<s>[INST] What are the values of the parameters $a,b \in \mathbb{R}$ such that $\lim\limits_{x \to \infty}(\sqrt{x^2+x+1}+\sqrt{x^2+2x+2}-ax-b)=0.$ I have to find the values of $a$ and $b$ (with $a,b \in \mathbb{R}$) such that the following is true:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
This is what I did:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
$$\lim\limits_{x \to \infty} \bigg (x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-ax-b \bigg )=0$$
$$\lim\limits_{x \to \infty} x\bigg (\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + \sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-a-\dfrac{b}{x} \bigg )=0$$
So, we'd have something like:
$$\infty \cdot(2-a) = 0$$
$(*)$If $a \in (-\infty, 2)$ $\Rightarrow$ $(2-a) > 0$, which means:
$$\infty \cdot(2-a) = \infty$$
$(*)$If $a\in (2, +\infty)$ $\Rightarrow$ $(2-a) < 0$, which means:
$$\infty \cdot(2-a) = -\infty$$
So the only option left for the limit to have any chance of being true is:
$$a=2$$
which would result in the indeterminate form:
$$\infty \cdot (2-a) = \infty \cdot 0$$
And now that we have $a=2$, we need to find the value of $b$ for the limit to be true. Substituting $a$ with $2$ in the initial limit we get:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x-b)=0$$
Since $b$ is a constant we can pull it out of the limit and get:
$$b=\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x)$$
And this is where I got stuck. I tried a bunch of methods and tricks for finding this limit and I got nowhere. That leads me to think that either I made some mistake/mistakes along the way, or I simply don't know how to solve this limit. So, how can I find $b$?
 [/INST] Note
$$x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}=x(1+\frac1{2x}+O(\frac1{x^2}))=x+\frac12+O(\frac1x)$$
$$x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}=x+1+O(\frac1x)$$
Therefore,
$$\lim\limits_{x \to \infty} \bigg (x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-ax-b \bigg )$$
$$=\lim\limits_{x \to \infty} (2-a)x+(\frac32-b) + O(\frac1x)$$
Thus, to make the limit zero, we need to have $a=2$ and $b=\frac32$.
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