Utilizand criteriul de convergenta cu ε sa se arate ca:
doar subpunctele: 1,2,3, 6, 8 ,9
Răspunsuri la întrebare
Răspuns:
ε∈(0,1]
Explicație pas cu pas:
1,n->∞lim2n/(n+3)=2∈∈∈∈l2n/(n+3)-2l<ε
l[2n-2(n+3)]/(n+3)l<ε
l(2n-2n-6)/(n+3)l<ε
l-6l/(n+3)<ε
6/(n+3)<ε
Fie n=9
6/(9+3)<ε
6/12<ε
1/2<ε
DEci pt n≥9 toti termenii sirului vor fi mai mici decat 1/2
_____________________________________________________
2)n->∞ lim(5n+2)/(3n+4)=5/3
l(5n+2)/(3n+4)-5/3l≤ε
l3(5n+2)/3(3n+4)-5(3n+4)/3(3n+4)l<ε
l(15n+6-15n-20)/(9n+12)l<ε
l-14l/(9n+12)<ε
14/(9n+12)<ε⇄²²Fie n=1ε14/(9*1+12)=14/21=2/3<1
Deci incepand cu primul termen lan-al<2/3=>3/5 limita
----------------------------------------
3)n->∞lim n²/(n²+1)=1
ln/ln²/(n²+1)-1l≤ε
ln²/(n²+1)-(n²+1)/(n²+1)l≤ε
l(n²-n²-1)/(n²+1)l<ε
l-1l/(n²+1)≤ε
1/(n²+1)≤ε
Fie n=1
1/(1²+1)=1/2 =ε
Deci inceand cu n≥1∞lan-1l≤1
-----------------------------------------------l
Ex 8 n→∞lim(n³+1)=∞
ln³+1-∞l<ε
l-∞l=+∞>ε=> lim an=+∞
9)n→+∞lim(-n⁴+6)= -∞
l-n⁴+6-∞l<ε
l-∞l≤ε
+∞>ε undeε este un numar pozitiv oricat de mare
=> limita srului e -∞