Ytukyg

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 better method for the problem?

Set $1/x=h$ to find $$lim_{hto0^+}dfrac{sqrt{a+bh+ch^2}-(alpha+beta h)}h$$

$$=lim_{hto0^+}dfrac{(a+bh+ch^2)-(alpha+beta h)^2}hcdotlim_{hto0^+}dfrac1{sqrt{a+bh+ch^2}-(alpha+beta h)}$$

$(a+bh+ch^2)-(alpha+beta h)^2=a-alpha^2+h(b-2alphabeta)+h^2(c-beta^2)$

As the denominator $to0,$ $$a-alpha^2=0$$

as the denominator is $O(h)$ $$b-2alphabeta=0$$

Another approach could be Taylor series
$$sqrt{a x^2+2 b x+c}=x sqrt{a+frac{2 b}{x}+frac{c}{x^2} }$$ So, for large $x$,
$$sqrt{a x^2+2 b x+c}=sqrt{a} x+frac{b}{sqrt{a}}+frac{a c-b^2}{2 a^{3/2}
x}+Oleft(frac{1}{x^2}right)$$

You want $$ lim _{xto infty }ax^2+2bx+c -(alpha x +beta )^2=0$$

That implies $$ a= alpha ^2, c=beta ^2, b=alpha beta$$