نأخذ نفس المثال السابق
f
(
x
)
=
1
x
{\displaystyle f(x)={\frac {1}{x}}}
R
{\displaystyle \mathbb {R} }
0
{\displaystyle 0}
x
{\displaystyle x}
f
(
x
)
{\displaystyle f(x)}
لانهاية فان صورته تؤول نحو الصفر .
f
(
x
)
=
1
x
{\displaystyle f(x)={\frac {1}{x}}}
و نكتب:
lim
x
→
+
∞
1
x
=
0
{\displaystyle \lim _{x\to +\infty }{\frac {1}{x}}=0}
lim
x
→
+
∞
1
x
=
0
+
{\displaystyle \lim _{x\to +\infty }{\frac {1}{x}}=0^{+}}
لانهاية :
lim
x
→
−
∞
1
x
=
0
{\displaystyle \lim _{x\to -\infty }{\frac {1}{x}}=0}
lim
x
→
+
∞
1
x
=
0
+
{\displaystyle \lim _{x\to +\infty }{\frac {1}{x}}=0^{+}}
lim
|
x
|
→
+
∞
1
x
=
0
{\displaystyle \lim _{|x|\to +\infty }{\frac {1}{x}}=0}
f
{\displaystyle f}
l
{\displaystyle l}
+
∞
{\displaystyle +\infty }
∀
ε
>
0
,
∃
A
>
0
,
∀
x
∈
D
,
x
>
A
⟹
|
f
(
x
)
−
l
|
≤
ε
{\displaystyle \forall \varepsilon >0,\exists A>0,\forall x\in {\mathcal {D}},x>A\Longrightarrow |f(x)-l|\leq \varepsilon }
f
{\displaystyle f}
l
{\displaystyle l}
−
∞
{\displaystyle -\infty }
∀
ε
>
0
,
∃
A
<
0
,
∀
x
∈
D
,
x
<
A
⟹
|
f
(
x
)
−
l
|
≤
ε
{\displaystyle \forall \varepsilon >0,\exists A<0,\forall x\in {\mathcal {D}},x<A\Longrightarrow |f(x)-l|\leq \varepsilon }
D
{\displaystyle {\mathcal {D}}}
