
2
+
2
⋅
1
4
+
4
⋅
1
8
+
.
.
.
+
2
j
−
1
⋅
1
2
j
+
.
.
.
+
1
n
{\displaystyle x_{n}>1+{1 \over 2}+2\cdot {1 \over 4}+4\cdot {1 \over 8}+...+2^{j-1}\cdot {1 \over 2^{j}}+...+{1 \over n}}
Considere
j
=
m
a
x
{
y
∈
N
:
2
y
<
n
<
2
y
+
1
:
n
∈
N
}
{\displaystyle j=max\{y\in \mathbb {N} :2^{y}<n<2^{y+1}:n\in \mathbb {N} \}}
t
=
n
−
2
j
⇒
2
j
=
n
−
t
⇒
j
l
o
g
2
=
l
o
g
(
n
−
t
)
⇒
j
=
l
o
g
2
(
n
−
t
)
≈
l
o
g
2
n
{\displaystyle t=n-2^{j}\Rightarrow 2^{j}=n-t\Rightarrow jlog2=log(n-t)\Rightarrow j=log_{2}(n-t)\approx log_{2}n}
x
n
>
1
+
j
⋅
1
2
+
1
2
j
+
1
+
1
2
j
+
2
+
.
.
.
+
1
n
>
1
+
j
2
=
2
+
j
2
{\displaystyle x_{n}>1+j\cdot {1 \over 2}+{1 \over 2^{j+1}}+{1 \over 2^{j+2}}+...+{1 \over n}>1+{j \over 2}={2+j \over 2}}
j
=
l
o
g
2
n
,
x
n
>
2
+
l
o
g
2
n
2
{\displaystyle j=log_{2}n,x_{n}>{2+log_{2}n \over 2}}
Qualquer que seja
m
>
0
,
m
∈
Q
,
∃
n
∈
N
,
{\displaystyle m>0,m\in \mathbb {Q} ,\exists \;n\in \mathbb {N} ,}
2
+
l
o
g
2
n
2
>
m
{\displaystyle {2+log_{2}n \over 2}>m}
x
n
>
m
{\displaystyle x_{n}>m}
x
n
{\displaystyle x_{n}}
Como
x
n
+
1
=
x
n
+
1
n
+
1
⇒
x
n
<
x
n
+
1
,
∀
n
∈
N
{\displaystyle x_{n+1}=x_{n}+{1 \over n+1}\Rightarrow x_{n}<x_{n+1},\forall \;n\in \mathbb {N} }
x
n
{\displaystyle x_{n}}
Portanto
x
n
{\displaystyle x_{n}}
A sequência
x
n
{\displaystyle x_{n}}
Toda sequência de Cauchy é limitada. Como
x
n
{\displaystyle x_{n}}
x
n
{\displaystyle x_{n}}
Sejam
lim
x
n
=
a
,
lim
y
n
=
b
{\displaystyle \lim x_{n}=a,\lim y_{n}=b}
lim
x
n
+
y
n
=
a
+
b
{\displaystyle \lim x_{n}+y_{n}=a+b}
Como
lim
x
n
=
a
,
lim
y
n
=
b
{\displaystyle \lim x_{n}=a,\lim y_{n}=b}
∀
ϵ
>
0
,
∃
n
1
,
n
2
∈
N
{\displaystyle \forall \;\epsilon >0,\exists \;n_{1},n_{2}\in \mathbb {N} }
n
>
n
1
⇒
|
x
n
−
x
n
1
|
<
ϵ
2
{\displaystyle n>n_{1}\Rightarrow |x_{n}-x_{n_{1}}|<{\epsilon \over 2}}
n
>
n
2
⇒
|
y
n
−
y
n
2
|
<
ϵ
2
{\displaystyle n>n_{2}\Rightarrow |y_{n}-y_{n_{2}}|<{\epsilon \over 2}}
Tome
n
0
=
m
a
x
{
n
1
,
n
2
}
{\displaystyle n_{0}=max\{n_{1},n_{2}\}}
n
>
n
0
⇒
n
>
n
1
{\displaystyle n>n_{0}\Rightarrow n>n_{1}}
n
>
n
2
{\displaystyle n>n_{2}}
Assim
|
x
n
+
y
n
−
(
a
+
b
)
|
=
|
(
x
n
−
a
)
+
(
y
n
−
b
)
|
≤
|
(
x
n
−
a
)
|
+
|
(
y
n
−
b
)
|
<
ϵ
2
+
ϵ
2
=
ϵ
{\displaystyle |x_{n}+y_{n}-(a+b)|=|(x_{n}-a)+(y_{n}-b)|\leq |(x_{n}-a)|+|(y_{n}-b)|<{\epsilon \over 2}+{\epsilon \over 2}=\epsilon }
Sejam
lim
x
n
=
a
,
lim
y
n
=
b
{\displaystyle \lim x_{n}=a,\lim y_{n}=b}
lim
x
n
⋅
y
n
=
a
⋅
b
{\displaystyle \lim x_{n}\cdot y_{n}=a\cdot b}
Temos que
lim
x
n
=
a
⇒
lim
(
x
n
−
a
)
=
0
{\displaystyle \lim x_{n}=a\Rightarrow \lim(x_{n}-a)=0}
y
n
{\displaystyle y_{n}}
lim
(
x
n
−
a
)
⋅
y
n
=
0
{\displaystyle \lim(x_{n}-a)\cdot y_{n}=0}
∀
ϵ
>
0
,
∃
n
1
∈
N
{\displaystyle \forall \;\epsilon >0,\exists n_{1}\in \mathbb {N} }
n
>
n
1
⇒
|
(
x
n
−
a
)
⋅
y
n
|
<
ϵ
2
{\displaystyle n>n_{1}\Rightarrow |(x_{n}-a)\cdot y_{n}|<{\epsilon \over 2}}
Tome
lim
y
n
=
b
{\displaystyle \lim y_{n}=b}
∀
ϵ
>
0
,
ϵ
∈
Q
,
∃
n
2
∈
N
{\displaystyle \forall \;\epsilon >0,\epsilon \in \mathbb {Q} ,\exists n_{2}\in \mathbb {N} }
n
>
n
2
⇒
|
y
n
−
b
|
<
ϵ
2
|
a
|
⇒
|
y
n
−
b
|
⋅
|
a
|
<
ϵ
2
{\displaystyle n>n_{2}\Rightarrow |y_{n}-b|<{\epsilon \over 2|a|}\Rightarrow |y_{n}-b|\cdot |a|<{\epsilon \over 2}}
Tome
n
0
=
max
{
n
1
,
n
2
}
{\displaystyle n_{0}=\max\{n_{1},n_{2}\}}
n
>
n
0
⇒
n
>
n
1
{\displaystyle n>n_{0}\Rightarrow n>n_{1}}
n
>
n
2
{\displaystyle n>n_{2}}
Assim
|
x
n
⋅
y
n
−
(
a
⋅
b
)
|
=
|
x
n
⋅
y
n
−
a
⋅
y
n
+
a
⋅
y
n
−
b
a
|
=
|
(
x
n
−
a
)
⋅
y
n
+
(
y
n
−
b
)
⋅
a
|
≤
{\displaystyle |x_{n}\cdot y_{n}-(a\cdot b)|=|x_{n}\cdot y_{n}-a\cdot y_{n}+a\cdot y_{n}-ba|=|(x_{n}-a)\cdot y_{n}+(y_{n}-b)\cdot a|\leq }
≤
|
(
x
n
−
a
)
⋅
y
n
|
+
|
(
y
n
−
b
)
|
⋅
|
a
|
<
ϵ
2
+
ϵ
2
=
ϵ
{\displaystyle \leq |(x_{n}-a)\cdot y_{n}|+|(y_{n}-b)|\cdot |a|<{\epsilon \over 2}+{\epsilon \over 2}=\epsilon }
Sejam
(
x
n
)
,
(
y
n
)
{\displaystyle (x_{n}),(y_{n})}
(
x
n
+
y
n
)
{\displaystyle (x_{n}+y_{n})}
Como
(
x
n
)
,
(
y
n
)
{\displaystyle (x_{n}),(y_{n})}
∀
ϵ
>
0
,
ϵ
∈
Q
{\displaystyle \forall \;\epsilon >0,\epsilon \in \mathbb {Q} }
∃
n
1
∈
N
{\displaystyle \exists \;n_{1}\in \mathbb {N} }
m
,
n
>
n
1
⇒
|
x
m
−
x
n
|
<
ϵ
/
2
⇒
x
n
−
ϵ
/
2
<
x
m
<
x
n
+
ϵ
/
2
⇒
x
m
∈
(
x
n
−
ϵ
/
2
,
x
n
+
ϵ
/
2
)
{\displaystyle m,n>n_{1}\Rightarrow |x_{m}-x_{n}|<\epsilon /2\Rightarrow x_{n}-\epsilon /2<x_{m}<x_{n}+\epsilon /2\Rightarrow x_{m}\in (x_{n}-\epsilon /2,x_{n}+\epsilon /2)}
∃
n
2
∈
N
{\displaystyle \exists \;n_{2}\in \mathbb {N} }
m
,
n
>
n
2
⇒
|
y
m
−
y
n
|
<
ϵ
/
2
⇒
y
n
−
ϵ
/
2
<
y
m
<
y
n
+
ϵ
/
2
⇒
y
m
∈
(
y
n
−
ϵ
/
2
,
y
n
+
ϵ
/
2
)
{\displaystyle m,n>n_{2}\Rightarrow |y_{m}-y_{n}|<\epsilon /2\Rightarrow y_{n}-\epsilon /2<y_{m}<y_{n}+\epsilon /2\Rightarrow y_{m}\in (y_{n}-\epsilon /2,y_{n}+\epsilon /2)}
Tome
n
0
=
max
{
n
1
,
n
2
}
⇒
n
0
≥
n
1
,
n
0
≥
n
2
{\displaystyle n_{0}=\max\{n_{1},n_{2}\}\Rightarrow n_{0}\geq n_{1},n_{0}\geq n_{2}}
Assim
∀
n
>
n
0
⇒
|
x
n
−
x
n
0
|
<
ϵ
/
2
{\displaystyle \forall \;n>n_{0}\Rightarrow |x_{n}-x_{n_{0}}|<\epsilon /2}
|
y
n
−
y
n
0
|
<
ϵ
/
2
{\displaystyle |y_{n}-y_{n_{0}}|<\epsilon /2}
Assim
|
(
x
n
+
y
n
)
−
(
x
n
0
+
y
n
0
)
|
=
|
(
x
n
−
x
n
0
)
+
(
y
n
−
y
n
0
)
|
≤
|
x
n
−
x
n
0
|
+
|
y
n
−
y
n
0
|
<
ϵ
/
2
+
ϵ
/
2
=
ϵ
{\displaystyle |(x_{n}+y_{n})-(x_{n_{0}}+y_{n_{0}})|=|(x_{n}-x_{n_{0}})+(y_{n}-y_{n_{0}})|\leq |x_{n}-x_{n_{0}}|+|y_{n}-y_{n_{0}}|<\epsilon /2+\epsilon /2=\epsilon }
Sejam
(
x
n
)
,
(
y
n
)
{\displaystyle (x_{n}),(y_{n})}
(
x
n
⋅
y
n
)
{\displaystyle (x_{n}\cdot y_{n})}
Como
(
x
n
)
,
(
y
n
)
{\displaystyle (x_{n}),(y_{n})}
∀
ϵ
>
0
,
ϵ
∈
Q
{\displaystyle \forall \;\epsilon >0,\epsilon \in \mathbb {Q} }
∃
n
1
∈
N
{\displaystyle \exists \;n_{1}\in \mathbb {N} }
m
,
n
>
n
1
⇒
|
x
m
−
x
n
|
<
ϵ
2
{\displaystyle m,n>n_{1}\Rightarrow |x_{m}-x_{n}|<{\epsilon \over 2}}
∃
n
2
∈
N
{\displaystyle \exists \;n_{2}\in \mathbb {N} }
m
,
n
>
n
2
⇒
|
y
m
−
y
n
|
<
ϵ
2
{\displaystyle m,n>n_{2}\Rightarrow |y_{m}-y_{n}|<{\epsilon \over 2}}
Como
x
n
{\displaystyle x_{n}}
∃
c
>
0
,
c
∈
Q
{\displaystyle \exists \;c>0,c\in \mathbb {Q} }
|
x
n
|
<
c
,
∀
n
∈
N
{\displaystyle |x_{n}|<c,\forall \;n\in \mathbb {N} }
Como
y
n
{\displaystyle y_{n}}
∃
d
>
0
,
d
∈
Q
{\displaystyle \exists \;d>0,d\in \mathbb {Q} }
|
y
n
|
<
d
,
∀
n
∈
N
{\displaystyle |y_{n}|<d,\forall \;n\in \mathbb {N} }
Tome
n
0
=
max
{
n
1
,
n
2
}
⇒
n
0
≥
n
1
,
n
0
≥
n
2
{\displaystyle n_{0}=\max\{n_{1},n_{2}\}\Rightarrow n_{0}\geq n_{1},n_{0}\geq n_{2}}
Assim
∀
m
,
n
>
n
0
⇒
|
x
n
−
x
m
|
<
ϵ
2
d
{\displaystyle \forall \;m,n>n_{0}\Rightarrow |x_{n}-x_{m}|<{\epsilon \over 2d}}
|
y
n
−
y
m
|
<
ϵ
2
c
{\displaystyle |y_{n}-y_{m}|<{\epsilon \over 2c}}
Assim
|
(
x
n
⋅
y
n
)
−
(
x
m
⋅
y
m
)
|
=
|
(
x
n
⋅
y
n
)
−
(
x
m
⋅
y
n
)
+
(
x
m
⋅
y
n
)
−
(
x
m
⋅
y
m
)
|
=
{\displaystyle |(x_{n}\cdot y_{n})-(x_{m}\cdot y_{m})|=|(x_{n}\cdot y_{n})-(x_{m}\cdot y_{n})+(x_{m}\cdot y_{n})-(x_{m}\cdot y_{m})|=}
=
|
(
x
n
−
x
m
)
⋅
y
m
+
x
m
⋅
(
y
n
−
y
m
)
|
=
|
x
n
−
x
m
|
⋅
|
y
m
|
+
|
x
m
|
⋅
|
y
n
−
y
m
|
=
{\displaystyle =|(x_{n}-x_{m})\cdot y_{m}+x_{m}\cdot (y_{n}-y_{m})|=|x_{n}-x_{m}|\cdot |y_{m}|+|x_{m}|\cdot |y_{n}-y_{m}|=}
≤
ϵ
2
d
⋅
d
+
ϵ
2
c
⋅
c
<
ϵ
2
+
ϵ
2
=
ϵ
{\displaystyle \leq {\epsilon \over 2d}\cdot d+{\epsilon \over 2c}\cdot c<{\epsilon \over 2}+{\epsilon \over 2}=\epsilon }
Como
(
x
n
)
,
(
y
n
)
{\displaystyle (x_{n}),(y_{n})}
x
n
y
n
{\displaystyle {x_{n} \over y_{n}}}
Como
(
x
n
)
,
(
y
n
)
{\displaystyle (x_{n}),(y_{n})}
∀
ϵ
>
0
,
ϵ
∈
Q
{\displaystyle \forall \;\epsilon >0,\epsilon \in \mathbb {Q} }
∃
n
1
∈
N
{\displaystyle \exists \;n_{1}\in \mathbb {N} }
m
,
n
>
n
1
⇒
|
x
m
−
x
n
|
<
ϵ
2
e
{\displaystyle m,n>n_{1}\Rightarrow |x_{m}-x_{n}|<{\epsilon \over 2e}}
∃
n
2
∈
N
{\displaystyle \exists \;n_{2}\in \mathbb {N} }
m
,
n
>
n
2
⇒
|
y
m
−
y
n
|
<
ϵ
2
c
e
e
{\displaystyle m,n>n_{2}\Rightarrow |y_{m}-y_{n}|<{\epsilon \over 2cee}}
Tome
n
0
=
max
{
n
1
,
n
2
}
⇒
n
0
≥
n
1
,
n
0
≥
n
2
{\displaystyle n_{0}=\max\{n_{1},n_{2}\}\Rightarrow n_{0}\geq n_{1},n_{0}\geq n_{2}}
Como
x
n
{\displaystyle x_{n}}
∃
c
>
0
,
c
∈
Q
{\displaystyle \exists \;c>0,c\in \mathbb {Q} }
|
x
n
|
<
c
,
∀
n
∈
N
{\displaystyle |x_{n}|<c,\forall \;n\in \mathbb {N} }
Como
y
n
{\displaystyle y_{n}}
∃
d
>
0
,
d
∈
Q
{\displaystyle \exists \;d>0,d\in \mathbb {Q} }
|
y
n
|
<
d
,
∀
n
∈
N
{\displaystyle |y_{n}|<d,\forall \;n\in \mathbb {N} }
Mas em
y
n
{\displaystyle y_{n}}
|
y
n
|
>
1
e
⇒
1
|
y
n
|
<
e
,
∀
n
∈
N
{\displaystyle |y_{n}|>{1 \over e}\Rightarrow {1 \over |y_{n}|}<e,\forall \;n\in \mathbb {N} }
Assim
∀
m
,
n
>
n
0
⇒
|
x
n
−
x
m
|
<
ϵ
2
e
{\displaystyle \forall \;m,n>n_{0}\Rightarrow |x_{n}-x_{m}|<{\epsilon \over 2e}}
|
y
n
−
y
m
|
<
ϵ
2
c
e
e
{\displaystyle |y_{n}-y_{m}|<{\epsilon \over 2cee}}
Assim
|
x
n
y
n
−
x
m
y
m
|
=
|
(
x
n
⋅
y
m
)
−
(
x
m
⋅
y
n
)
y
n
⋅
y
m
|
=
|
(
x
n
⋅
y
m
−
x
m
⋅
y
m
)
+
(
x
m
⋅
y
m
−
x
m
⋅
y
n
)
y
n
⋅
y
m
|
{\displaystyle {\bigg |}{x_{n} \over y_{n}}-{x_{m} \over y_{m}}{\bigg |}={\bigg |}{(x_{n}\cdot y_{m})-(x_{m}\cdot y_{n}) \over y_{n}\cdot y_{m}}{\bigg |}={\bigg |}{(x_{n}\cdot y_{m}-x_{m}\cdot y_{m})+(x_{m}\cdot y_{m}-x_{m}\cdot y_{n}) \over y_{n}\cdot y_{m}}{\bigg |}}
=
|
(
x
n
−
x
m
)
⋅
y
m
+
x
m
⋅
(
y
m
−
y
n
)
y
n
⋅
y
m
|
=
|
x
n
−
x
m
y
n
|
+
|
x
m
⋅
(
y
n
−
y
m
)
y
n
⋅
y
m
|
=
{\displaystyle ={\bigg |}{(x_{n}-x_{m})\cdot y_{m}+x_{m}\cdot (y_{m}-y_{n}) \over y_{n}\cdot y_{m}}{\bigg |}={\bigg |}{x_{n}-x_{m} \over y_{n}}{\bigg |}+{\bigg |}{x_{m}\cdot (y_{n}-y_{m}) \over y_{n}\cdot y_{m}}{\bigg |}=}
=
|
x
n
−
x
m
|
⋅
|
1
y
n
|
+
|
x
m
|
⋅
|
y
n
−
y
m
|
⋅
|
1
y
n
|
⋅
|
1
y
m
|
=
{\displaystyle =|x_{n}-x_{m}|\cdot {\bigg |}{1 \over y_{n}}{\bigg |}+|x_{m}|\cdot |y_{n}-y_{m}|\cdot {\bigg |}{1 \over y_{n}}{\bigg |}\cdot {\bigg |}{1 \over y_{m}}{\bigg |}=}
<
ϵ
2
e
⋅
e
+
c
⋅
ϵ
2
c
e
e
⋅
e
⋅
e
=
ϵ
2
+
ϵ
2
=
ϵ
{\displaystyle <{\epsilon \over 2e}\cdot e+c\cdot {\epsilon \over 2cee}\cdot e\cdot e={\epsilon \over 2}+{\epsilon \over 2}=\epsilon }
![{\displaystyle [(x_{n})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35cd428abd515542d3427c9b6ef45e6d42529b66)
