<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] Set $1/x=h$ to find $$\lim_{h\to0^+}\dfrac{\sqrt{a+bh+ch^2}-(\alpha+\beta h)}h$$
$$=\lim_{h\to0^+}\dfrac{(a+bh+ch^2)-(\alpha+\beta h)^2}h\cdot\lim_{h\to0^+}\dfrac1{\sqrt{a+bh+ch^2}-(\alpha+\beta h)}$$
$(a+bh+ch^2)-(\alpha+\beta h)^2=a-\alpha^2+h(b-2\alpha\beta)+h^2(c-\beta^2)$
As the denominator $\to0,$ $$a-\alpha^2=0$$
as the denominator is $O(h)$ $$b-2\alpha\beta=0$$
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