Verified answer

f(x) = (5x^2)/(x^2+4)

When x-->±infinity f(x) = 5

Horizontal asymptote

y = 5

Let y = F(x)=(5x^2)/(x^2+4)

dy/x = 10x/(x^2+4) -{5x^2/(x^2+4) this can be true only for x = infinity henec ^2}*2x = 0 gives

10x*(x^2 +4) = 10x^3 or

(x^2+4) = x^2

Lim (5x^2)/(x^2+4) as x tends to infinity = 5/(1+4/X^2) = 5

So asymptote will be parallel to x axis and it will be the line y = 5