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 ⇒∞ (x ²+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

Answer: Lim x ⇒∞ √ (x²+x+1) - √ (x²-x-1) = 1