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Quarto anno
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Per limite si intende il valore,
l
{\displaystyle l\,}
f
(
x
)
{\displaystyle f(x)}
a
{\displaystyle a\,}
lim
x
→
a
f
(
x
)
=
l
{\displaystyle \lim _{x\rightarrow a}f(x)=l}
e si legge: Il limite per
x
{\displaystyle x}
a
{\displaystyle a\,}
f
(
x
)
{\displaystyle f(x)}
l
{\displaystyle l\,}
.
Sia
a
{\displaystyle a\,}
l
{\displaystyle l\,}
∞
{\displaystyle \infty }
Il valore
a
{\displaystyle a\,}
f
(
x
)
{\displaystyle f(x)\,}
a
{\displaystyle a\,}
Sulla base del concetto di limite si possone definire la continuità, la derivata e gli asintoti di una funzione, aspetti fondamentali per studiarne l'andamento.
Se
a
{\displaystyle a\,}
f
(
x
)
{\displaystyle f(x)\,}
lim
x
→
a
f
(
x
)
=
l
{\displaystyle \lim _{x\rightarrow a}f(x)=l}
l
{\displaystyle l\,}
a
{\displaystyle a\,}
x
{\displaystyle x\,}
a
{\displaystyle a\,}
a
{\displaystyle a\,}
f
(
x
)
{\displaystyle f(x)\,}
l
{\displaystyle l\,}
Quindi, chiamando
I
l
{\displaystyle I_{l}\,}
I
a
{\displaystyle I_{a}\,}
l
{\displaystyle l\,}
a
{\displaystyle a\,}
∀
I
l
∃
I
a
|
∀
x
∈
(
I
a
−
a
)
→
f
(
x
)
∈
I
l
{\displaystyle \forall I_{l}\ \exists I_{a}\ |\ \forall x\in (I_{a}-a)\rightarrow f(x)\in I_{l}}
Per verificare il limite bisogna verificare che:
∀
ϵ
>
0
∃
δ
|
∀
x
|
x
−
a
|
<
δ
∧
x
≠
x
0
→
|
f
(
x
)
−
l
|
<
ϵ
{\displaystyle \forall \epsilon >0\ \exists \delta \ |\ \forall x\ |x-a|<\delta \land x\neq x_{0}\rightarrow |f(x)-l|<\epsilon }
Dalla precedente definizione si ricavano i casi in cui il valore a cui tende
x
{\displaystyle x\,}
Un limite finito per
x
{\displaystyle x\,}
:
lim
x
→
a
f
(
x
)
=
l
{\displaystyle \lim _{x\rightarrow a}f(x)=l}
Un limite infinito per
x
{\displaystyle x\,}
:
lim
x
→
a
f
(
x
)
=
∞
{\displaystyle \lim _{x\rightarrow a}f(x)=\infty }
Un limite finito per
x
{\displaystyle x\,}
:
lim
x
→
∞
f
(
x
)
=
l
{\displaystyle \lim _{x\rightarrow \infty }f(x)=l}
Un limite infinito per
x
{\displaystyle x\,}
:
lim
x
→
∞
f
(
x
)
=
∞
{\displaystyle \lim _{x\rightarrow \infty }f(x)=\infty }
La funzione
f
(
x
)
{\displaystyle f(x)\,}
infinito per
x
{\displaystyle x\,}
c
{\displaystyle c\,}
lim
x
→
c
f
(
x
)
=
∞
{\displaystyle \lim _{x\rightarrow c}f(x)=\infty }
Dati due infiniti
f
(
x
)
{\displaystyle f(x)\,}
g
(
x
)
{\displaystyle g(x)\,}
lim
x
→
∞
{\displaystyle \lim _{x\rightarrow \infty }}
f
(
x
)
g
(
x
)
=
n
{\displaystyle {\frac {f(x)}{g(x)}}=n\,}
n
≠
0
{\displaystyle n\neq 0\,}
lim
x
→
∞
{\displaystyle \lim _{x\rightarrow \infty }}
f
(
x
)
g
(
x
)
=
0
{\displaystyle {\frac {f(x)}{g(x)}}=0\,}
g
(
x
)
{\displaystyle g(x)\,}
f
(
x
)
{\displaystyle f(x)\,}
lim
x
→
∞
{\displaystyle \lim _{x\rightarrow \infty }}
f
(
x
)
g
(
x
)
=
∞
{\displaystyle {\frac {f(x)}{g(x)}}=\infty \,}
f
(
x
)
{\displaystyle f(x)\,}
g
(
x
)
{\displaystyle g(x)\,}
La funzione
f
(
x
)
{\displaystyle f(x)\,}
infinitesimo per
x
{\displaystyle x\,}
c
,
{\displaystyle c,}
lim
x
→
c
f
(
x
)
=
0
{\displaystyle \lim _{x\rightarrow c}f(x)=0}
Dati due infinitesimi
f
(
x
)
{\displaystyle f(x)\,}
g
(
x
)
{\displaystyle g(x)\,}
lim
x
→
0
{\displaystyle \lim _{x\rightarrow 0}}
f
(
x
)
g
(
x
)
=
n
{\displaystyle {\frac {f(x)}{g(x)}}=n\,}
n
≠
0
{\displaystyle n\neq 0\,}
lim
x
→
0
{\displaystyle \lim _{x\rightarrow 0}}
f
(
x
)
g
(
x
)
=
0
{\displaystyle {\frac {f(x)}{g(x)}}=0\,}
f
(
x
)
{\displaystyle f(x)\,}
g
(
x
)
{\displaystyle g(x)\,}
lim
x
→
0
{\displaystyle \lim _{x\rightarrow 0}}
f
(
x
)
g
(
x
)
=
∞
{\displaystyle {\frac {f(x)}{g(x)}}=\infty \,}
g
(
x
)
{\displaystyle g(x)\,}
f
(
x
)
{\displaystyle f(x)\,}
teorema
Chiamando
α
(
x
)
{\displaystyle \alpha (x)\,}
x
→
c
{\displaystyle x\rightarrow c}
lim
x
→
c
f
(
x
)
=
l
{\displaystyle \lim _{x\rightarrow c}f(x)=l}
allora:
f
(
x
)
=
l
+
α
(
x
)
{\displaystyle f(x)=l+\alpha (x)\,}
e viceversa.
dimostrazione
.
. q.e.d.
Se
α
(
x
)
{\displaystyle \alpha (x)\,}
β
(
x
)
{\displaystyle \beta (x)\,}
α
(
x
)
+
β
(
x
)
{\displaystyle \alpha (x)+\beta (x)\,}
Se
α
(
x
)
{\displaystyle \alpha (x)\,}
β
(
x
)
{\displaystyle \beta (x)\,}
α
(
x
)
⋅
β
(
x
)
{\displaystyle \alpha (x)\cdot \beta (x)\,}
teorema
Il limite del prodotto di due funzioni è uguale al prodotto dei limiti.
Se
lim
x
→
a
f
(
x
)
=
l
{\displaystyle \lim _{x\rightarrow a}f(x)=l}
lim
x
→
a
g
(
x
)
=
m
{\displaystyle \lim _{x\rightarrow a}g(x)=m}
lim
x
→
a
(
f
(
x
)
⋅
g
(
x
)
)
=
l
⋅
m
{\displaystyle \lim _{x\rightarrow a}(f(x)\cdot g(x))=l\cdot m}
Dimostrazione
Dobbiamo dimostrare che:
1.
|
f
(
x
)
⋅
g
(
x
)
−
l
⋅
m
|
<
ϵ
{\displaystyle \left|f(x)\cdot g(x)-l\cdot m\right|<\epsilon }
Tenendo conto del teorema su limiti e infinitesimi e cioè che:
f
(
x
)
=
l
+
α
(
x
)
{\displaystyle f(x)=l+\alpha (x)\,}
e
g
(
x
)
=
m
+
β
(
x
)
{\displaystyle g(x)=m+\beta (x)\,}
con
α
(
x
)
{\displaystyle \alpha (x)\,}
β
(
x
)
{\displaystyle \beta (x)\,}
x
→
a
{\displaystyle x\rightarrow a}
L'espressione 1. diventa:
2.
|
(
l
+
α
(
x
)
)
⋅
(
m
+
β
(
x
)
)
−
l
⋅
m
|
<
ϵ
{\displaystyle \left|(l+\alpha (x))\cdot (m+\beta (x))-l\cdot m\right|<\epsilon }
Svolgendo i calcoli si ottiene:
|
l
⋅
m
+
l
⋅
α
(
x
)
+
m
⋅
β
(
x
)
+
α
(
x
)
⋅
β
(
x
)
−
l
⋅
m
|
<
ϵ
{\displaystyle \left|l\cdot m+l\cdot \alpha (x)+m\cdot \beta (x)+\alpha (x)\cdot \beta (x)-l\cdot m\right|<\epsilon }
e:
3.
|
l
⋅
α
(
x
)
+
m
⋅
β
(
x
)
+
α
(
x
)
⋅
β
(
x
)
|
<
ϵ
{\displaystyle \left|l\cdot \alpha (x)+m\cdot \beta (x)+\alpha (x)\cdot \beta (x)\right|<\epsilon }
Ora:
i prodotto di valori finiti per infinitesimi (
l
⋅
α
(
x
)
{\displaystyle l\cdot \alpha (x)}
m
⋅
β
(
x
)
{\displaystyle m\cdot \beta (x)}
il prodotto tra due infinitesimi (
α
(
x
)
⋅
β
(
x
)
{\displaystyle \alpha (x)\cdot \beta (x)}
la somma tra questi infinitesimi è un infinitesimo. Quindi il contenuto del modulo è minore di qualunque
ϵ
{\displaystyle \epsilon }
x
→
a
{\displaystyle x\rightarrow a}
q.e.d.
Limite di una funzione polinomiale
modifica
Limiti notevoli di funzioni goniometriche
modifica
lim
x
→
0
sin
x
x
=
[
0
0
]
1
{\displaystyle \lim _{x\to 0}{\frac {\sin {x}}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}
lim
x
→
0
sin
ϵ
(
x
)
ϵ
(
x
)
=
[
0
0
]
1
{\displaystyle \lim _{x\to 0}{\frac {\sin {\epsilon (x)}}{\epsilon (x)}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}
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lim
x
→
0
tan
x
x
=
[
0
0
]
1
{\displaystyle \lim _{x\to 0}{\frac {\tan {x}}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}
lim
x
→
0
1
−
cos
x
x
=
[
0
0
]
0
{\displaystyle \lim _{x\to 0}{\frac {1-\cos {x}}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}0}
lim
x
→
0
1
−
cos
x
x
2
=
[
0
0
]
1
2
{\displaystyle \lim _{x\to 0}{\frac {1-\cos {x}}{x^{2}}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}{\frac {1}{2}}}
Limiti notevoli di funzioni esponenziali
modifica
lim
x
→
∞
(
1
+
1
x
)
x
=
[
∞
0
]
e
{\displaystyle \lim _{x\to \infty }{\left(1+{\frac {1}{x}}\right)^{x}}{\stackrel {\left[\infty ^{0}\right]}{=}}e}
lim
x
→
∞
(
1
+
1
ϵ
(
x
)
)
ϵ
(
x
)
=
[
∞
0
]
e
{\displaystyle \lim _{x\to \infty }{\left(1+{\frac {1}{\epsilon (x)}}\right)^{\epsilon (x)}}{\stackrel {\left[\infty ^{0}\right]}{=}}e}
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lim
x
→
∞
(
1
−
1
x
)
x
=
[
∞
0
]
1
e
{\displaystyle \lim _{x\to \infty }{\left(1-{\frac {1}{x}}\right)^{x}}{\stackrel {\left[\infty ^{0}\right]}{=}}{\frac {1}{e}}}
lim
x
→
0
(
1
+
x
)
1
x
=
[
1
∞
]
e
{\displaystyle \lim _{x\to 0}{\left(1+x\right)^{\frac {1}{x}}}{\stackrel {\left[1^{\infty }\right]}{=}}e}
lim
x
→
0
(
1
+
ϵ
(
x
)
)
1
ϵ
(
x
)
=
[
1
∞
]
e
{\displaystyle \lim _{x\to 0}{\left(1+\epsilon (x)\right)^{\frac {1}{\epsilon (x)}}}{\stackrel {\left[1^{\infty }\right]}{=}}e}
lim
x
→
0
(
1
−
x
)
1
x
=
[
1
∞
]
1
e
{\displaystyle \lim _{x\to 0}{\left(1-x\right)^{\frac {1}{x}}}{\stackrel {\left[1^{\infty }\right]}{=}}{\frac {1}{e}}}
lim
f
(
x
)
g
(
x
)
=
lim
e
g
(
x
)
⋅
ln
f
(
x
)
{\displaystyle \lim {f(x)^{g(x)}}=\lim {e^{g(x)\cdot \ln {f(x)}}}}
Limiti notevoli di funzioni contenenti logaritmi o esponenziali
modifica
lim
x
→
0
ln
(
1
+
x
)
x
=
[
0
0
]
1
{\displaystyle \lim _{x\to 0}{\frac {\ln {(1+x)}}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}
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lim
x
→
0
ln
(
1
+
ϵ
(
x
)
)
ϵ
(
x
)
=
[
0
0
]
1
{\displaystyle \lim _{x\to 0}{\frac {\ln {(1+\epsilon (x))}}{\epsilon (x)}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}
lim
x
→
0
e
x
−
1
x
=
[
0
0
]
1
{\displaystyle \lim _{x\to 0}{\frac {e^{x}-1}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}
Errore del parser (Errore di conversione. Server ("https://wikimedia.org/api/rest_") riporta: "Cannot get mml. TeX parse error: Undefined control sequence \emph"): {\displaystyle \lim _{x\to 0}{\frac {{\emph {a}}^{x}-1}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}\ln {\emph {a}}}
lim
x
→
0
e
ϵ
(
x
)
−
1
ϵ
(
x
)
=
[
0
0
]
1
{\displaystyle \lim _{x\to 0}{\frac {e^{\epsilon (x)}-1}{\epsilon (x)}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}
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I limiti nello studio di funzioni
modifica
![{\displaystyle \lim _{x\to 0}{\frac {\sin {x}}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/104f936475600f7aa5e5f48e0ca402cb047e99c6)
![{\displaystyle \lim _{x\to 0}{\frac {\sin {\epsilon (x)}}{\epsilon (x)}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8595323e8b4f3129f76348a3067db6c8182ba41)
![{\displaystyle \lim _{x\to 0}{\frac {\tan {x}}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a318c630c80a44815080d5751fef031e4433e6a5)
![{\displaystyle \lim _{x\to 0}{\frac {1-\cos {x}}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abbd8f0013905d168e3cc537587ff9b69e42558b)
![{\displaystyle \lim _{x\to 0}{\frac {1-\cos {x}}{x^{2}}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}{\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1de537e0b79ecd4e4cdcf8c9140b60c4894ef50)
![{\displaystyle \lim _{x\to \infty }{\left(1+{\frac {1}{x}}\right)^{x}}{\stackrel {\left[\infty ^{0}\right]}{=}}e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad97727f9c0e80cd465d12e0bf0cdfe11130a146)
![{\displaystyle \lim _{x\to \infty }{\left(1+{\frac {1}{\epsilon (x)}}\right)^{\epsilon (x)}}{\stackrel {\left[\infty ^{0}\right]}{=}}e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b21d5c91aa43cc5bc65a1c9bf8c3802de69a0cc)
![{\displaystyle \lim _{x\to \infty }{\left(1-{\frac {1}{x}}\right)^{x}}{\stackrel {\left[\infty ^{0}\right]}{=}}{\frac {1}{e}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f99b2e9d50731dfe5b027974bd629bacc75acbc)
![{\displaystyle \lim _{x\to 0}{\left(1+x\right)^{\frac {1}{x}}}{\stackrel {\left[1^{\infty }\right]}{=}}e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2b13735a772b7fd778fe47b8c6fa13a63912517)
![{\displaystyle \lim _{x\to 0}{\left(1+\epsilon (x)\right)^{\frac {1}{\epsilon (x)}}}{\stackrel {\left[1^{\infty }\right]}{=}}e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ba7f6d3fc7bc4c7531e61666801ac08f8ef8a53)
![{\displaystyle \lim _{x\to 0}{\left(1-x\right)^{\frac {1}{x}}}{\stackrel {\left[1^{\infty }\right]}{=}}{\frac {1}{e}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a28f353cea8bd6c1de1d9317cff8fae64e09a04)
![{\displaystyle \lim _{x\to 0}{\frac {\ln {(1+x)}}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76b2a0f123666616e6d9470e3a9d1eabbb444bac)
![{\displaystyle \lim _{x\to 0}{\frac {\ln {(1+\epsilon (x))}}{\epsilon (x)}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8543da4bc35eb906bdb956db00a0167335ebcf30)
![{\displaystyle \lim _{x\to 0}{\frac {e^{x}-1}{x}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e403ca8b1e3d882ee1750f8c187a1db2df84d28d)
![{\displaystyle \lim _{x\to 0}{\frac {e^{\epsilon (x)}-1}{\epsilon (x)}}{\stackrel {\left[{\frac {0}{0}}\right]}{=}}1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ad1525143e6000402091fe64c8a54252831d50b)
