<s>[INST] Find $\alpha$,$\beta$ if $\lim_{x→∞}[ \sqrt {ax^2+2bx+c} - \alpha x -\beta] = 0$ 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;
$$\sqrt {ax^2+2bx+c} = \sqrt {(x+\frac {b-√D}{a})(x+\frac {b+√D}{a})}$$
$$=x\sqrt {(1+\frac {b-√D}{ax})(1+\frac {b+√D}{ax})}$$
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?
 [/INST] 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 $x$,
$$\sqrt{a x^2+2 b x+c}=\sqrt{a} x+\frac{b}{\sqrt{a}}+\frac{a c-b^2}{2 a^{3/2}
   x}+O\left(\frac{1}{x^2}\right)$$
</s>