Calculul limitelor unor iruri cu

SOLUII

1) Indicaie : se reduce la calculul integralei :

2) Indicaie : se reduce la calculul integralei :

3) Indicaie : se scrie

4) Indicaie : se reduce la calculul integralei :

5) Indicaie : se reduce la calculul integralei :

6) Indicaie : se reduce la calculul integralei :

7) Indicaie : se reduce la calculul integralei :

8) Indicaie : se reduce la calculul integralei : .9) Indicaie : se reduce la calculul integralei : .10) Indicaie : se reduce la calculul integralei : .11) Soluie :se logaritmeaz suma i se ajunge la

12) se logaritmeaz suma i se ajunge la

13) Indicaie : se reduce la calculul integralei :

14) Soluie : se scrie

15) Soluie : voi folosi definiia integralei Riemann pentru funcia

Voi folosi definiia integralei Riemann pentru funcia alegnd diviziunea i sistemul de puncte intermediare . Atunci suma dat este suma Riemann asociat tripletului cu deci limita este .16) Soluie :

pentru c suma reprezint suma Riemann asociat funciei continue , diviziunii cu i sistemul de puncte intermediare

cu .

17) Soluie :

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

18) Soluie :

Avem

Pentru a calcula ultima integral folosim

EMBED Equation.3

19) Soluie : folosesc inegalitile aplcate succesiv pentru numerele pt. :

20) Indicaie : se folosete faptul c i definiia cu a limitei unei funcii ntr-un punct.21)

Indicaie : se folosete faptul c i definiia cu a limitei unei funcii ntr-un punct.

22) Soluie : folosesc inegalitile aplicate succesiv pentru numerele pt. :

23) Soluie:

EMBED Equation.3

EMBED Equation.3

Avem

Apoi

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3 =

Fiecare integral se transform cu substituia x+j=u n integrala

unde F este o primitiv a funciei .

Calculez F=

EMBED Equation.3

24) Soluie :

Fie i .Punctele formeaz o diviziune a intervalului .

.

ns deoarece f monoton pe avem .

ntr-adevr dac spre exemplu f cresctoare avem:

Un raionament similar se face dac f descresctoare.

Atunci

25) Soluie :

pentru c

Aplicm lema , considernd funcia i avem

26) Soluie :

pentru c avem

Aplicm lema , considernd funcia i avem

cci k>1.27) Soluie : a)

Voi afla primitivacu substituia

EMBED Equation.3 Fie i

Avem Apoi :

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

c) se utilizeaz inegalitile n care lum pe rnd , n-fixat :

Sumnd dup toi i nmulind apoi relaia obinut cu rezult :

Dac facem obinem :

EMBED Equation.3

Rmne s calculm cu substituia

28) Soluie : a) prin derivare

b) folosim inegalitile

Alegem n aceste inegaliti pe rnd care verific pentru i apoi summ :

(1)

Dar

Apoi

Prin trecere la limit n inegalitile (1) se obine c limita din enun este egal cu .c)folosim inegalitile

Alegem n aceste inegaliti pe rnd care verific pentru adic

i apoi summ , rezult,dup nsumare c :

ns dac trecem la limit i inem cont c

obinem enunul.29) a) prin derivare.b) se folosesc inegalitile demonstrate la punctul a).c) se folosesc inegalitile demonstrate la punctul a).d) se folosesc inegalitile demonstrate la punctul a)

e) folosim inegalitile

Alegem n aceste inegaliti pe rnd care verific pentru, adic

i apoi summ , rezult,dup nsumare c :

ns dac trecem la limit i inem cont c

30) Soluie : a) de exemplu pentru a demonstra inegalitatea din dreapta considerm funcia cu f(0)=0

Cum f(0)=0 i f- descresctoare rezult c .

Punem n inegalitile (1) succesiv i apoi nsumm dup :

Facem i obinem :

Se obine :

.folosim inegalitile

Alegem n aceste inegaliti pe rnd care verific pentru adic

i apoi summ , rezult,dup nsumare c :

ns dac trecem la limit i inem cont c .

.31) b) Soluie :

n care alegem i summ dup :

Dac facem avem :

.

c) logaritmm produsul i apoi folosim inegalitile

Alegem n aceste inegaliti pe rnd care verific pentru adic

i apoi summ , rezult,dup nsumare c :

ns dac trecem la limit i inem cont c

32) Indicaie : se scrie

33) Indicaie : se folosesc inegalitile i apoi faptul c

34) a) Artai c au loc inegalitile:

Indicaie: a) inegalitatea din dreapta este inegalitatea lui Bernoulli cu exponentul subunitar , cea din stnga se obine prin derivare.c)Se pot folosi inegalitile demonstrate la punctele anterioare sau faptul c

35) Indicaie: se folosete criteriul cletelui.36) Soluie . Limita se rescrie astfel :

Demonstrez inegalitile :

Pentru cea din stnga avem folosind notaiile :

EMBED Equation.3 Deoarece

EMBED Equation.3 Alegem succesiv , (n fixat) n aceste inegaliti i obinem :

Dac nsumm dup k i apoi trecem la limit cnd obinem :

.Deoarece f este continu pe va fi integrabil pe i funcia este continu pe ca produs a dou funcii continue , deci este integrabil pe adic .Conform criteriului cletelui rezult cerina din enun .

37) Soluie :

a) evident

b) nlocuim a cu i se obine c.c.t.d.

c)

EMBED Equation.3 e) prin integrarea pe [0,1] relaiei avem :

i apoi innd cont c se obine c.c.t.d.

f) se obine din e) pt. p=2 :

g)

Prima limit este obinut din (1) pt. p=2 i are valoarea iar a doua din (1) pt. p=1 i are valoarea

38)

Indicaie : se folosete

39) Soluie :

Considerm funcia cu .

Aplicnd definiia limitei unei funcii ntr-un punct rezult :

a.. din s rezulte pt.

EMBED Equation.3 Pt. fiecare avem pentru

( am folosit ). Atunci i avem :

EMBED Equation.3

Dac facem se obine :

.

Cum a fost ales arbitrar putem face i rezult :

.

40) .Indicaie : se folosete .41) Soluie : voi folosi faptul c i apoi definiia limitei unei funcii ntr-un punct : a.. dac

avem :

Pt. fiecare avem dac .

Atunci i avem :

de unde prin trecere la limit i innd seama c se obine :

De aici , dac facem obinem .

42) Soluie : voi folosi faptul c i apoi definiia limitei unei funcii ntr-un punct : a.. dac

avem : .Deoarece funcia feste integrabil Riemann pe rezult c f este mrginit pe deci exist .Pt. fiecare avem dac .

Atunci i avem prin nmulirea relaiilor cu n ipoteza (i) avem :

de unde prin trecere la limit i innd seama c se obine :

Cum a fost arbitrar ales , dac facem obinem .43) Soluie : a) avem de unde

iar i atunci

Avem

b) Avem

Aplicnd teorema lui Lagrange asupra funciei f:[0,1]R rezult :

( se observ mai sus c avem o sum Riemann ataat diviziunii i sistemului de puncte intermediare ).

c)

Apoi

44) Soluie :

a) ,

EMBED Equation.3 b)

c)

- convergent.

d) fie

e)

f) cci .

g)

45) Soluie 1: fie i fie unde :

cu sistemul de puncte intermediare .Atunci suma dat este suma Riemann asociat lui cu deci limita este .

Soluie 2 :

Avem

unde

46) Soluie 1 : s observm c de unde

Considerm funcia continu i diviziunea a intervalului , cu sistemul de puncte intermediare .Atunci suma dat este suma Riemann asociat lui cu deci limita este .Soluie 2 :

Avem

unde

47) Soluie:

Avem

unde

48) Indicaie : se reduce la calculul integralei .

49) Indicaie : se reduce la calculul integralei .50) Soluie : avem

fie funciile i

Avem f continu pe iar g continu pe cu derivata continu pe .

51) Soluie : dac notm

. Aplicnd lema Cesaro-Stolz obinem : .52 )Soluie : avem

53) Soluie :

obinem c .PAGE 20

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