Limite de Functii - Clasa a 11-A

Clasa a XI-a -1 Elemente de analiza matematica – Limite de functii

 Definitie vecinatate a unui punct : - Fixam un punct a ∈R ; - Se numeste vecinatate a punctului a orice multime V ⊂ R care contine un interval deschis centrat in a , adica : in acest caz exista r > 0 astfel incat ( a − r , a + r ) ⊂ V .

 Definitie punct de acumulare ( punct limita ) : - Fie A o submultime nevida din R : A ⊂ R ; - Un punct a ∈R se numeste punct de acumulare (sau punct limita) pentru multimea A daca

(∀)

V ∈Va

(= multimea vecinatatilor punctului a ) sa rezulte ( V − { a} ) ∩ A ≠ Φ .

- aceasta definitie spune ca un punct a ∈R este punct de acumulare pntru multimea A daca orice vecinatate V a punctului a mai contine si alte puncte din A , diferite de a , adica exista :

x ∈V ∩ A cu x ≠ a

 Definitie punct izolat : - Fie A o submultime nevida din R : A ⊂ R ; - Un punct V a punctului

x0 ∈ A

se numeste punct izolat al multimii A daca exista cel putin o vecinatate

x0 a s t f ienl c a Vt ∩ A = { x0}

.

 Observatie : 

Orice punct al unei multimi A este fie punct de acumulare , fie punct izolat .

Limite de functiii


Fixam o functie ,

x0 ∈ R

f : D →R

, ( D ⊂ R ) si un punct

x0

punct de acumulare a lui D

.

 Definitie criteriul cu vecinatati : - Functia f are limita in punctul

x0

, egala cu  si scriem :

f ( x) =  lim →

x

x0

daca pentru orice vecinatate V a lui  exista o vecinatate U a lui orice :

x0

astfel incat pentru

x ∈ D ∩ U \ { x0} ⇒ f ( x) ∈ V

 Toreme de caracterizare a limitei unei functii intr-un punct : 

Criteriul : ε - δ . - Fie f : D → R , ( D ⊂ R ) , o functie si - Functia f are limita in punctul

daca si numai daca

orice



x0

x = x0

punct de acumulare a lui D , , egala cu ∈R si scriem :

x0 ∈ R

;

lim f ( x ) = 

x → x0

(∀) ε > 0 exista numarul real δ ( ε ) > 0 , depinzand de ε , astfel incat pentru

x ∈ D \ { x0}

, cu proprietatea

x − x0 < δ

sa rezulte : f ( x ) − <ε .

Criteriul : cu siruri . - Fie f : D → R , ( D ⊂ R ) , o functie si

x0

punct de acumulare a lui D ,

x0 ∈ R

;

Limite de functiii


- Functia f are limita in punctul

lim f ( x ) = 

x → x0

daca pentru orice sir

x = x0

, egala cu ∈R , finit sau infinit , si scriem :

( an ) , a ∈ D \ { x } , a → x n≥ 0 n

0 n 0

avem :

f ( an ) → 

.

 Observatie : 

Daca exista , limita unei functii intr-un punct este unica .

 Definitie : Limita la stanga : - Fie f : D → R , ( D ⊂ R ) ,

x0

punct de acumulare pentru multimea :

' D = D ∩ ( − ∞ , x0 ) = { x ∈ D x < x0} si s ∈ R

- Functia f are limita la stanga in punctul

V a lui

s

x0

, exista o vecinatate U a lui

x < x0 - Vom folosi notatiile :

l s = f ( x0 − 0 ) = l im f ( x ) → x x0 x < x0

,

x0

egala cu

s

daca oricare ar fi vecinatatea

, astfel incat pentru orice :

x ∈ U ∩ D \ { x0} ⇒ f ( x) ∈ V

.

 Definitie : Limita la dreapta : - Fie f : D → R , ( D ⊂ R ) ,

x0

punct de acumulare pentru multimea :

' D = D ∩ ( x0 ,+ ∞) = { x ∈ D x > x0} si d ∈ R

Limite de functiii


- Functia f are limita la dreapta in punctul vecinatatea V a lui

d

x0

egala cu

x0

, exista o vecinatate U a lui

x > x0 - Vom folosi notatiile :

l d = f ( x0 + 0 ) = lim f ( x ) → x x0 x > x0

,

d

daca oricare ar fi

, astfel incat pentru orice :

x ∈ U ∩ D \ { x0} ⇒ f ( x) ∈ V

.

 TEOREMA : de caracterizare a limitei unei functii intr-un punct cu

ajutorul limitelor laterale

- Fie f : D → R , ( D ⊂ R ) , existe limitele laterale in

x0

x0

punct de acumulare pentru multimea D astfel incat sa

( deci exista

f ( x0 − 0 )

,

f ( x0 + 0 )

) ;

- Atunci urmatoarele afirmatii sunt echivalente : 1). 2).

f

are limita in punctul

x0

;

f ( x0 − 0 ) = f ( x0 + 0 )

In aceste conditii :

f ( x ) = f ( x0 − 0 ) = f ( x0 + 0 ) lim →

x

x0

- Aceasta teorema spune ca o functie are limita intr-un punct daca si numai daca exista limitele laterale cu proprietatea ca sunt si egale .

 Observatii : 1). Daca f : ( a, b ) → R si are limite in punctele a ,b in punctul a vorbim de limita laterala la dreapta , iar in punctul b de limita laterala la stanga . 2). Limitele laterale se folosesc in urmatoarele situatii : - in punctele in care o functie definita pe ramuri isi schimba expresia ; - daca trecand la limita obtinem :

a ; 0

- daca domeniul de definitie este restrictiv , de exemplu :

f ( x ) = ln(1− x2 )

.

Limite de functiii




Fie f , g : D →R si



Daca :

x0

un punct de acumulare pentru D ;

f ( x ) = 1 lim →

x

si

x0

g ( x ) = 2 ,  ,  ∈ R , c ∈ R lim 1 2 x →x

atunci functia

0

:

1)

(f

x0

+ g ) are limita in

si

lim( f + g )( x ) = 1 + 2 = lim f ( x ) + lim g ( x )

x → x0

x → x0

x → x0

.

( Limita sumei este egala cu suma limitelor ) Caz execeptat : ( ∞ − ∞) daca

2)

1 = ∞ , 2 = − ∞ s a u 1 = − ∞ , 2 = ∞

x0

( c ⋅ f ) are limita in

si

limc( f )( x ) = c ⋅ 1 = c ⋅ lim f ( x )

x → x0

x → x0

.

( O constanta iasa in afara limitei )

3)

(f

x0

⋅ g ) are limita in

si

lim( f ⋅ g )( x ) = 1 ⋅ 2 = lim f ( x ) ⋅ lim g ( x )

x → x0

x → x0

x → x0

.

( Limita produsului este egala cu produsul limitelor ) Caz execeptat : ( 0 ⋅ ∞) daca

4)

f  g 

   are limita in 

1 = 0 , 2 = ± ∞s a u 1 = ± ∞, 2 = 0

x

0 si

f ( x) lim f → x x    ( x ) = 1 = lim x →x  g  2 lim g ( x ) → 0

., daca

2 ≠ 0

0

x

x0

Limite de functiii


( Limita catului este egala cu catul limitelor ) Cazuri execeptate :

(f ) g

5)

are limita in

f ( x) > 0 .

Cazuri exceptate :

6)

f

±∞ daca ∞

(1

∞

1 = ∞ , 2 = − ∞ s a u 1 = − ∞ , 2 = ∞

x0

si lim

x →x 0

( f )( x ) = g

2 1

   lim g ( x )  →x 0   x  

  =  lim f ( x ) x →x0 

, daca

, 00 , ∞0 )

are limita in

x0

si

f ( x ) = 1 . lim f ( x ) = lim →

x →x 0

x

x0

( Limita modulului este egala cu modulul limitei )

 Criteriul : MAJORARII . - Fie f , g : D →R doua functii si x0 un punct de acumulare pentru D si V o vecinatate a lui

x0

.

g ( x ) = 0 atunci : - Daca f ( x ) − ≤ g ( x ) , ( ∀) x ∈V ∩ D , x ≠ x0 si daca xlim → x0

f ( x) =  lim →

x

x0

 Consecinte : 1)

Daca f ( x ) ≥ g ( x ) si

2)

Daca f ( x ) ≤ g ( x ) si

lim g ( x ) = + ∞ atunci lim f ( x ) = + ∞ .

x → x0

x → x0

g ( x) = − ∞ lim →

x x0

atunci

f ( x) = − ∞ lim →

x

x0

.

 Trecerea la Limita in Inegalitati . - Fie f , g : D →R doua functii si a lui x0 . - Daca f ( x ) ≤ g ( x ) ,

x0

un punct de acumulare pentru D si V o vecinatate

(∀ ) x∈ V ∩ D , x ≠ x

0

si daca f si g au limite in punctul

x0 ∈ R

atunci : Limite de functiii


lim f ( x ) ≤ lim g ( x )

x → x0

x → x0

 Corolar :

a∈V a

Fie f : D → R ,

,(

V a = multimea vecinatatilor punctului

a ) , f are limita in

a

si

V ∈V a

.

lim f ( x ) ≥ 0



Daca f ( x ) ≥ 0 , ( ∀) x ∈V ∩ D , x ≠ a atunci



Daca f ( x ) ≤ 0 , ( ∀) x ∈V ∩ D , x ≠ a atunci



Daca α ≤ f ( x ) ≤ β , ( ∀) x ∈V ∩ D , x ≠ a atunci

x→ a

f ( x) ≤ 0 lim x→ a

. .

α ≤ lim f ( x ) ≤ β x→ a

.

 TEOREMA : CLESTELUI .  Fie trei functii

f , g , h : D →R ,

a

un punct de acumulare pentru D ,

vecinatate a lui a .  Daca : 1). f ( x ) ≤ g ( x ) ≤ h( x ) , ( ∀) x ∈V ∩ D , x ≠ a . 2).

a∈V a

si V o

f ( x ) = limh( x ) =  lim x→ a x→ a

atunci g are limita in Schematic :

a

si mai mult :

g( x) =  lim x→ a

f ( x ) ≤g ( x ) ≤h( x ) ↓ 

↓ 

↓

.

.



 TEOREMA : ( criteriu) . Aceasta teorema este un alt rezultat important care permite calculul limitei unui produs de functii :

 Fie

f , g : D →R , doua functii si

a∈V a

,(

V a = multimea vecinatatilor punctului

a

), Limite de functiii


a

punct de acumulare , si 1). 2).

V ∈V a

f ( x ) ≤M

, ( ∀) x ∈V ∩ D , M > 0 ( f marginita pe o vecinatate a lui

g( x) = 0 lim x→ a

 Atunci :

.cu proprietatile :

a

);

.

lim f ( x ) ⋅ g ( x ) = 0 . x→ a

 Limita produsului dintre o functie marginita si o functie de limita zero este zero !!!

In cele enuntate si discutate anterior acestui capitol , am vazut cateva operatii cu limite de functii .Pentru ca ele sa devina operabile este nevoie de cunoasterea procedurii de calcul a limitelor principalelor functii. Vom discuta si calcula limita functiei , in general , in doua cazuri : 1). Cand 2). Cand



a a

este punct de acumulare finit ; este punct de acumulare infinit ( daca exista ) .

1 Limita :

. Limite de functiii


 

- Fie f : R → R , f ( x ) = c , c ∈ R ; - Atunci :

f ( x) = c lim x→ a

 

,

( ∀) a ∈R

2 Limita :

.

- Fie functia polinomiala : f : R → R

f ( x ) = an xn + an−1 xn−1 + .....+ ak xk + ak −1 xk −1 + ..... + a1 x + a0 unde : 

ak ∈ R , k = 0, n , an ≠ 0

.

- Avem cazurile :

1). Daca a este un punct de acumulare finit atunci :

f ( x) = f ( a) lim x→ a Deci limita unei functii polinomiale intr-un punct de acumulare finit

a

, se obtine inlocuind

x

cu

a

.

2). Daca a este un punct de acumulare infinit atunci :

f ( x ) = lim a n xn = a n ⋅ ( ±∞) lim x →a x →a Deci limita unei functii polinomiale la

±∞

n

este aceeasi cu limita termenului de grad maxim .

Limite de functiii

Clasa a XI-a - 10 Elemente de analiza matematica – Limite de functii

1).

2 ( x + 2x − 3 ) = lim x→ 1

...................................................................................................................... 2).

2 ( 5x + 7 x ) = lim x→ −2

........................................................................................................................... 3).

3 lim( 2 x + 2 x − 7 ) = x→ 7

................................................................................................................... 4).

2 2 ( 3x + 6 x − 3 ) = lim x→ 0

................................................................................................................... 5).

6 3 3 ( 5 x + 2 x − 3x ) = lim x→ 0

................................................................................................................. 6).

2 ( − 2x + 2x − 6 ) = lim x→ −3

................................................................................................................ 7).

4 ( x + 6x ) = lim x→ −1

............................................................................................................................. 8).

3 x 2 lim( 2 x e − x + 5 ) = x→ 3

................................................................................................................. 9).

3 2 lim( x + x + x ) = x→ ∞

....................................................................................................................... 10).

3 2 ( 2 x + x + 3x − 8 ) = lim x→ ∞

........................................................................................................... Limite de functiii


11).

2 ( 7 − 2x − 7x ) = lim x→ ∞

................................................................................................................... 12).

4 2 ( x + 5x + x ) = lim x → −∞

..................................................................................................................... 13).

3 2 ( x − x + 5x + 7 ) = lim x → −∞

............................................................................................................. 14).

4 lim ( − 3x + 6 x − 1 ) = x→ ∞

................................................................................................................. 15).

3 ( x + 2005 ) = lim x→ ∞

......................................................................................................................... 16).

3 2 ( − x + 3x + 10 x ) = lim x → −∞

.............................................................................................................. 17).

3 2 ( − x + 6x + 3 ) = lim x → −∞

.................................................................................................................. 18).

5 2 ( − 3x + 4 x + x + 1 ) = lim x→ ∞

.........................................................................................................

Limite de functiii






3 Limita :

.

- Fie functia rationala : f ( x ) =

P( x ) , f : R − { x Q ( x) = 0} → R Q( x )

unde P si Q sunt functii polinomiale :

P( x ) = ak xk + a k −1 xk −1 + ..... + ai xi + ai −1 xi −1 + ..... + a1 x + a0 Q( x ) = bl xl + bl −1 xl −1 + ..... + b j x j + b j −1 x j −1 + ..... + b1 x + b0 unde : 

ai , b j ∈ R , i = 0, k , j = 0, l , a k , bl ≠ 0

.

- Distingem cazurile :

1). Subcazul 1 :

Daca a este un punct de acumulare finit cu proprietaea ca Q ( a ) ≠ 0 ( deci a nu este radacina pentru numitor ) atunci : lim f ( x ) =lim x →a

x →a

P( x ) P( a ) = = f (a ) Q( x ) Q( a )

Subcazul 2 : Daca a este un punct de acumulare finit cu proprietaea ca Q ( a ) = 0 ( deci a este radacina pentru numitor ) atunci : f ( x ) =lim lim x →a x →a

P( x ) P(a ) P( a ) = = Q( x ) Q( a ) 0

caz de nedeterminare !!!

Discutia pt. acest subcaz 2 este mai complexa . Eliminarea acestiu caz de nedeterminare o vom discuta in capitolele ce vor urma( cazurile de nedeterminare ale limitelor de functii ) . O modalitate de a scapa de nedeterminare este ca sa descompunem polinoamele in factori primi si prin reducerea termenilor asemenea sa ajungem la rezultatul final , dar aceasta numai in conditiile in care si P( a ) = 0 .

2). Daca a este un punct de acumulare infinit atunci :

Limite de functiii


    ( ) f x = lim  x→ ± ∞    

ak k − l ⋅ ( ± ∞ ) , p e n tk r( gu r a dn uu lm a rlau)t i>o l r( gu r a dn uu lm i t io) r u l u bl ak , p e n tk r= ul bl 0 , p e n tk r< ul ( g r a dn uu lm a rlau t
x −1 = lim 2 x→1 2x − x − 1 2

1). ........................................................................................................................

x +1 = lim 2 x →1 x −1

2). ...................................................................................................................................

2 x + 2x − 1 = lim 2 x→ − 1 x + 8x + 1

3). ........................................................................................................................

x lim x→ −1 x

4).

−1 = −3

.................................................................................................................................. 5).

2 2x + x + 5 = lim 2 x→ ∞ 3x − 2 x + 1

......................................................................................................................

Limite de functiii


x +5 = lim 3 2 x → −∞ 4x − x + 1 3

6).

........................................................................................................................ 2 2x − x − 5 = lim x→ ∞ x −3

7). ........................................................................................................................

4 36x + 5 x + 7 = lim 2 x → −∞ 3x − x + 1

8).

.................................................................................................................. 5 4 2x + x = lim 2 x→ ∞ x −8

9).

...............................................................................................................................

x −2 = lim 2 x→ 1 x + x +1

10). ...........................................................................................................................

(1 − x )( x + 4 ) = 2

11).

lim x→ ∞

3 3x − 2

.................................................................................................................. 12).

2 x − 2x + 1 = lim x→1 x −1

........................................................................................................................ 13).

2 x + x −1 = lim 3 x→ − 1 x +1

...........................................................................................................................

Limite de functiii


14).

2 x + 2 x − 15 = lim 2 x→ 3 x + 8 x + 15

...................................................................................................................... 2 x +1 = lim 2 x→2 ( 2 − ) x 4

15).

...............................................................................................................................

( x −1 ) 2

lim

16).

x→ 1

2

x − 2x + 1 2

=

........................................................................................................................

15 4 = lim3 x 3 2 x→ 2 x − − 2 2 2 x +x −

17).

....................................................................................................................... 18).

2 x − 25 = lim 2 x→ 1 x + x − 20

....................................................................................................................... 4 x +x = lim 4 x→ 0 x −x

19). ..................................................................................................................................

4 3 x −x = lim 4 x→ 0 x

20). .................................................................................................................................. 21).

2 2x − 4x = lim 2 x→ ∞ 5x − 8

............................................................................................................................. Limite de functiii


3 4 4x − x = lim 2 x → −∞ x − 7x

22). ...............................................................................................................................

2 2x − x = lim x→ ∞ x +1

23). ...............................................................................................................................

3 2 2x + x − 5 = lim 3 x → −∞ 15x − 23

24). ........................................................................................................................

2 3x − 5 x + 1 = lim α x→ ∞ x +1

25).

...................................................................................................................... 2 2x + x = lim 2 x→ ∞ 3x − x

26). ...............................................................................................................................

x −1 = lim 2 x → −∞ x +x

27). .................................................................................................................................

3 2 x +x = lim 2 x → −∞ x +4

28). .................................................................................................................................

lim x→ ∞

29).

( x + 2)( 2 x + 1) = 2 3x − x

....................................................................................................................



4 Limita :





- Distingem urmatoarele cazuri :

I.

Cazul radicalilor de ordin par ( n = 2k ) : avem functia radical , unde se impune

conditia de existenta a radicalului de ordin par ,

f : [ 0, ∞ ) → [ 0, ∞ ) , f ( x ) = 2 k x , k ∈ N * cu subcazurile : Subcazul 1 : Daca a ∈[0,+∞ ) ,

lim x →a

a

Subcazul 2 : Daca

2k

a

punct de acumulare finit , atunci :

x = lim 2 k a x →a

punct de acumulare infinit , a = + ∞ , atunci :

2k x = lim 2 k + ∞ = +∞ lim x →+∞ x →+∞

a

Subcazul 3 : Daca

punct de acumulare infinit ,

a = −∞

, atunci :

2k 2k lim x = lim − ∞ = nu exista

x → −∞

x → −∞

deoarece nu putem calcula radical de ordin par dintr-un numar negativ !!!

II.

Cazul determinantilor de ordin impar ( n = 2k + 1 ) : avem functia radical , caz in

care nu avem de pus nici o conditie de existenta a radicalului ,

f : R → R , f ( x ) = 2 k +1 x , k ∈ N * cu subcazurile :

a

Subcazul 1 : Daca

lim x →a Subcazul 2 : Daca

lim x →+∞ Subcazul 3 : Daca

lim

x →−∞

punct de acumulare finit , atunci : 2 k +1

x = lim 2 k +1 a x →a

a


2 k +1

x = lim 2 k +1 + ∞ = +∞ x →+∞

a 2 k +1


a = −∞

, atunci :

x = lim 2 k +1 − ∞ = −∞ x →−∞

Limite de functiii


1).

3 x −x = lim x→ 3

............................................................................................................................... 2).

− x2 − 1 = lim x → −∞

...........................................................................................................................

x −4 = 3). lim .............................................................................................................................. x→ 2 3 x x>2 2

4).

x

lim

x +4 2

x →−∞

=

.............................................................................................................................. 5).

2 x + x +1 = lim 4 x→ ∞ x +1

........................................................................................................................... 6).

2 x − x +1 = lim x→ 2

........................................................................................................................ 7).

3 x= lim x→ 2

........................................................................................................................................ 8).

5 x= lim x → − 32

....................................................................................................................................... 9).

5 x= lim x→∞

........................................................................................................................................ 10).

x= lim x→0

........................................................................................................................................ 11).

(

)

2 4 x +5 −x = lim x→ 2

..................................................................................................................... Limite de functiii


x x + 16 = lim x→ 0

12).

.......................................................................................................................... 13).

lim x + x + x =

...................................................................................................................

3 3 3 lim x − x − x =

...................................................................................................................

x→ ∞

14).

x → −∞

15). lim x x →∞

x = x +1

...............................................................................................................................

( x − 2 ) 3 x5 − 2 x3 = lim x→ 3

16).

.................................................................................................................



5 Limita :

.

f : R → ( 0,+ ∞) , f ( x ) = bx , b > 0 , b ≠ 1



- Fie functia exponentiala :




.

Daca b >1 atunci distingem urmatoarele subcazuri :

I.

Subcazul 1 : Daca

a

punct de acumulare finit , atunci : x a b =b lim x →a

Subcazul 2 : Daca

a


x x +∞ b = lim b = b = +∞ lim x →a x →+∞

Distingem la acest subcaz urmatoarea situatie : n

x = 0 , ∀ n∈ Z lim x x→ + ∞ b

functia exponentiala este mai mare decat functia polinomiala !!!

Subcazul 3 : Daca

a


a = −∞

, atunci :

Limite de functiii

Clasa a XI-a - 20 Elemente de analiza matematica – Limite de functii x x −∞ b = lim b = b = 0 lim x →a x →−∞

Daca 0 < b < 1 atunci distingem urmatoarele subcazuri :

II.

Subcazul 1 : Daca

a

punct de acumulare finit , atunci : x a b =b lim x →a

Subcazul 2 : Daca

a


x x +∞ b = lim b = b = 0 lim x →a x →+∞

Subcazul 3 : Daca

a


a = −∞

, atunci :

x x − ∞ b = lim b = b = +∞ lim x →a x →− ∞

x

1).

1   = lim x→ 2 2  

...................................................................................................................................... x

2).

1   = lim x→ ∞ 5  

...................................................................................................................................... 3).

x lim 6 = x →3

...........................................................................................................................................

Limite de functiii


4).

x lim 5 =

x →−∞

.......................................................................................................................................... x

5).

 5 lim   = x → −∞ 7  

..................................................................................................................................... 6).

lim x →∞

(

)

x

10 =

................................................................................................................................... x

7).

 1    = lim x→ ∞  3

.................................................................................................................................. 8).

lim e

2 x + 5 x +1

x → −2

=

................................................................................................................................. 9).

lim e

2 x + 5 x +1

x→ + ∞

=

................................................................................................................................. 10).

lim e

3 x + x +2

x → −∞

=

..................................................................................................................................

11).

 1 lim   x→ + ∞  3 

4 2 x + x +9 x

=

........................................................................................................................... 12).

lim ( 0.04 )

x → −∞

3 x + x +2

=

......................................................................................................................... 1− x

13).

ex = lim x →0 6

........................................................................................................................................ Limite de functiii


lim5 10

14).

x →−

2 x +5 x +1

=

2

....................................................................................................................................

lim a

15).

4 2 x + 3 x − x +1

x → −∞

=

stiinnd ca : 0 < a < 1

? ...................................................................................

lim e

16).

x →−∞

−x2

=

........................................................................................................................................





6 Limita :

- Fie functia logaritmica :

. f : ( 0,+ ∞) → R , f ( x ) = lo gb x , cu b > 0 , b ≠ 1

conditiile de

existenta ale logaritmilor . 


I.

Daca 0 < b < 1 atunci distingem urmatoarele subcazuri : Subcazul 1 : Daca a = 0 punct de acumulare finit , atunci : f ( x ) = lim logb x = + ∞ lim x→ 0 x→ 0 x>0

x>0

Subcazul 2 : Daca a ∈( 0,+∞ ) punct de acumulare finit , atunci :

f ( x ) = lim log b x = log b a lim x →a x →a Subcazul 3 : Daca

a


log b x = −∞ lim f ( x ) = xlim →+∞

x →+∞

II.

Daca b >1 atunci distingem urmatoarele subcazuri : Subcazul 1 : Daca a = 0 punct de acumulare finit , atunci : f ( x ) = lim logb x = −∞ lim x→0 x→0 x> 0

x>0

Limite de functiii


Subcazul 2 : Daca a ∈( 0,+∞ ) punct de acumulare finit , atunci :

f ( x ) = lim log b x = log b a lim x →a x →a Subcazul 3 : Daca

a


lim f ( x ) = lim log b x = +∞

x →+∞

x →+∞

Limita logaritmului este egala cu logaritmul limitei .

1).

lim1 log 1 x = x→ 4

2

.................................................................................................................................... 2).

log 1 x = lim x →3 3

.................................................................................................................................... 3).

log 1 x = lim x →∞ 2

.................................................................................................................................... lg x = 4). lim ....................................................................................................................................... x →0 x >0

log 2 x = 5). lim x →0 5 x >0

.................................................................................................................................... 6).

lim ln 3 x = x→∞

.................................................................................................................................... 7).

lim ln x = x→e2

.......................................................................................................................................

Limite de functiii


log 7 x = 8). lim x →0 x >0

....................................................................................................................................



7 Limita : •

Limita functiei sinus :



- Fie functia : sin : R → [ − 1,1] .




I.

.

Daca

a

este un punct de acumulare finit , a ∈ R , atunci :

sinx = sina lim x→ a Deci limita functiei sin intr-un punct de acumulare finit a ∈ R se obtine inlocuind pe

II.

Daca

a

x cu a

este un punct de acumulare infinit , a = ± ∞ , atunci functia sinus nu are

limita !!! Limite de functiii


•

Limita functiei cosinus :



- Fie functia : cos : R → [ − 1,1] .




I.

Daca

a

este un punct de acumulare finit , a ∈ R , atunci :

c o sx = c o sa lim x→ a Deci limita functiei cos intr-un punct de acumulare finit a ∈ R se obtine inlocuind pe

II.

Daca

a

x cu a

este un punct de acumulare infinit , a = ± ∞ , atunci functia cosinus nu are

limita !!

•

Limita functiei tangenta : π  k ∈Z → R . 2 



 - Fie functia : tg : R − ( 2k + 1)






I.

Daca

a

apartine domeniului de definitie atunci :

tgx = tga lim x→ a Se poate lua :

sinx sina sinx lim x→ a tgx = lim = = = tga lim x→ a x → a c o sx c o sx c o sa lim x→ a

Deci limita functiei tg intr-un punct de acumulare din domeniul de definitie se obtine inlocuind pe

x cu a Limite de functiii


Daca a =

II.

π 2

, atunci : π π

l i m

t g

x

•

l i m

t g

x

= − ∞

x→ 2 x> 2

Limita functiei cotangenta :



- Fie functia : ctg : R − {kπ k ∈Z } → R .




I.

π π

,

= + ∞

x→ 2 x< 2

Daca

a

apartine domeniului de definitie atunci :

ctgx = ctga lim x→ a Se poate lua :

c o sx c o sa c o sx lim x→ a c tgx = lim = = = c tga lim x→ a x → a sinx sinx sina lim x→ a

Deci limita functiei ctg intr-un punct de acumulare din domeniul de definitie se obtine inlocuind pe

x cu a II.

Daca a = 0 , atunci : c t g x l i m x→ 0 x< 0

= − ∞

,

c t g x = + ∞ l i m x→ 0 x> 0

Limite de functiii




8 Limita : •

.

Limita functiei arcsinus : π π , .  2 2 



 - Fie functia : arcsin : [ − 1,1] → −



- Daca a ∈[ −1,1] , atunci :

arcsinx = arcsina lim x→ a

•

Limita functiei arccosinus :



- Fie functia : arccos : [ − 1,1] → [ 0, π ] .



- Daca a ∈[ −1,1] , atunci : Limite de functiii


a rc c oxs = a rc c oas lim x→ a

•

Limita functiei arctangenta : π π ,  .  2 2



 - Fie functia : arctg : R →  −




I.

Daca

a

apartine domeniului de definitie , a ∈ R , atunci :

a rctgx = a rc tga lim x→ a Daca a = ± ∞ , atunci :

II.

lim arctg x =

x → +∞

•

π

,

2

π 2

Limita functiei arccotangenta :



- Fie functia : arcctg : R → ( 0,π ) .




I.

lim arctg x = −

x →- ∞

Daca

a

apartine domeniului de definitie , a ∈ R , atunci :

a r c c tgx = a r c c tga lim x→ a II.

Daca a = ± ∞ , atunci :

lim a r c c tgx = 0

x→ +∞

,

lima r c c tgx = π

x → -∞

Limite de functiii




9 Limite :

. ( cu functii trigonometrice )

I.

lim x →0

sin x =1 x

Generalizand : lim x →a

II.

lim x →0

x →a

lim x →0

x →a

lim x →0

arcsin u ( x ) =1 daca u( x)

u( x ) = 0 lim x→ a

tg x =1 x

Generalizand : lim

IV.

u( x) = 0 lim x→ a

arcsin x =1 x

Generalizand : lim

III.

sin u ( x ) =1 daca u( x )

tg u ( x ) =1 daca u( x )

u( x) = 0 lim x→ a

arctg x =1 x

Generalizand : lim x →a

arctg u ( x ) =1 daca u( x)

u( x) = 0 lim x→ a

Limite de functiii


Exercitiul nr. 1 :

•

Calculati limitele urmatoare : 1).

sin x = lim x→ 0

...................................................................................................................................... 2).

sin x = lim x→ 2

...................................................................................................................................... 3).

limπ sin x = x→

6

...................................................................................................................................... 4).

sinx = lim x→ + ∞

..................................................................................................................................... 5).

2 lim( 3x − 5 x + 1 ) = x→ −3

.................................................................................................................... 6).

2 ( 3x − x + 1 ) = lim x → −∞

......................................................................................................................

Limite de functiii


7).

limc o sx =

x→ − 2

...................................................................................................................................... 8).

limπ cosx = x→

4

.....................................................................................................................................

9).

limc o s( x − 3) =

x→ − 2

............................................................................................................................ 10).

limc o sx =

x→ − ∞

.................................................................................................................................... 11).

cos( x3 + 2 x2 − 3 ) = lim x→ + ∞

.............................................................................................................. 12).

limπ tg x = x→

3

....................................................................................................................................... 13).

tg x = lim x→ 0

....................................................................................................................................... 14). 15). 16).

t g lim π

x =

t g lim 5π

x =

3 x→ 2 3 x< 2

π

x→ 2 5 x> 2

π

...................................................................................................................................... .....................................................................................................................................

ctg3x = lim x→ 5

................................................................................................................................... 17).

limπ ctg 3x = x→

6

................................................................................................................................... 18).

c t g l i mπ

π

x→ 3 x> 3

6x =

...................................................................................................................................

ctg ( 3 x2 − 2 x ) = 19). lim x →0 x <0



20).

ctg ( 5 x2 − 7 x ) = lim x→ −1

.................................................................................................................... 21).

a rc s inx = lim x→ 1

................................................................................................................................ 22).

arcsin( x2 − 32x ) = lim x→ 0

............................................................................................................... 23).

lim arcsinx = x→

2 2

.............................................................................................................................. 24).

a rc s inx = lim x → -1

................................................................................................................................ 25).

a rc c oxs= lim x→ 1

............................................................................................................................... 26).

lim1 arccosx = x→

2

............................................................................................................................... 27).

a rc c oxs= lim x→ 0

............................................................................................................................... 28).

lim arcc o xs = x→

2 2

.............................................................................................................................

29).

lim a r c c tgx = x→

2 2

.............................................................................................................................

Limite de functiii


30).

a r c c txg= lim x → −1

..............................................................................................................................

31).

a r c c tgx = lim x→ 3

...............................................................................................................................

32).

a r c c t( gx − 2) = lim x→ −∞

....................................................................................................................

33).

3 lim a r c c tg( 5 x − 2 3) =

x→ +∞

..............................................................................................................

34).

lim a rc tgx = x→

3 2

...............................................................................................................................

35).

a r c tgx = lim x→ 1

.................................................................................................................................

36).

lima r c tg( x − 2) = x→ 5

.......................................................................................................................

37).

lima r c t(g3x − 3) =

x→ +∞

....................................................................................................................

38).

arctg( 3x3 − 2 x ) = lim x→ −∞

...............................................................................................................

Limite de functiii


Exercitiul nr. 2 :

•

Calculati limitele urmatoare :

sin 5 x = x

1). lim x →0

……………………………………………………………………………………...

sin 5 x = sin x

2). lim x →0

……………………………………………………………………………………... 3).

sin( 5 x − 5) = 2 − x 1

lim x →1

……………………………………………………………………………….. 4).

sin 4 x = sin 2 x

limπ x→

2

……………………………………………………………………………………... 5). lim x →0

sin 5 x + sin x = x

……………………………………………………………………………..

( 1 − sinx ) (1 − sin x ) = 2

6).

lim x→ 0

co sx 4

……………………………………………………………………. 7). lim x →0

sin 5 x = sin 10 x

……………………………………………………………………………………. 8). lim x →0

sin mx = sin nx

…………………………………………………………………………………….. 9). lim x sin x →∞

1 = x

……………………………………………………………………………………..

Limite de functiii


10).

lim2 x→ 3

sin ( 3 x − 2 ) = 2 3x + x − 2

………………………………………………………………………………

11).

sin x3 = lim 2 x → 0 sin x

………………………………………………………………………………………

sin π ( x + 1) = 3 x →−1 x +1

12). lim

……………………………………………………………………………….

13).

2 2 sin x + sin 2 x = lim 2 x→ 0 x

…………………………………………………………………………...

14).

sin( x2 + x − 2 ) = lim 2 x → − 2 sin( x + 3x + 2 )

………………………………………………………………………...

15).

lim x→ 0

sin( x sin( 2 x sin3x ) ) x

3

=

……………………………………………………………………..

16). lim x →1

sin πx = sin 4πx

…………………………………………………………………………………….

17). lim x →0

tg 3 x = x

……………………………………………………………………………………… 18). lim x →0

tg 5 x = tg 10 x

…………………………………………………………………………………….. Limite de functiii


19.).

tg ( x − 1) = lim 2 x →1 − x 1

…………………………………………………………………………………. 20). lim x →0

arcsin 2 x = x

…………………………………………………………………………………. 21).

arcsin( x − 3) = lim 2 x→ 3 − x 9

……………………………………………………………………………... 22).

arc sin( x2 + 3x − 1 0 ) = lim 2 x→ 2 x −4

…………………………………………………………………… 23). lim x →0

sin 2 x = arcsin x

…………………………………………………………………………………...

24).

s in( x2 − 1 ) = lim 2 x → − 1 a rc s in (x − x − 2 )

……………………………………………………………………… 25).

tg x3 = lim 2 x → 0 arcsin x

………………………………………………………………………………….. 26). lim x →0

arctg 3x = arctg 7 x

………………………………………………………………………………….. 27). lim

x →0

tg 3 x = tg 9 x

……………………………………………………………………………………… 28).

lim x→ −1

tg ( x + 1) = 2 x −1

………………………………………………………………………………….. Limite de functiii


29). lim x →0

sin ( tg 2 x ) tg ( arcsin 3 x )

………………………………………………………………………………..

30).

t g( a r c s (inx 2 − 1 ) ) = lim 2 x → 1 s in( a r c tg ( x − 3x + 2 ) )

……………………………………………………………….. 31). lim x →0

sin 5 x = 4x

……………………………………………………………………………………... 32). lim x →0

2tg x − sin x = x

……………………………………………………………………………...

sin x = lim 2 x →0 tg 2 x 2

33).

……………………………………………………………………………………...

34). lim π x→

4

tg x − 1 = sin x − cos x

……………………………………………………………………………… 35). lim x →0

tg x − cos x + 1 = x

…………………………………………………………………………... 36). lim π x→

6

1 − 2 sin x = π − 6x

………………………………………………………………………………… 37). lim x →0

arcsin 4 x = sin 5 x

…………………………………………………………………………………. 38). lim x →0

arcsin ( tgx ) = arctg ( sin x )

……………………………………………………………………………… Limite de functiii


39). lim x →0

x − sin 2 x = x + sin 3 x

………………………………………………………………………………… 40). lim x →0

x − tg 5 x = x + sin 7 x

…………………………………………………………………………………

sin 5 x = lim 2 x→ 0 x 2

41).

…………………………………………………………………………………….. 2

42).

x = lim x→ 0 1 − cos x

………………………………………………………………………………... 43).

limπ x→

4

tg x − 1 = ctg x − 1

………………………………………………………………………………….. 44). lim π x→

3

tg x − 3 = 3ctg x − 1

………………………………………………………………………………. 45).

lim x →1 x <1

2 arcsin x − π = sin πx

46). lim x →2

…………………………………………………………………………….

arctg x − arctg 2 = tg x − tg 2

……………………………………………………………………….. 47).

limπ x→

4

sin 2 x − tg x = cos 2 x

……………………………………………………………………………... 48).

1 − cos 3 x = lim x →0 x sin 2 x

………………………………………………………………………………….

Limite de functiii




10 Limita :

. a



- Fie functiile : f : E → F , g : F → R , E , F ⊂R ,



- Putem vorbi de compunerea functiei g cu f : h = g  f : E → R .



- Ne intereseaza in ce conditii functia compusa are limita in punctul

punct de acumulare pentru E .

a

.

Aceasta problema este rezolvata de urmatoarea Teorema :

TEOREMA . Daca : 1).

f ( x) = b lim x→ a

2).

f ( x ) ≠ b , pentru orice

3).

g( y) =  lim y→ b

atunci :

;

x ≠ a , x ∈E ;

lim g ( f ( x ) ) = lim g ( y ) x →a

y →b

Limite de functiii


π  limπ sin  x −  = 2  x→ 2

1).

..........................................................................................................................

π  limπ tg  x −  = 4  x→ 4

2).

........................................................................................................................... 1

lim e( x −1) =

3).

2

x →1

.....................................................................................................................................



11 Limita :

.



- Fie functiile : f , g : E →R , E ⊂ R , iar



- Presupunem ca f ( x ) > 0 , x ∈ E ;



- In acest caz puterea :



- Are loc urmatoarea Teorema :

( f ( x) )

g( x)

a

punct de acumulare pentru E ;

este definita .

TEOREMA .  1).

Daca :

lim f ( x ) = 1 x→ a

2).

g ( x ) = 2 lim x→ a

3).

1



2

; ;

are sens

atunci : functia

( f ( x) )

g( x)

are limita in

a

si mai mult

Limite de functiii


( f ( x)) lim x →a

g( x)

lim g ( x )

= lim f ( x )    x→a 

x →a

( Limita se distribuie in baza si in exponent )

Cazuri exceptate :

•

( 0 ) , cand :  = 0 ,  = 0 ; ( ∞ ) , cand :  = ± ∞ ,  = 0 ; (1 ) , cand :  = 1 ,  = ± ∞ ; 0

1).

1

0

2).

1

∞

3).

•

I.

2

2

1

2

Cazuri particulare : Daca :

1).

g( x) = c

lim ( f ( x ) )

⇒

x →a

exceptand cazul cand :

1 = 0

2). Pentru c =

1 , n

* n ∈N

c

c

= lim f ( x )    x →a 

si c ≤ 0

, n≥2

avem :

n f ( x ) = n lim f ( x ) lim x→ a x→ a

( Limita radicalului este egala cu radicalul limitei )

II.

Daca f ( x ) = b > 0 , atunci :

b lim x→ a

g( x)

= blim g ( x ) x→ a

Limite de functiii


Daca f ( x ) = x > 0 , g ( x ) = r ∈ R ,

III. atunci :

r r ln x x =e

si limita , cand exista :

r r ln x r ln x x = lim e = elim lim x →a x →a x →a

1).

5 lim x →∞

x −2

=

........................................................................................................................................ 2).

lim 3 x →∞

x −2 2 x −5

=

....................................................................................................................................... 3 2x +x

3).

 x +5   lim  x →2  2 x −1 

=

........................................................................................................................

2 x

4).

3

 2 + 3x    = lim 2 x→ 0  4 −x 

.......................................................................................................................... Limite de functiii

Clasa a XI-a - 43 Elemente de analiza matematica – Limite de functii 3

5).

 3x2 − 2  lim  2  x →∞  6 x +5 

2x +x 3 2 5x +x

=

.....................................................................................................................

6).

2 5x + 1 = lim 2 x→ ∞ 2x − 1

.............................................................................................................................

 3x  lim  2  x →−∞  3 x −1  2

7).

x +1 x

=

......................................................................................................................... 8).

2 x − 4x + 2 = lim x →0 x−5

....................................................................................................................

9).

1 x

 4x   = lim  2 x →∞  1 +3 x  2

............................................................................................................................ 3

10).

1   lim x→∞ 3  

4 x + x −5 3 2 5x +x

=

............................................................................................................................

11).

2   x   lim 2  x →∞   2 x −3 

3x −1 3x

=

.....................................................................................................................

Limite de functiii

Clasa a XI-a - 44 Elemente de analiza matematica – Limite de functii 2 x −4

12).

 2 x +1  x −2 =   lim x →2 3 x − 5  

......................................................................................................................... 13).

x 2 e (x + x − 6 ) = lim x→ 2 x−2

...................................................................................................................

14).

1 = lim 1 x→ 0 x> 0 1 + e x

15).

3x e −1 = lim x →0 x

.................................................................................................................................

................................................................................................................................

Limite de functiii


12 Limite : 

A

.

Limita remarcabila

: x

 1 1+  = e lim x →∞  x 

.

- Trecand la limita in baza si exponent se obtine nedeterminarea :

acestei formule poate fi eliminate . 

- Daca punem : y =

1 , atunci cand x

x→ ∞

∞

1

care cu ajutorul

rezulta y →0 si avem :

1 y

lim (1+ y ) = e y →0



- Mai general avem , folosind si teorema de la limite de functii compuse :

 1   lim 1+ x→ a  u( x ) 

u( x)

=e

daca :

u( x ) = ± ∞ lim x→ a

sau Limite de functiii


(1+u ( x ) ) lim x →a

1 u( x)

=e

daca :

u( x) = 0 lim x→ a

x

1).

 x +1    = lim x →∞  x −1 

……………………………………………………………………………………. x

2).

 2  1+  = lim x →∞  5x 

............................................................................................................................... 2

3).

(1−sin x ) 3 x = lim x →∞ 2

.......................................................................................................................... 2 x +1

4).

 3 x +1    lim x →∞ 3 x +1  

=

........................................................................................................................... x

5).

 x    = lim x →∞  x −1 

.................................................................................................................................

Limite de functiii


6).

1− x 1−x

 x +4   lim  x →1 3 x + 2  

=

.........................................................................................................................

x

7).

4

 x −1   2  = lim x→ ∞  x  2

............................................................................................................................. x

8).

 x2 + 2 x + 3   2  = lim x→ ∞  x − 3x +1 

.................................................................................................................... 9).

limπ ( sin x ) x→

 1     2 x −π 

=

2

.......................................................................................................................... 10).

lim (13 −4 x )

1 x −3

x →3

=

........................................................................................................................... x

11).

 x +1    = lim x →∞  x+2 

................................................................................................................................ x

12).

 x +1  lim  2  = x →∞  x +2 

...........................................................................................................................

 x +x   lim  2 x → −∞  x − 3 x +1  2

13).

−x

=

................................................................................................................. Limite de functiii




B

Limita remarcabila lim x →0



ln (1 + x ) =1 . x

- Folosind relatia de mai sus si teorema de la limite functii compuse avem :

lim x →a

1).

:

ln (1 + u ( x ) ) =1 . u( x )

daca :

u( x ) = 0 lim x→ a

[ ln( 3x + 1) − ln( x − 5) ] = lim x→ ∞

....................................................................................................... 2). lim x →0

ln (1 + ex ) = x

............................................................................................................................. 3). lim x →0

ln (1 + 10 x ) = x

.......................................................................................................................... 4). lim x →0

ln (1 + sin x ) = ln (1 + sin 2 x )

...................................................................................................................... 5).

x[ ln( x + 2) − ln( x + 1) ] = lim x→ ∞

...................................................................................................... 6). lim x →0

ln (1 + arcsin x ) = sin 3 x

................................................................................................................... 7). lim x →0

ln (1 + tg 3 x ) = 6x

.........................................................................................................................

Limite de functiii


8).

lim x→ 0

ln( cos x ) 3x

2

=

.............................................................................................................................. 9).

2 1 +1 2 x ln = lim x →0 x x +1

......................................................................................................................

10).

ln( x3 + e2 x ) lim 5 x = x → ∞ ln( x +e)

........................................................................................................................ 11).

( ln x ) lim x →e

1

( x −e ) ( x −3 e )

=

...................................................................................................................... 12). lim x →0

ln (1 + sin ( x − 1) ) = ln[1 + arcsin 2( x − 1) ]

...................................................................................................... 13). lim x →0

ln (1 + 2 x ) = ln (1 + 4 x )

.......................................................................................................................... 14). lim x →0

ln (1 + arctg 2 x ) = ln (1 + tg 4 x )

................................................................................................................ 15).

limπ x→

2

ln (1 + 2 cos x ) = cos x

...................................................................................................................

ln [1 + tg ( x + 1) ] = x →−1 ln [1 + arcsin 3( x + 1) ]

16). lim

...................................................................................................... 17). lim x →0

tgx = ln (1 + 2 x )

.......................................................................................................................... Limite de functiii

Clasa a XI-a - 50 Elemente de analiza matematica – Limite de functii 2 ln x

18).

1   1 −   lim x →∞  ln x 

=

........................................................................................................................

x = 2 + 3 x ln (1 + x )

19). lim x →0

............................................................................................................. 20). lim x →0

ln (1 + 2 x ) − ln (1 + 3 x ) = .................................................................................................... ln (1 + 4 x ) − ln (1 + 5 x ) 1

21).

( cos x ) x = lim x →0 2

.............................................................................................................................. tgx x = ....................................................................................................................................... 22). lim x →0 x >0

[ln (1+x )] = 23). lim x→ 0 x

x >0

...........................................................................................................................

lim [ cos ( x sin x ) ] arcsin 1

24).

x →0

2

x

=

........................................................................................................... x

25).

1 1  cos + sin   = lim x →∞ x x 

.................................................................................................................



C

Limita remarcabila

:

x a −1 = ln a , a > 0 lim x→ 0 x



.

- Folosind relatia de mai sus si teorema de la limite functii compuse avem :

lim x→ a

a

u( x)

−1 = ln a , a > 0 u( x )

.

daca :

u( x) = 0 lim x→ a Limite de functiii




- Daca

a=e

avem : x e − 1 = ln e = 1 lim x→ 0 x

sau :

u( x)

lim

e

x→a

1).

.

−1 = ln e = 1 u( x )

.

daca :

u( x) = 0 lim x→ a

3x e −1 = lim x →0 6x

.................................................................................................................................. 2).

3x e −1 = lim 2 x x→ 0 e −1

.................................................................................................................................. 3).

3x 4x e −e = lim 2x x x→ 0 e −e

................................................................................................................................

2 −8 = lim x →3 x − 3 3

4).

..................................................................................................................................

5).

2x e −1 = lim x →0 x

...................................................................................................................................

Limite de functiii


−1 = 3x

sin 2 x

6).

lim

e

x →0

................................................................................................................................ 7).

sin 2 x sin x e −e = lim x → 0 sin 2 x − sin x

...................................................................................................................... tg 2 ( x −1 )

8).

lim x →1

−1 = 3x − 3

e

............................................................................................................................. arcsin 4 ( x − 2 )

9).

lim

e

−1

x−2

x →2

=

........................................................................................................................ 10).

arctg2 ( x − 2 ) −1 e = lim arctg( x − 2 ) x→ 0 −1 e

........................................................................................................................ x

11).

e 3 −1 = lim x →0 x

...................................................................................................................................

12).

3x 5

e −1 = lim 2 x x→0 e 3 −1

................................................................................................................................. 13).

x −x e −e = lim x →0 5x

.................................................................................................................................

−1 = 2x

tg 3 x

14).

lim x →0

e

................................................................................................................................

Limite de functiii


15).

arcsin2 x − earcsinx e = lim x → 0 arcsin2 x − arcsin 3x

........................................................................................................ 16).

lim x →1

e

arcsin 3 ( x −1 )

−1 = 2( x − 1)

........................................................................................................................ 17).

tgx tg 2 x e −e = lim x x→ 0 e −1

.............................................................................................................................



D

Limita remarcabila lim x→ 0

(1+ x ) x

r

:

−1

= r , ∀r∈R

x −1 = a , ∀ a ∈ R* lim x →1 x − 1

.

a

1).

lim x →0

(1+2 x )

5

3x

−1

.

=

.........................................................................................................................

Limite de functiii


(1+x )

5

lim x →0

2).

3 x 2

−1 =

............................................................................................................................

(1+ 2 x )

−1 = lim 2 x→ 0 ( + 2 + 1 x x ) −1

3).

5

..................................................................................................................



E

Alte limite remarcabile

:

1

x lim x = lim x x = 1 . x →0 x →∞

0 lim x ln x = x→ 0 x> 0

.

x >0

x

e lim n = ∞ x→ ∞ x

lim x→ ∞

.

ln x x

n

=0

Vom prezenta in cele ce urmeaza cateva tehnici de calcul a limitelor de functii pentru a usura rezolvarea acestora . I. n ( a0 xn + a1 xn−1 + ..... ) f ( x ) = lim a0 x f ( x ) lim x →∞ x →∞

.

II. lim x →∞

(

n

)

a1 n n −1 x + a1 x + ..... + f ( x ) = + lim ( x + f ( x ) ) x →∞ n

III. Limite de functiii


lim x →∞

(

n

)

n n −1 x + a1 x + ..... ⋅ f ( x ) = lim ( x ⋅ f ( x ) ) x →∞

IV. ln ( xn + a1 x n −1 + ..... ) ⋅ f ( x ) = n lim ln x ⋅ f ( x ) lim x →∞ x →∞

V. ...

lim sin x ⋅ f ( x ) = lim x ⋅ f ( x ) x →0

x →0

VI. tg ⋅ x ⋅ f ( x ) = lim x ⋅ f ( x ) lim x →0 x →0

VII. lim arcsin x ⋅ f ( x ) = lim x ⋅ f ( x ) x →0

x →0

VIII. arctg ⋅ x ⋅ f ( x ) = lim x ⋅ f ( x ) lim x→0 x→0

IX. lim ln x ⋅ f ( x ) = lim ( x − 1) ⋅ f ( x ) x →1

x →1

X. x ( a − 1 ) ⋅ f ( x ) = ln a ⋅ lim x ⋅ f ( x ) lim x →0 x →0

XI.

( a x − 1 ) ⋅ f ( x ) = a ⋅ lim x ⋅ f ( x) lim x →1 x →1 XII. ln x ⋅ f ( x ) = 0 lim x→ 0

daca exista o vecinatate U

a lui

x0

ca functia

1 f ( x ) sa fie marginita pe U ∩ E , unde f : E → R . x Limite de functiii


Asa cum am vazut in capitolele precedente la calculul limitelor de functii apar si cazuri de nedeterminare care ne obliga sa gasim o alta metoda de rezolvare decat cele clasice pentru aflarea limitei acestor functii , daca exista . In continuare vom prezenta cazurile de nedeterminare intalnite precum si tehnica de lucru pentru eliminarea acestor nedeterminari .

1 Limite :

:

0 0





a). Limite de functii rationale in puncte finite a : Explicitarea nedeterminarii se va realiza prin simplificarea cu * k ∈N

( x −a )

k

,

.

lim x →3

1).

x −3 = 2 x − 4 x − 21

..................................................................................................................... 2 x −4 = lim 2 x→ − 2 x −x −6

2). ..........................................................................................................................

3 x − 3x + 2 = lim 4 x→ 1 x − 4x + 3

3). ........................................................................................................................

4 2 x + 2x − 2 = lim 2 x→1 x − 3x + 2

4). .......................................................................................................................

3 2 x + 3x − 4 = lim x→1 ( 2 x − 2) 2

5).

........................................................................................................................ 6).

m x −1 = lim n x →1 x −1

.................................................................................................................................. Limite de functiii




b). Limite de functii rationale in compunere cu functia modul : In acest caz se va explicita modulul :

1).

lim x→ 0

x = x

.......................................................................................................................................... 2).

lim x→ 1

x −1 = x −1

.................................................................................................................................... 2 x +x = lim x →0 x

3). ................................................................................................................................ 4).

2 x + −2 x = lim x →0 x

............................................................................................................................ 5).

lim x→ 2

x −2

2

x −2

=

...................................................................................................................................

 

c). Limite de functii definite prin cat de expresii irationale :

- Distingem cazurile : Limite de functiii


I.

Sub radicali de ordine diferite figureaza aceeasi expresie .

Se scimba variabila , notandu-se radicalul de ordin egal cu cel mai mic multiplu comun al ordinelor radicalilor cu alta variabila , cand se ajunge la limita unei functii rationale .

1).

lim x →1

x −1 = x −1

lim

x −8 = x −4

.................................................................................................................................. 2).

x →64 3

................................................................................................................................. 4

3).

lim x →1

x −1 = x −1

..................................................................................................................................

4).

lim x→ 1 3

( x − 1)

2

x − 2 x+1 2

=

................................................................................................................... 4

5).

lim x →0

x +1 −1 = x +1 −1

............................................................................................................................

II.

Sub radicali figureaza expresii diferite. Se amplifica numitorul si (sau) numaratoru; cu expresia conjugata.

1).

lim x →0

1+ x − 1− x = x

.................................................................................................................

Limite de functiii


lim x →5

2).

x −5 = 2 x −1 − 3

......................................................................................................................... 3

3). lim x →0

1 + x −1 = x

............................................................................................................................ 4).

2 1+x +x −1 = x

lim x →0

...................................................................................................................... 2 x + x − x= lim x→ ∞

5). ........................................................................................................................

3 x + x + x= lim x → −∞

6). ......................................................................................................................

x − x2 − 2 x = lim x→ ∞

7). ......................................................................................................................

2 x − x − 2x = lim x → −∞

8). ..................................................................................................................... 9).

lim x→7

2− x−3 = 2 x − 49

.......................................................................................................................... 2 7+ x − 4 = lim 2 x3 x − 5x + 6

10). ........................................................................................................................ 11).

2 2 x − 2x + 6 − x + 2x − 6 = lim 2 x→ 3 x − 4x + 3

......................................................................................

Limite de functiii

Clasa a XI-a - 61 Elemente de analiza matematica – Limite de functii 2 2 x + 2 x + 12 − 3 2 x + x + 6 = lim x→3 x−3 3

12).

.................................................................................... 13).

x −2 = x +5 −3

lim x →4

..........................................................................................................................

3− x +6 = lim 3 2 x →3 x −1 − 2

14). .......................................................................................................................... 15).

lim x →4

3 − x +5 = 1− 5 − x

.......................................................................................................................... 2 x +4 −2= lim 2 x→ 0 x +9 −3

16).

.......................................................................................................................

x + x +6 = x →−2 x + 2−x

17).

lim

..........................................................................................................................

1 + x − 1+ x2 = lim x →0 1 + x −1

18). ...............................................................................................................

3

19)

lim

x →0 3

x +1 −1 = 2 x +1 −1

........................................................................................................................ 20). lim x →1

2 x − 1 − 3 3x − 2 = .......................................................................................................... x −1 3 2 x − 5 x + 3 − x + 3x − 9 = lim 2 x→ 2 x + x−6 3

21).

........................................................................................... Limite de functiii


x + 4 − 3 x + 22 = ............................................................................................................ 4 x + 11 − 2

22). lim x →5

2 1+ x − 4 1 − 2 x = lim 2 x→ 0 x+ x 3

23). ..............................................................................................................

 

d). Limite de functii trigonometrice :

- Pentru a elimina nedeterminarea se utilizeaza limitele :

lim

u ( x ) →0

sin u ( x ) tg u ( x ) arcsin u ( x ) arctg u ( x ) = lim = lim = lim =1 u ( x ) →0 u ( x ) →0 u ( x ) →0 u( x ) u( x ) u( x ) u( x )

sin αx = βx

1). lim x →0

................................................................................................................................... unde : α, β ∈ R, β ≠ 0 . 2

2).

sin 3 x = lim 2 x →0 5x

.................................................................................................................................

3).

lim x→ 0

1 − cos x x

2

=

............................................................................................................................... 4).

limπ x→ 4

tgx −1 = 2 2 − sin x cos x

................................................................................................................... Limite de functiii


3tgx − 3 = 3x − π

5). lim π x→

3

............................................................................................................................ 6).

limπ x→

4

sin x − cos x = cos 2 x

........................................................................................................................ 2

7).

sin x = lim x →π 1 + cos x

.............................................................................................................................. 8).

lim x →π

1 − cos x 2

x cos x

=

............................................................................................................................... 9).

lim

cos x − 1 2

sin x

x→0

=

........................................................................................................................... 10). lim x →0

sin 2 x = x + 1 −1

............................................................................................................................ 11). lim x →0

1 − cos x = x 1 + x −1

(

)

....................................................................................................................... 12). lim x →0

x − sin 2 x = x + sin 3 x

............................................................................................................................

1 + sin x − cos x = x →0 1 − sin x − cos x

13). lim

................................................................................................................. 14).

limπ x→

4

cos x − sin x = 2 cos x − 2 sin 2 x

............................................................................................................

Limite de functiii

Clasa a XI-a - 64 Elemente de analiza matematica – Limite de functii 2 6 cos x + 7 sin x − 8 = limπ 2 x→ 2 cos x − 5 sin x + 1 6

15).

...................................................................................................... 16). lim π x→

2

1 − sin x = 1 + sin 3 x

............................................................................................................................. 17). lim x →0

e). Limite de functii trigonometrice :

 

1 + sin x − 1 − sin x = ..................................................................................................... x

- Pentru a elimina nedeterminarea se utilizeaza limitele :

lim u ( x ) →0

1). lim x→ 0

ln( 1 + 2 x ) 4x

x> 0

2). lim x →0 x >0

3).

2

ln (1 + u ( x ) ) =1 u( x)

,

lim ( )

u x →0

a

u( x)

−1 = ln a u( x )

, a >0 .

= ............................................................................................................................

ln (1 + 4 x ) = ............................................................................................................................ x

lim x→ 2

ln( 2 x − 3) = 2 − x 4

............................................................................................................................ 4).

ln(1 + 3x − x2 ) = lim x→ 0 x

.................................................................................................................. 5). lim x →0

ln (1 + arctgx ) = x



arctgx = ln (1 + x )

6). lim x →0

...............................................................................................................................

(

)

ln x + x2 + 1 = .................................................................................................................. 7). lim x→ 0 ( ) ln cos x x> 0 sin 2 x

8).

lim

e

x →0

− esin x = x

........................................................................................................................... 1 x

9).

1

e − e x +1 = lim x →∞ 1 3x

2

............................................................................................................................... 10). lim x →0

ln (1 + x ) − ln (1 − x ) = .............................................................................................. arctg (1 + x ) − arctg (1 − x )

2 Limite : 

:

∞ ∞

.

a). Limite de functii rationale : Explicitarea nedeterminarii se va realiza prin raportul termenilor de grad maxim .

1).

3 2 −5x + 6 x − 3 = lim 2 x→ ∞ 2 x + 3x − 5

................................................................................................................... 2).

3 4x − 5x = lim x →−∞ − x +1

............................................................................................................................. Limite de functiii


3).

3 2 5x + 2 x − 3 = lim 5 2 x→ ∞ − 3x + x

..................................................................................................................... 4).

4 2 − 2x + 2x − 1 = lim 4 x → −∞ 4 x + 5x − 5

.................................................................................................................. 5).

lim x →∞

6 x −1 = 3 2 3x − 6 x

............................................................................................................................. 6).

2 6x − 3 = lim 5 x → −∞ 8x + 2 x − 1

......................................................................................................................



b). Limite de functii irationale , exponentiale , logaritmice : Explicitarea nedeterminarii se va realiza prin raportul termenilor de grad maxim .

1).

2 x + 3x + 1 = lim x →∞ − 2x + 5

.....................................................................................................................

Limite de functiii


2).

lim

x →−∞

2 2x + x = 3x + 1

........................................................................................................................... 3).

ln (1+e3 x ) = lim x →∞ x

..............................................................................................................................

ln (1 + x ) = x →∞ x

4). lim

...............................................................................................................................

5).

ln( x3 − 5 x + 3 ) = lim 5 2 x → ∞ ln( x − 6x − 6 )

................................................................................................................. 6).

x 2x e +e lim 2 x 3 x = x→ ∞ e +e

................................................................................................................................

3 Limite :

:

∞ -∞





a). Limite de functii rationale : Explicitarea nedeterminarii se va realiza prin aducerea la acelasi numitor .

1).

3   1 −  = lim 3 x →1 1 − x 1− x  

.................................................................................................................... 2).

2  1 − 2  lim x→ 2 x−2 x −4

 = 

............................................................................................................... 3).

2   1 −  = lim 2 3 x →3 − −  x 9 x 27 

..........................................................................................................

1 1  = 4). lim  − x →0  x x( x +1)  .................................................................................................................... 5).

6   1 − 2   = ............................................................................................................... lim x→ 3 x − 3 − x 9   x> 3

6).

 3x2 ( 2 x − 1 )( 3x2 + x + 2 )  − = lim 2   x→ ∞ 2 x + 1 4x  

...................................................................................... 7).

 x+2  x−4 + lim  2− = 2 x →1 ( ) + − +  x 5 x 4 3 x 3x 2 

.................................................................................. Limite de functiii




b). Limite de functii irationale : Explicitarea nedeterminarii se va realiza prin amplificare cu conjugata .

1).

(

)

2 x − x +1 = lim x→ ∞

...................................................................................................................... 2).

(

)

2 2 x + x +1 − x +1 = lim x→ ∞

................................................................................................... 3).

(

)

2 2 x + x +1 − x +1 = lim x → −∞

.................................................................................................. 4).

(

)

2 2 x + 2x − x + 4 = lim x→ ∞

...................................................................................................... 5).

(

)

2 2 x + 2x − x + 4 = lim x → −∞

...................................................................................................... 6).

(

)

2 2x + x + 1 = lim x → −∞

.................................................................................................................. 7).

(

)

x x2 + 1 − x = lim x→ ∞

................................................................................................................... Limite de functiii


8).

(

)

3 x + 3 1− x = lim x→∞

........................................................................................................................ 9).

(

)

2 lim x 4 x + 7 + 2 x =

x → −∞

............................................................................................................. 10).

9 x − x2 − 4 = lim x→ ∞ x

..................................................................................................................... 11).

lim x + x + x − x = x→ ∞

12).

13).



1).

(

)

limx x x + 1 + x − 1 − 2 x =

x→ − ∞

limx

x→ − ∞

14).

.........................................................................................................

3 2

(

..........................................................................................

)

x+ 1− 2 x− 1+ x+ 3 =

(

)

x x+ x + x− x = lim x→ ∞

.......................................................................................

.................................................................................................

c). Limite de functii exponentiale , logaritmice :

[ ln( 2 x + 1) − ln( x + 2) ] = lim x→ ∞

2).

x[ ln( x + 1) − ln( x + 2) ] = lim x→ ∞

3).

x 3x e − 2e = lim 2x 3x x→ ∞ e − 3e

..................................................................................................... .....................................................................................................

..............................................................................................................................

Limite de functiii


4 Limite :

1).

:

∞ •0

.

x c tg x= lim x→ 0

.................................................................................................................................... 2). lim x sin x →∞

π x

=

.................................................................................................................................. 3).

1 lim 2 ln( cosx ) = x→ 0 x

.......................................................................................................................... 4).

π  limπ  x − tgx = 2 x→  2

......................................................................................................................... 5). lim (1 − x ) tg x →1

πx 2

=

..........................................................................................................................

6).

(

1 x

lim x e − e 2

x →∞

1 x +1

)=

.......................................................................................................................... 7).

lim x →∞

(

)

4 3 2 x + x + 1 − x sin

1 = x

....................................................................................................

Limite de functiii


8).

lim ( e

sin a

x →1

− esin ax )tg

πx 2

=

................................................................................................................

x  π − arctg  = ........................................................................................................... x +1 4

9). lim x x →∞

 

10). lim x arctg x →∞

x +1 x  − arctg  = .......................................................................................... x+2 x +2

5 Limite :

:

∞

1

.

Explicitarea nedeterminarii se va realiza utilizand :

lim

u ( x ) →0

1).

lim ( 6 −x )

1 x −5

x →5

(1+u ( x ) )

1 u( x)

=e

=

................................................................................................................................ x

2).

 3 1+  = lim x →∞  x

................................................................................................................................. x

3).

 x    = lim x →∞  x +1 

.................................................................................................................................

Limite de functiii


4).

 x −1   lim  x →∞  x +3 

x+2

=

..............................................................................................................................

x

5).

2

 x +1   2  = lim x→ ∞  x −2  2

............................................................................................................................

6).

x

 x + x + x +1   3  = lim 2 x→∞  x −5x +3  3

2

.................................................................................................................

7).

 1 − 1−x 1+ lim  x →0  x 

2

1 x

  =  

.................................................................................................................

8).

1 x

 a +b   = lim  x →0  2  x

x

............................................................................................................................. unde : a, b > 0

 a +b lim  x →e  a +b ln x

9).

ln x

  

1 ln x −1

=

................................................................................................................... unde : a, b > 0 10).

( ln x ) lim x →e

1 2

2 x −3 xe + 2 e

=

.......................................................................................................................

Limite de functiii


11).

1 x

lim (1+sin x ) = x →0

............................................................................................................................. 12).

1 x

lim ( cos x ) = x →0

.................................................................................................................................

13).

 sin x    lim x →0 x  

sin x x −sin x

=

......................................................................................................................... 14).

lim ( x )

tg

x →1

πx 2

=

.................................................................................................................................... 15).

(

lim 1 +3tg x x →0 2

)

2

ctg x

=

....................................................................................................................

16).

 π  tg + x   lim  x →0   4 

1 sin x

=

.................................................................................................................... 17).

limπ x→

(

3tgx

)

tg 3 x

=

6

............................................................................................................................

 x + 2 x +1   2  lim x→ 0  5x + 8x  2

18).

3 ctg 6 x

=

...............................................................................................................

6 Limite :

:

0

0

.

Explicitarea nedeterminarii se va realiza utilizand : Limite de functiii

Clasa a XI-a - 75 Elemente de analiza matematica – Limite de functii x l n l i m x→ 0

x = 0

x> 0

si scrierea

f = e g ln f g

x x = ........................................................................................................................................... 1). lim x →0 x >0

(sin x ) = 2). lim x →0 x

x >0

.................................................................................................................................. sin x x = ....................................................................................................................................... 3). lim x →0 x >0

( x −1) 4). lim x→ 1

tg ( x −1 )

=

x >1

.............................................................................................................................

(arcsin 5). lim x→ 0

x)

sin x

=

x >0

.........................................................................................................................

7 Limite :

:

∞

0

.

Limite de functiii


1).

lim x

1 x

=

x →∞

........................................................................................................................................ 2).

lim x

sin

1 x

x →∞

=

....................................................................................................................................... 3).

lim ( x x→ 0

)

2 tgx

=

..................................................................................................................................... 4).

1 x

( x +1) = lim x →∞

...................................................................................................................................

Limite de functiii


Sa se calculeze urmatoarele limite , discutand dupa valorile parametrilor reali corespunzatori : 1).

(

)

2 x − x + 1 + mx = lim x→ ∞

.......................................................................................................... unde : m ∈ R . 2).

(

)

2 x + x + 1 + mx = lim x→ ∞

.......................................................................................................... unde : m ∈ R . 3).

(

)

a x+ 1+ b x+ 2 + c x+ 3 = lim x→ ∞

………………………………………………………..

unde : a, b, c ∈R .

Sa se determine

1).

2).

( lim(

)

astfel incat sa fie indeplinite egalitatile :

2 x + x − ax − b = 0 lim x→ ∞

x→ ∞

lim x→ 0

)

4 3 2 x + 2 x − ax − bx − c = 0

e − cos x = 3 2 2 x ax

3).

a, b, c ∈R

2

Limite de functiii


4).

x

 x + ax +1   2  =e lim x→ ∞  x + 3x − 2  2

Limite de functiii

Limite de Functii - Clasa a 11-A

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