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 $0$

HINT
I would start with noticing that:
\begin{align*}
\lim_{x\to\infty}(\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{(4 - a^{2})x^{2} + (2 - 2ab)x + 1 - b^{2}}{\sqrt{4x^{2} + 2x + 1} + ax + b}\\\\
& = \lim_{x\to\infty}\frac{(4 - a^{2})x + (2 - 2ab) + \frac{(1 - b^{2})}{x}}{\sqrt{4 + \frac{2}{x} + \frac{1}{x^{2}}} + a + \frac{b}{x}}
\end{align*}
Based on such expression, can you set the corresponding system of equations in terms of $a$ and $b$?
