Pui conditia ca limit la stanga lui a sa fie egala cu limita la dreapta
Ls:x→a x<a lim f(x)=lim(x³+a³)=a³+a³=2a³
Ld x→a x>a limf(x)=lim(x+1)=a+1
2a³=a+1
2a³-a-1=0 <=>
(a³-a)+(a³-1)=0
a(a²-1)+(a-1)(a²+a+1)=0
a(a-1)(a+1)+(a-1)(a²+a+1)=0
(a-1)[a(a+1)+a²+a+1]=0
(a-1=0 a1=1
2a²+2a+1=0
calculezi determinantul Δ=-4<0 ecuatia nu admite solutii reale
deci a=1
