Limit x approaches to infinity √x^2+x+1 -√x^2-x-1

Lim x ⇒∞  √(x²+x+1)-√(x²-x-1)=∞-∞  (IDT)
we have to multiply and divide by de conjugate of √(x²+x+1)-√(x²-x-1)

Lim x ⇒∞  √(x²+x+1)-√(x²-x-1)=
Lim x ⇒∞  [√(x²+x+1)-√(x²-x-1)][√(x²+x+1)-√(x²-x-1)] / [√(x²+x+1)+√(x²-x-1)]=
Lim x ⇒∞ (  x²+x+1-x²+x+1) / √(x²+x+1)+√(x²-x-1)]=
lim x ⇒∞ (2x+2) / [√(x²+x+1)+√(x²-x-1)]=∞/∞  (IDT)

We have to divide all terms by "x",
√x²=x
Then:

lim x ⇒∞ (2x /x  +  2/x) / [√(x²/x² + x/x²+ 1/x²)+√(x²/x² - x/x³ - 1/x)²]=
=(2+0) / [(√(1+0+0) + √(1-0-0)]=
=2/2=1

 Lim x ⇒∞  √(x²+x+1)-√(x²-x-1)=1

38 dollars is your answer.