Ytukyg

up vote 2 down vote favorite 1 Here is my approach. Consider; $$ax^2 + 2bx + c = 0$$ or; $$ x_{±} = frac {-b±sqrt {b^2-ac}}{a} = frac {-b±sqrt D}{a}$$ Hence; $$sqrt {ax^2+2bx+c} = sqrt {(x+frac {b-√D}{a})(x+frac {b+√D}{a})}$$ $$=xsqrt {(1+frac {b-√D}{ax})(1+frac {b+√D}{ax})}$$ For large values of $x$ we may apply the binomial approximation, so that; $$sqrt {ax^2+2bx+c} ≈ x(1+frac {b-√D}{2ax})(1+frac {b+√D}{2ax})$$ $$=x + frac {b}{a} + frac {c}{4ax}$$ As $x→∞$ the final term in the above expression vanishes. Hence; $$lim_{x→∞}[ sqrt {ax^2+2bx+c} - alpha x - beta ] = 0,$$ gives; $$x+frac {b}{a} - alpha x - beta = 0,$$ or; $$(1-alpha)x + (frac {b}{a} - beta) = 0$$ As $x→∞$ , $1-alpha$ must be $0$ for the former term to vanish, hence, $$alpha = 1, beta = frac {b}{a}$$ But I doubt it is hardly correct. Is there any