What are the values of the parameters $a,b \in \mathbb{R}$ such that $\lim\limits_{x \to \infty}(\sqrt{x^2+x+1}+\sqrt{x^2+2x+2}-ax-b)=0.$

Question

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 $b$?

Answer 1

\begin{align*} &\sqrt{x^{2}+x+1}-x+\sqrt{x^{2}+2x+2}-x\\ &=\dfrac{x+1}{\sqrt{x^{2}+x+1}+x}+\dfrac{2x+2}{\sqrt{x^{2}+2x+2}+x}\\ &\rightarrow \dfrac{1}{2}+\dfrac{2}{2}\\ &=\dfrac{3}{2}. \end{align*}

Answer 2

Hint: Intuitively, for large $x$, we have $\sqrt{x^2+x+1} = \sqrt{(x+\tfrac{1}{2})^2+\tfrac{3}{4}} \approx x+\tfrac{1}{2}$ and $\sqrt{x^2+2x+2} = \sqrt{(x+1)^2+1} \approx x+1$. But how can we make these observations rigorous?

Using the identity $\sqrt{a}-\sqrt{b} = \dfrac{a-b}{\sqrt{a}+\sqrt{b}}$, we get that$$\sqrt{x^2+x+1} - (x+\tfrac{1}{2}) = \dfrac{(x^2+x+1)-(x+\tfrac{1}{2})^2}{\sqrt{x^2+x+1} + (x+\tfrac{1}{2})} = \dfrac{\tfrac{3}{4}}{\sqrt{x^2+x+1} + (x+\tfrac{1}{2})}$$ and $$\sqrt{x^2+2x+2} - (x+1) = \dfrac{(x^2+2x+2)-(x+1)^2}{\sqrt{x^2+2x+2} + (x+1)} = \dfrac{1}{\sqrt{x^2+2x+2} + (x+1)}.$$ Both of these expressions tend to $0$ as $x \to \infty$. Do you see how to use these results to solve the problem?

Answer 3

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 $b=\frac32$.

Answer 4

HInt

$$F=\lim_{x\to\infty}{\sqrt{x^2+px+q}-x}=\lim_{h\to0}\dfrac{\sqrt{1+ph+qh^2}-1}h$$

Rationalize the numerator to find $$F=\lim\dfrac{1+ph+qh^2-1}{h(?)}=\dfrac p{1+1}$$

Set $p=q=1$

and $p=q=2$