Limita unei funcţii într-un punct [ ] În calculul diferenţial şi calculul integral un concept important este cel de limită a unei funcţii într-un punct.
Conceptul este folosit în studiul continuităţii , derivatei , integralei şi alte studii.
Considerăm o funcţie
f
:
A
⊂
R
1
→
R
1
.
{\displaystyle f: A \subset \mathbb R^1 \rightarrow \mathbb R^1. \!}
f
(
x
)
{\displaystyle f(x) \!}
x se apropie de o valoare reală fixată a .
Pentru aceasta se presupune că f(x) este definit pentru orice x care se apropie de a .
Cu alte cuvinte vom presupune că domeniul de definiţie A conţine o mulţime de forma
(
a
−
r
,
a
)
∪
(
a
,
a
+
r
)
{\displaystyle (a-r, a) \cup (a, a+r) \!}
r
>
0.
{\displaystyle r>0. \!}
Altă formulare :
Fie
f
:
D
⊆
R
→
R
,
x
0
{\displaystyle f: D \subseteq \mathbb R \rightarrow \mathbb R, \; \; x_0 \!}
f are limită în
x
0
⇔
l
s
(
x
0
)
=
l
d
(
x
0
)
,
{\displaystyle x_0 \; \Leftrightarrow \; l_s (x_0) = l_d(x_0), \!}
l
s
{\displaystyle l_s \!}
l
d
{\displaystyle l_d \!}
limitele laterale
f are limita L în punctul a se notează:
lim
x
→
a
f
(
x
)
=
L
{\displaystyle \lim_{x \to a} f(x) = L \!}
f
(
x
)
→
x
→
a
L
.
{\displaystyle f(x) \underset {x \to a} \rightarrow L .\!}
Observaţii [ ] 1 . Valoarea funcţiei f în punctul a , dacă există, nu intervine în definiţia limitei.
Valoarea f(a) poate să nu verifice inegalitatea din definiţia limitei.
2 . Fiind dată funcţia f şi numărul L , inegalitatea
|
f
(
x
)
−
L
|
<
ϵ
{\displaystyle |f(x) -L|< \epsilon \!}
L
−
ϵ
<
f
(
x
)
<
L
+
ϵ
{\displaystyle L- \epsilon < f(x) < L+ \epsilon \!}
ϵ
{\displaystyle \epsilon\!}
L ; cât de aproape vrem să fie L .
3 . Numărul
δ
{\displaystyle \delta \!}
ϵ
{\displaystyle \epsilon\!}
δ
(
ϵ
)
,
{\displaystyle \delta(\epsilon), \!}
δ
′
<
δ
(
ϵ
)
,
{\displaystyle \delta' < \delta(\epsilon), \!}
Proprietăţi [ ] 1 . Dacă
lim
x
→
x
0
f
(
x
)
{\displaystyle \lim_{x \to x_0} f(x) \!}
2 . Dacă
lim
x
→
x
0
f
(
x
)
=
l
{\displaystyle \lim_{x \to x_0} f(x) =l \!}
atunci
lim
x
→
x
0
|
f
(
x
)
|
=
|
l
|
.
{\displaystyle \lim_{x \to x_0} |f(x)| =|l|. \!}
Reciproca nu este valabilă.
3 . Dacă
lim
x
→
x
0
|
f
(
x
)
|
=
0
⇒
lim
x
→
x
0
f
(
x
)
=
0.
{\displaystyle \lim_{x \to x_0} |f(x)| =0 \; \Rightarrow \; \lim_{x \to x_0} f(x) =0. \!}
4 . Fie
f
,
g
:
D
⊆
R
→
R
,
{\displaystyle f, g : D \subseteq \mathbb R \rightarrow \mathbb R, \!}
cu proprietatea că:
∃
U
{\displaystyle \exists U \!}
x
0
∈
D
{\displaystyle x_0 \in D \!}
f
(
x
)
≤
g
(
x
)
∀
x
∈
D
∩
U
∖
{
x
0
}
{\displaystyle f(x) \le g(x) \; \forall x \in D \cap U \setminus \{x_0\} \!}
şi dacă există
lim
x
→
x
0
f
(
x
)
,
lim
x
→
x
0
g
(
x
)
{\displaystyle \lim_{x \to x_0} f(x), \; \lim_{x \to x_0} g(x) \!}
atunci:
lim
x
→
x
0
f
(
x
)
<
lim
x
→
x
0
g
(
x
)
.
{\displaystyle \lim_{x \to x_0} f(x) < \lim_{x \to x_0} g(x). \! }
5 . Dacă
f
(
x
)
≤
g
(
x
)
≤
h
(
x
)
,
∀
x
∈
D
∩
U
∖
{
x
0
}
{\displaystyle f(x) \le g(x) \le h(x), \; \forall x \in D \cap U \setminus \{ x_0 \} \!}
∃
lim
x
→
x
0
f
(
x
)
=
lim
x
→
x
0
h
(
x
)
=
l
{\displaystyle \exists \lim_{x \to x_0} f(x) = \lim_{x \to x_0} h(x) =l \!}
atunci:
∃
lim
x
→
x
0
g
(
x
)
=
l
.
{\displaystyle \exists \lim_{x \to x_0} g(x) = l. \!}
6 . Dacă
|
f
(
x
)
−
l
|
≤
g
(
x
)
,
∀
x
∈
D
∩
U
∖
{
x
0
}
{\displaystyle |f(x)-l| \le g(x), \; \forall x \in D \cap U \setminus \{ x_0 \} \!}
şi
lim
g
(
x
)
=
0
{\displaystyle \lim g(x) = 0 \!}
lim
f
(
x
)
=
l
.
{\displaystyle \lim f(x) = l. \!}
7 . Dacă
lim
f
(
x
)
=
0
{\displaystyle \lim f(x) =0 \!}
∃
M
>
0
{\displaystyle \exists M>0 \!}
|
g
(
x
)
|
≤
M
,
{\displaystyle |g(x)| \le M, \!}
atunci
lim
f
(
x
)
⋅
g
(
x
)
=
0.
{\displaystyle \lim f(x) \cdot g(x)=0. \!}
8 . Dacă
f
(
x
)
≥
g
(
x
)
{\displaystyle f(x) \ge g(x) \!}
lim
g
(
x
)
=
+
∞
,
{\displaystyle \lim g(x) = + \infty, \!}
atunci
lim
f
(
x
)
=
+
∞
.
{\displaystyle \lim f(x) = + \infty. \!}
Dacă
f
(
x
)
≤
g
(
x
)
{\displaystyle f(x) \le g(x) \!}
lim
g
(
x
)
=
−
∞
,
{\displaystyle \lim g(x) = - \infty, \!}
atunci:
lim
f
(
x
)
=
−
∞
.
{\displaystyle \lim f(x) = - \infty. \!}
Operaţii cu funcţii [ ] Dacă există
lim
f
(
x
)
=
l
1
,
lim
g
(
x
)
=
l
2
{\displaystyle \lim f(x) = l_1, \; \lim g(x) = l_2 \!}
l
1
+
l
2
,
l
1
−
l
2
,
l
1
⋅
l
2
,
l
1
l
2
,
l
1
l
2
,
l
1
,
{\displaystyle l_1+l_2, \; l_1-l_2, \; l_1 \cdot l_2, \; \frac {l_1}{l_2}, \; l_1^{l_2}, \; \sqrt{l_1}, \!}
1 .
lim
(
f
(
x
)
±
g
(
x
)
)
=
l
1
±
l
2
.
{\displaystyle \lim (f(x) \pm g(x)) = l_1 \pm l_2. \!}
2 .
lim
(
f
(
x
)
g
(
x
)
)
=
l
2
⋅
l
2
.
{\displaystyle \lim (f(x) g(x)) = l_2 \cdot l_2. \!}
3 .
lim
f
(
x
)
g
(
x
)
=
l
1
l
2
.
{\displaystyle \lim \frac {f(x)}{g(x)} = \frac {l_1}{l_2}. \!}
4 .
lim
f
(
x
)
g
(
x
)
=
l
1
l
2
.
{\displaystyle \lim f(x)^{g(x)} = l_1^{l_2}. \!}
5 .
lim
f
(
x
)
=
l
1
.
{\displaystyle \lim \sqrt {f(x)} = \sqrt {l_1}. \!}
P
(
x
)
=
a
0
x
n
+
a
1
x
n
−
1
+
⋯
+
a
n
,
a
0
≠
0
{\displaystyle P(x) = a_0 x^n + a_1 x^{n-1} + \cdots + a_n, \; a_0 \neq 0 \!}
lim
x
→
±
∞
=
a
0
(
±
∞
)
n
.
{\displaystyle \lim_{x \to \pm \infty} = a_0 (\pm \infty)^n. \!}
lim
x
→
∞
q
x
=
{
0
,
d
a
c
a
q
∈
(
−
1
,
1
)
1
,
d
a
c
a
q
=
1
∞
,
d
a
c
a
q
>
1
n
u
e
x
i
s
t
a
,
d
a
c
a
q
≤
−
1
{\displaystyle \lim_{x \to \infty} q^x = \begin{cases} 0, & daca \; q \in (-1, 1) \\ \\ 1, & daca \; q=1 \\ \\ \infty, & daca \; q>1 \\ \\ nu \; exista, & daca \; q \le -1 \end{cases} \!}
lim
x
→
∞
a
0
⋅
x
p
+
a
1
⋅
x
p
−
1
+
⋯
+
a
p
b
0
⋅
x
q
+
b
1
⋅
x
q
−
1
+
⋯
+
b
q
=
{
0
,
d
a
c
a
p
<
q
a
0
b
0
,
d
a
c
a
p
=
q
∞
,
d
a
c
a
p
>
q
s
i
a
0
b
0
>
0
−
∞
,
d
a
c
a
p
>
q
s
i
a
0
b
0
<
0.
{\displaystyle \lim_{x \to \infty} \frac {a_0 \cdot x^p + a_1 \cdot x^{p-1} + \cdots + a_p}{b_0 \cdot x^q + b_1 \cdot x^{q-1} + \cdots + b_q} = \begin{cases} 0, & daca \; p<q \\ \\ \frac {a_0}{b_0}, & daca \; p=q \\ \\ \infty, & daca \; p>q \; si \; \frac {a_0}{b_0} >0 \\ \\ - \infty, & daca \; p>q \; si \ \frac {a_0}{b_0} <0. \end{cases} \!}
a
>
1
{\displaystyle a>1 \!}
lim
x
→
∞
a
x
=
∞
{\displaystyle \lim_{x \to \infty} a^x = \infty \!}
lim
x
→
−
∞
a
x
=
0
{\displaystyle \lim_{x \to - \infty} a^x = 0 \!}
a
∈
(
0
,
1
)
{\displaystyle a \in (0, 1) \!}
lim
x
→
∞
a
x
=
0
{\displaystyle \lim_{x \to \infty} a^x = 0 \!}
lim
x
→
−
∞
a
x
=
∞
{\displaystyle \lim_{x \to - \infty} a^x = \infty \!}
a
>
1
{\displaystyle a>1 \!}
lim
x
→
∞
log
a
x
=
∞
{\displaystyle \lim_{x \to \infty} \log_a x = \infty \!}
lim
x
→
0
log
a
x
=
−
∞
{\displaystyle \lim_{x \to 0} \log_a x = - \infty \!}
a
∈
(
0
,
1
)
{\displaystyle a \in (0, 1) \!}
lim
x
→
∞
log
a
x
=
−
∞
{\displaystyle \lim_{x \to \infty} \log_a x = - \infty \!}
lim
x
→
0
log
a
x
=
∞
{\displaystyle \lim_{x \to 0} \log_a x = \infty \!}
lim
x
→
0
sin
x
x
=
1
{\displaystyle \lim_{x \to 0} \frac {\sin x}{x} = 1 \!}
lim
u
(
x
)
→
0
sin
u
(
x
)
u
(
x
)
=
1
{\displaystyle \lim_{u(x) \to 0} \frac {\sin u(x)}{u(x)} =1 \!}
lim
x
→
0
tan
x
x
=
1
{\displaystyle \lim_{x \to 0} \frac {\tan x}{x} = 1 \!}
lim
u
(
x
)
→
0
tan
u
(
x
)
u
(
x
)
=
1
{\displaystyle \lim_{u(x) \to 0} \frac {\tan u(x)}{u(x)} =1 \!}
lim
x
→
0
arcsin
x
x
=
1
{\displaystyle \lim_{x \to 0} \frac {\arcsin x}{x} = 1 \!}
lim
u
(
x
)
→
0
arcsin
u
(
x
)
u
(
x
)
=
1
{\displaystyle \lim_{u(x) \to 0} \frac {\arcsin u(x)}{u(x)} =1 \!}
lim
x
→
0
arctan
x
x
=
1
{\displaystyle \lim_{x \to 0} \frac {\arctan x}{x} = 1 \!}
lim
u
(
x
)
→
0
arctan
u
(
x
)
u
(
x
)
=
1
{\displaystyle \lim_{u(x) \to 0} \frac {\arctan u(x)}{u(x)} =1 \!}
lim
x
→
0
(
1
+
x
)
1
x
=
e
{\displaystyle \lim_{x \to 0} (1+x)^{\frac 1 x} = e \!}
lim
u
(
x
)
→
0
(
1
+
u
(
x
)
)
1
u
(
x
)
=
e
{\displaystyle \lim_{u(x) \to 0} (1+ u(x))^{\frac{1}{u(x)}} = e \!}
lim
x
→
∞
(
1
+
1
x
)
x
=
e
{\displaystyle \lim_{x \to \infty} \left ( 1 + \frac 1 x \right ) ^x = e\!}
lim
u
(
x
)
→
∞
(
1
+
1
u
(
x
)
)
u
(
x
)
=
0
{\displaystyle \lim_{u(x) \to \infty} \left ( 1+ \frac {1}{u(x)} \right )^{u(x)} = 0 \!}
lim
x
t
o
0
ln
(
1
+
x
)
x
=
1
{\displaystyle \lim_{x to 0} \frac {\ln (1+x)}{x} = 1 \!}
lim
u
(
x
)
→
0
ln
(
1
+
u
(
x
)
)
u
(
x
)
=
1
{\displaystyle \lim_{u(x) \to 0 } \frac {\ln (1+ u(x))}{u(x)} = 1 \!}
lim
x
→
0
a
x
−
1
x
=
ln
a
{\displaystyle \lim_{x \to 0 } \frac {a^x-1}{x} = \ln a \!}
lim
u
(
x
)
→
0
a
u
(
x
)
−
1
u
(
x
)
=
ln
a
{\displaystyle \lim_{u(x) \to 0} \frac {a^{u(x)} - 1}{u(x)} = \ln a \!}
lim
x
→
0
(
1
+
x
)
r
−
1
x
=
r
{\displaystyle \lim_{x \to 0} \frac {(1+x)^r -1}{x} = r \!}
lim
u
(
x
)
→
0
(
1
+
u
(
x
)
)
r
−
1
u
(
x
)
=
r
{\displaystyle \lim_{u(x) \to 0} \frac {(1+ u(x))^r-1}{u(x)} = r \!}
lim
x
→
∞
x
k
a
x
=
0
{\displaystyle \lim_{x \to \infty} \frac {x^k}{a^x} = 0 \!}
lim
u
(
x
)
→
∞
u
(
x
)
k
a
u
(
x
)
=
0
{\displaystyle \lim_{u(x) \to \infty} \frac {u(x)^k}{a^{u(x)}} =0 \!}
lim
x
→
∞
ln
x
x
k
=
0
{\displaystyle \lim_{x \to \infty} \frac {\ln x}{x^k} = 0 \!}
lim
u
(
x
)
→
∞
ln
u
(
x
)
u
(
x
)
k
.
{\displaystyle \lim_{u(x) \to \infty} \frac {\ln u(x)}{u(x)^k}. \!}
Exemple [ ] EXEMPLUL 1.
Să se arate că
lim
x
→
2
x
2
−
4
x
−
2
=
4.
{\displaystyle \lim_{x \to 2} \frac {x^2-4}{x-2}=4. \!}
Soluţie .
Fie
ϵ
>
0
{\displaystyle \epsilon>0 \!}
|
x
2
−
4
x
−
2
−
4
|
<
ϵ
{\displaystyle \left | \frac{x^2-4}{x-2} -4 \right | < \epsilon \!}
4
−
ϵ
<
x
2
−
4
x
−
2
<
4
+
ϵ
{\displaystyle 4- \epsilon < \frac{x^2-4}{x-2}< 4+ \epsilon\!}
pentru
x
≠
2.
{\displaystyle x \neq 2. \!}
4
−
ϵ
<
x
+
2
<
4
+
ϵ
{\displaystyle 4- \epsilon < x+2 < 4 + \epsilon \!}
2
−
ϵ
<
x
<
2
+
ϵ
,
{\displaystyle 2- \epsilon <x< 2+ \epsilon, \!}
δ
=
ϵ
.
{\displaystyle \delta = \epsilon. \!}
a
>
0
,
{\displaystyle a>0 ,\!}
f
(
x
)
x
{\displaystyle f(x) \sqrt x \!}
lim
x
→
a
x
=
a
.
{\displaystyle \lim_{x \to a} \sqrt x = \sqrt a.\!}
Soluţie :
Într-adevăr, dacă
ϵ
>
0
,
{\displaystyle \epsilon >0, \!}
|
x
−
a
|
<
ϵ
{\displaystyle |\sqrt x - \sqrt a| < \epsilon \!}
a
−
ϵ
<
x
<
a
+
ϵ
{\displaystyle \sqrt a - \epsilon < \sqrt x < \sqrt a + \epsilon \!}
care prin ridicare la pătrat devine:
a
−
2
a
⋅
ϵ
+
ϵ
2
<
x
<
a
+
2
a
⋅
ϵ
+
ϵ
2
.
{\displaystyle a-2 \sqrt a \cdot \epsilon + \epsilon^2 < x < a + 2 \sqrt a \cdot \epsilon + \epsilon^2. \!}
Pentru un a şi un
ϵ
{\displaystyle \epsilon\!}
δ
=
2
a
⋅
ϵ
+
ϵ
2
.
{\displaystyle \delta = 2 \sqrt a \cdot \epsilon + \epsilon^2. \!}
Limite laterale [ ] (Detalii la articolul Limite laterale ale unei funcții )
Limite infinite [ ] (Detalii la articolul Limită a unei funcții la infinit )
Limite notabile [ ] Resurse [ ] 
