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 $0$
 A: It should be a question to find a and b.
We need to eliminate $x^2$ term, otherwise limit is zero, so $a=\pm2$
$$=\lim_{x \to \infty}\frac{4x^2+2x+1-(ax+b)^2}{\sqrt{4x^2+2x+1}+(ax+b)} = \lim_{x \to \infty}\frac{x\left(2-2ab+\frac{1-b^2}{x}\right)}{x\left(\sqrt{\frac{4x^2+2x+1}{x^2}}+a+\frac{b}{x}\right)}=\frac{2-2ab}{2+a}=-\frac{1}{2}$$
Reject $a = -2$
When $a=2, b=1$
A: From
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
we have $$\lim_{x \to \infty}(\frac{\sqrt{4x^2+2x+1}-ax-b}{x}) = 0$$
hence
$$\lim_{x \to \infty}(\frac{\sqrt{4x^2+2x+1}}{x})=a=2$$
$$b=\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-2x) +\frac{1}{2}$$
and
$$\sqrt{4x^2+2x+1}-2x=\frac{2x+1}{\sqrt{4x^2+2x+1}+2x}$$
using L'hopital Rule we get $b=1$.
A: 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 $b$?
A: $$ L=\lim_{x\to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
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 $b=1$.
A: A variation. We obtain
\begin{align*}
\color{blue}{-\frac{1}{2}}&=\lim_{x\to\infty}\left(\sqrt{4x^2+2x+1}-ax-b\right)\\
&=\lim_{x\to\infty}\left(\sqrt{\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}}-ax-b\right)\\
&=\lim_{y:=2x+\frac{1}{2}\to\infty}\left(\sqrt{y^2+\frac{3}{4}}-a\left(\frac{y}{2}-\frac{1}{4}\right)-b\right)\\
&\,\,\color{blue}{=\lim_{y\to\infty}\left(y\sqrt{1+\frac{3}{4y}}-\frac{a}{2}y+\frac{a}{4}-b\right)}\tag{1}\\
\end{align*}

*

*Since the terms with factor $y$ in (1) need to cancel we obtain $\color{blue}{a=2}$.


*Taking $a=2$ the constant terms in (1) give $\frac{2}{4}-b=-\frac{1}{2}$ and $\color{blue}{b=1}$ follows.
