Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\lim_{x \to \infty}\frac{4x^2+2x+1-(ax+b)^2}{\sqrt{4x^2+2x+1}+(ax+b)} = \lim_{x \to \infty}\frac{x^2\left(4+\frac{2}{x}+\frac{1}{x^2}-a^2-\frac{2ab}{x}-\frac{b^2}{x^2}\right)}{x^2\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
Applying limit on the numerator and denominator
$$\frac{\lim_{x \to \infty}\left(4-a^2+\frac{2-2ab}{x}+\frac{1-b^2}{x^2}\right)}{\lim_{x \to \infty}\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
It can be seen that the denominator tends to $0$

$$ L=\lim_{x\to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
Let $x=\frac{1}{t}$, then
$$L=\lim_{t\to 0}\left( \frac{2}{t} (1+\frac{t}{2}+{\frac{t^2}{4}})^{\frac{1}{2}}-\frac{a}{t}-b\right)$$
Use $(1+z)^k=1+kz+O(z^2)$. Then
$$L=\lim_{t\to 0}\frac{2(1+\frac{t}{4}+O(t^2))-a-bt}{t}=\lim_{t\to 0} \frac{(2-a)+(\frac{1}{2}-b)t+O(t^2)}{t}.$$ For $L$ to be finite, we demand $2-a=0$, then $L=(\frac{1}{2}-b)=\frac{-1}{2}.$ Thus we get $a=2$ and $b=1$.
