Verified answer

***** Method 1 *****

(For students in US, please see Method 2)

y = 2x / √(x² + x + 1)

y² = 4x² / (x² + x + 1)

y²x² + yx + y = 4x²

(y² - 4)x² + yx + y = 0

x = {-y ± √[(y² - 4(y² - 4)(y)]} / [2(y² - 4)]

x→infinite when (y² - 4)→0

i.e. x→infinite when y→± 2

So, horizontal asymptotes are y = -2 and y = 2 .

***** Method 2 *****

(Recommended for US students)

lim(x→+∞) {2x/√(x² + x + 1)}

= lim(x→+∞) {2/√[(x² + x + 1)/x²]}

= lim(x→+∞) {2/√[1 + (1/x) + (1/x²)]}

= 2/√[1 + (0) + (0)]

= 2

lim(x→-∞) {2x/√(x² + x + 1)}

= lim(x→-∞) {-2/√[(x² + x + 1)/x²]}

= lim(x→-∞) {-2/√[1 + (1/x) + (1/x²)]}

= -2/√[1 + (0) + (0)]

= -2

So, horizontal asymptotes are y = -2 and y = 2 .

You look at the limiting behavior as x->+-infinity.

This will be easier to do if you divide numerator and denominator by x.

2x/sqrt(x^2 + x + 1) = 2/[sqrt(x^2 + x + 1)/x]

= 2/sqrt[(x^2 + x + 1)/x^2] x becomes x^2 when you bring it under the radical sign.

= 2/sqrt[1 + (1/x) + (1/x^2)]

As x->+-infinity, those 1/x and 1/x^2 terms go to 0.